Dr. Trefor Bazett · YouTube · 9:40
This 180-second introductory lecture segment motivates double integration through Riemann sums. It first compares area under a curve in single-variable calculus with volume under a surface in multivariable calculus, then shows how rectangles under a curve correspond to boxes under a surface. The speaker explains that refining the subdivision improves the approximation and uses a highlighted box to show local error where the box lies partly above and partly below the surface. In the final portion, the presentation formalizes one subrectangle in the domain with side lengths and and introduces a sample point inside it to determine the height of the box. The clip stops before writing the full summation or limit definition. This 180-second lecture segment introduces the construction of volume under a surface over a rectangular region using double Riemann sums. It begins by stating that the exact choice of sample point (,) inside each subrectangle will not matter in the limit, then visualizes one representative box whose height is . The lecturer formalizes the method in four steps: partition [a,b] × [c,d] into rectangles with area ; choose one point in each rectangle; approximate volume by ; and define the exact volume as the limit of these sums as ||P|| → 0. This 180-second clip introduces the double-integral idea geometrically as volume under a surface. In the first half, the lecturer writes a four-step definition: partition [a,b]×[c,d] into rectangles with area , choose a sample point (,) in each rectangle, approximate volume by , and define the exact volume as the limit as ||P||→0. He explains ||P||→0 informally as the largest rectangle shrinking to zero area while the number of rectangles grows without bound. The second half works a concrete example: approximate the volume under ²−y² above [-2,2]×[-2,2] using four equal subsquares. A 3D plot shows four red boxes under the surface, and a 2D plot labels the subsquares ,,, with yellow dots at their bottom-left corners. The board then builds the sum , simplifies to 2², and rewrites the approximation as []2². The clip ends before the numerical function values are computed. This 40-second clip shows a short worked example of approximating the volume under the surface over the rectangle [-2,2] × [-2,2] using a four-term double Riemann sum. The slide partitions the rectangle into four equal squares, chooses the lower-left corner of each square as the sample point, and evaluates the sum []·. The speaker emphasizes that this is an approximation to the volume under the surface. After the math slide, the video cuts to a presenter outro asking viewers to comment, like, and watch more multivariable calculus videos.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The segment opens with the title "INTEGRATING MULTIVARIABLE FUNCTIONS," signaling that the topic is integration in a multivariable setting.
The lecturer poses the main question: how to find the quantity under a surface. The screen contrasts a 2D curve with a 3D surface, establishing the analogy between area under a curve and volume under a surface.
He recalls the single-variable method: divide the region under the curve into rectangles, compute each rectangle's area, and add them. The left panel fills with red rectangles to visualize this Riemann-sum approximation.
The same strategy is transferred to multivariable calculus. Instead of rectangles in the plane, the approximation uses boxes standing under the surface, shown as a red tiling in the right panel.
The lecturer then explains that using more subdivisions and making the pieces smaller improves the approximation. The animation refines both the rectangle and box tilings at the same time, reinforcing the limiting idea.
To expose the source of error, the view isolates one box under the curved surface. The speaker notes that the box can rise above the surface in some places and fall below it in others, so each finite box both overestimates and underestimates locally.
He argues that as the number of boxes increases and their sizes shrink, these local errors become smaller and smaller. This is the intuitive bridge from finite sums to an exact integral.
The presentation then turns formal by focusing on the domain alone. A single red subrectangle is shown in the plane, and its side lengths are labeled and , with the subscript indicating one member of a larger partition.
Finally, the lecturer explains how to choose the height of the box above that subrectangle. Because the subrectangle is a whole region, one must select a specific sample point inside it and plug that point into the function. The clip ends while this setup is still being developed, before the full summation notation is written.
The segment opens with a key conceptual point: once the partition is refined enough, the exact placement of the sample point inside a subrectangle will not change the limiting volume. The 2D diagram supports this by showing one red rectangle with side labels and and a yellow point labeled inside it.
The view then shifts to the graph of the function in 3D. A vertical box rises from a base rectangle under the surface, and the lecturer identifies the height of that box as . This connects the planar partition data to a geometric volume element.
From that picture, the volume of one representative box is described as base times height: the base area is , and the height is , so the single-box contribution is .
The lecture then formalizes the construction in numbered steps. Step 1 restricts attention to a rectangular region and partitions it into little rectangles, with area for each subrectangle.
Step 2 adds the sampling rule: in each rectangle, choose one point . This is the point whose function value will determine the height of the corresponding box.
Step 3 builds the finite approximation. If the region is split into subrectangles, the total volume is approximated by summing all box volumes: . The lecturer stresses that this is only an approximation, not the exact answer.
Step 4 gives the exact definition by passing to a limit: . Here names the partition, and the limit is taken as the partition becomes arbitrarily fine. The clip ends before the lecturer fully defines .
The opening board lays out the definition in four ordered steps. First, the rectangular base [a,b]×[c,d] is partitioned into small rectangles whose areas satisfy . Second, one sample point (,) is chosen inside each rectangle. Third, the volume is approximated by the finite sum . Fourth, the exact volume is defined as the limit of these sums as ||P||→0.
The lecturer then interprets the notation ||P||→0 verbally rather than formally. He says it means the largest rectangle in the partition shrinks to zero area. Because that largest piece shrinks, every other piece shrinks too, and to keep covering the same fixed region the number of rectangles must grow without bound.
He closes the definitional portion by identifying this limiting process as the definition of volume. The key logical move is from a coarse finite approximation to an exact quantity obtained only after refining the partition indefinitely.
The lesson then switches to a concrete example. The function is ²−y² and the base region is the square [-2,2]×[-2,2]. The speaker explains the interval-product notation as a compact way to describe the bounds on x and y simultaneously.
For computational tractability, he chooses only four equal subregions rather than a much finer partition. The 3D graphic shows the curved surface and four red boxes standing over the subsquares. Each box has a different height because the surface height varies from one chosen sample point to another.
The lower-right 2D diagram isolates the domain bookkeeping. It labels the four subsquares ,,, and marks the chosen sample points with yellow dots. The rule stated on the board is that (,) is the bottom-left point of the k-th square.
Applying that rule gives the four sample points (-2,-2), (0,-2), (-2,0), and (0,0). The lecturer emphasizes that this is a convention for the approximation, not a uniquely required choice; another corner could have been used instead.
With the partition and sample points fixed, the general definition becomes the four-term sum Volume ≈ . Each summand is one box volume: height times base area .
Because the original square has side length 4 and is split into four equal subsquares, each subsquare has side length 2 in both directions. Therefore and for every k, so the area factor is the same in all four terms: .
This common factor allows the sum to be rewritten as Volume ≈ , and then expanded explicitly as []. The clip ends at this setup stage, before the individual function values are calculated.
The opening slide frames a concrete double-integration setup: approximate the volume under above using four equal regions.
The sample-point rule is made explicit: is chosen to be the bottom-left point of the square, so the four relevant points are , , , and .
Because the partition is uniform, each subrectangle contributes the same base area, and the displayed sum becomes .
Substituting the four chosen points into the surface function gives the heights , so the slide rewrites the approximation as .
Carrying out the arithmetic yields , but the notation and narration both stress that this is only an approximation produced by a finite Riemann sum.
The speaker’s closing phrase says “area under this particular surface,” while the displayed mathematics is clearly a volume approximation under a surface over a rectangular region.
The video then leaves the worked example and cuts to the presenter, who asks for comments, likes, and continued viewing from the multivariable calculus playlist.
The video introduces the central problem as finding the quantity under a surface in multivariable calculus, explicitly comparing it to the first-year calculus problem of finding area under a curve. The side-by-side graphics make the analogy concrete: a curve lives over an interval, while a surface lives over a planar domain.
In the single-variable case, the lecturer breaks the region under the curve into many rectangles, computes each rectangle's area, and adds them to obtain an approximation. This is presented as the prototype method that will be generalized.
The multivariable version replaces planar rectangles with three-dimensional boxes erected over subrectangles of the domain. Summing the volumes of these boxes gives an approximation to the volume under the surface.
The speaker states that using a larger number of smaller subdivisions yields a better approximation in both settings. The animation shows the rectangle and box grids becoming finer, illustrating the intuitive limiting process behind Riemann sums.
By zooming into one box, the video shows that a flat-topped box cannot match a curved surface exactly: part of the box may protrude above the surface and part may lie below it. The lecturer argues that these local errors shrink as the boxes get smaller.
To formalize the construction, the domain is subdivided into many small rectangles. One representative subrectangle is labeled with side lengths in the -direction and in the -direction, where indexes the chosen subrectangle.
Because a subrectangle is an entire region rather than a single input, the lecturer explains that one must choose a specific point inside that subrectangle and evaluate the function there to determine the height of the box above the subrectangle. Different choices of the point are allowed.
In the construction, each small rectangle in the partition carries a chosen point . The opening 2D diagram shows this point inside a red rectangle whose side lengths are labeled and . The lecturer states that in the limiting process, the exact interior choice of this point does not affect the final volume.
The first formal step is to partition a rectangular region into smaller rectangles. For the -th subrectangle, the area is the product of its side lengths in the - and -directions.
After partitioning the region, the next step is to select a sample point inside each subrectangle. That chosen point determines the height used for the corresponding box in the Riemann sum.
Over each subrectangle, build a rectangular box whose height is the function value at the chosen sample point. Its volume is base area times height, giving the elementary contribution used in the sum.
Adding the volumes of all representative boxes gives a finite approximation to the total volume under the surface. The lecturer explicitly treats this stage as approximate rather than exact.
The exact volume is defined by taking the limit of the finite sums as the partition becomes arbitrarily fine. In the displayed notation, this is written with , where denotes the partition. The clip introduces this definition but ends before fully explaining .
The video defines volume under a surface over a rectangle by a four-step procedure: partition [a,b]×[c,d] into subrectangles, choose one sample point in each subrectangle, form the sum of heights times base areas, and then take the limit as the partition is refined. This is presented as the conceptual basis of the double integral as volume.
Each small rectangle contributes base area , computed as width times height in the coordinate directions. This is the geometric factor multiplied by the function value to produce one box volume in the Riemann sum.
After partitioning the domain, one chooses a point inside each subrectangle. The function value at that point determines the height of the approximating box over that subrectangle.
The lecturer explains the partition norm informally: ||P||→0 means the largest rectangle in the partition shrinks to zero area. Consequently all rectangles shrink, and the number of rectangles grows without bound as the partition becomes finer and finer.
The worked example asks for the volume under the paraboloid ²−y² above the square region [-2,2]×[-2,2]. The interval-product notation records the independent bounds on x and y.
Instead of taking a very fine partition, the example uses only four equal subsquares so the sum can be computed explicitly in class. The 3D picture shows four boxes of different heights under the curved surface.
The board specifies that for the k-th square, the sample point is its bottom-left corner. This yields the four points (-2,-2), (0,-2), (-2,0), and (0,0). The speaker notes that other corners would also be permissible for an approximation.
Specializing the general definition to four subrectangles gives a finite sum of four box volumes. This is still an approximation, not yet the exact limiting volume.
Since each subsquare has side length 2 in both directions, every area factor equals . This uniformity lets the common factor be written as and treated consistently across all four terms.
Substituting the four chosen bottom-left points gives the fully displayed setup for the approximation. The clip stops here, before computing the numerical values of f at those points.
Type: definition. The clip defines a finite volume approximation by summing sampled heights times subrectangle areas: . Here the surface is over . Conditions: rectangular partition, one sample point per subrectangle, finite sum gives approximation rather than exact integral. Prerequisites: basic functions of two variables, rectangular regions. Related: lower-left sample rule, four equal regions, worked example. Time evidence: 0–20 s. To verify: none.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The speaker says the rectangle has a width that he will define to be delta x, and later says he calls it delta x k to denote the kth such little rectangle in the division of his domain.
The on-screen label below the red rectangle is .
Width of the th subrectangle in the domain partition.
Positive length associated with one subrectangle of the domain; exact numerical value not stated.
The speaker says the rectangle has a height change, a delta y, and later says he calls it delta y k to denote the kth such little rectangle in the division of his domain.
The on-screen label beside the red rectangle is .
Height change, or side length in the -direction, of the th subrectangle in the domain partition.
Positive length associated with one subrectangle of the domain; exact numerical value not stated.
The speaker says, "So let me choose a pair, a pair x k y k," and explains that this pair is just some point anywhere inside the rectangle.
A yellow point inside the red rectangle is labeled .
A chosen sample point inside the th subrectangle, used to determine the height of the corresponding box above that subrectangle.
A point in the plane lying inside the selected subrectangle; the speaker states that different choices are allowed.
At 00:00:00-00:00:16 the 2D plot labels a yellow point inside a red rectangle as .
At 00:01:43-00:01:52 step 2 reads "Choose a point in each rectangle".
The lecturer says he will choose an inside each little rectangle.
(,)
Sample point chosen inside the k-th subrectangle of the partition.
A point in the rectangular region [a,b] × [c,d].
At 00:00:00-00:00:16 the horizontal side of the red rectangle is labeled .
At 00:01:02-00:01:42 the formula appears.
The lecturer refers to the change in as one factor of the base area.
Width of the k-th subrectangle in the x-direction.
Positive length associated with the k-th partition cell.
At 00:00:00-00:00:16 the vertical side of the red rectangle is labeled .
At 00:01:02-00:01:42 the formula appears.
The lecturer refers to the change in as the other factor of the base area.
Height of the k-th subrectangle in the y-direction.
Positive length associated with the k-th partition cell.
At 00:01:02-00:01:42 the text shows .
The lecturer says the area of each rectangle is given by the product of the change in times the change in .
Area of the k-th small rectangle in the partition.
.
At 00:00:31-00:01:01 the vertical height on the surface is labeled .
At 00:01:53-00:02:33 the summand contains .
The lecturer says taking f of that point gives the height of the box.
The explicit formula for f is not stated in this clip; only its role as a surface height is shown.
f
Function whose graph forms the upper surface above the rectangular region.
Evaluated at sample points (,) to give heights.
At 00:00:31-00:01:01 the label marks the vertical height of the representative box.
At 00:01:53-00:02:33 and 00:02:34-00:03:00 the formulas use .
The lecturer says the height is given by the function value at ,.
Height of the representative rectangular box over the k-th subrectangle.
Real-valued function evaluation at the chosen sample point.
At 00:01:53-00:02:33 the summation index runs to n.
The lecturer says the region is partitioned into n different small little rectangles.
n
Number of subrectangles in the partition used for the finite Riemann sum.
Positive integer.
At 00:02:34-00:03:00 the limit is written with ||P|| → 0.
The lecturer says P is just the name I give for this partition where I took this big region and partitioned it into a bunch of little small rectangles.
The precise definition of ||P|| is not fully completed within the clip; the lecturer begins to explain it but the sentence is cut off.
P
Name of the partition of the rectangular region into subrectangles.
A partition of [a,b] × [c,d].
At 00:02:34-00:03:00 the formula shows lim_{||P||→0}.
The lecturer says the volume is defined by a limit as the length of P goes to zero.
The clip does not finish defining ||P|| explicitly; only the phrase 'length of P' is spoken before the explanation is cut off.
||P||
Norm or size measure of the partition P, described verbally as the length of P going to zero.
Nonnegative quantity associated with the partition.
The speaker introduces the topic as finding the area underneath of a surface and compares it to the single-variable problem of finding the area underneath of a curve.
The screen shows two panels titled "Single Variable Calculus" and "Multivariable Calculus," with a 2D curve on the left and a 3D surface on the right.
The video defines the central task as computing the quantity under a surface in multivariable calculus, by analogy with the first-year calculus task of computing the area under a curve. The visual contrast makes clear that the single-variable object is a curve over an interval, while the multivariable object is a surface over a planar region.
The setting is multivariable calculus.
The target object is a surface rather than a single-variable curve.
The speaker says the idea for area under a curve was to take the region and break it up into a bunch of different rectangles, compute each rectangle's area, and add them to get an approximation.
At about 00:30, the left graph fills with red rectangles under the blue curve.
The method presented is to subdivide the region under a curve into rectangles, use the easy-to-compute area of each rectangle, and sum those areas to approximate the total area under the curve. This is the one-dimensional prototype that the video then transfers to surfaces.
The region under the curve can be subdivided into rectangles.
Each rectangle has computable area.
The sum of rectangle areas is treated as an approximation.
The speaker says the situation in multivariable calculus is exactly the same except instead of little rectangles, he is adding up little boxes.
At about 00:30, the right panel shows a 3D surface covered by many red rectangular boxes.
The multivariable analogue replaces planar rectangles with three-dimensional boxes standing over subrectangles of the domain. Adding the volumes of these boxes approximates the volume under the surface.
The domain is subdivided into small rectangles.
Each subrectangle supports a box whose height is chosen from the function value at a sample point.
The total approximation is the sum of box volumes.
The speaker says that in both stories he can get a better and better approximation by taking a larger and larger number of individual subdivisions, and that dividing the domain into increasingly many increasingly small rectangles gives a better and better approximation.
From about 01:03 to 01:06, both the left rectangles and right boxes become visibly finer and more numerous.
The video states that increasing the number of subdivisions and making the pieces smaller yields a better approximation in both the single-variable and multivariable settings. The visual refinement supports the intuitive limiting idea behind Riemann sums.
The number of subdivisions increases.
The size of each subrectangle decreases.
The speaker says he will focus on the domain and specify some rectangle in the domain, with width delta x and height change delta y, and then says he calls them delta x k and delta y k to denote the kth such little rectangle in the division of his domain.
The labels and appear around the isolated red rectangle.
To formalize the construction, the video isolates one subrectangle in the domain and names its side lengths and . The subscript indicates that many such subrectangles are being used in a partition of the domain.
The domain has been subdivided into multiple rectangles.
indexes one chosen subrectangle.
The speaker says he needs to figure out what the height of the box is above the little rectangle, and to do that he plugs a specific point into the function. He chooses a pair that is just some point anywhere inside the rectangle, and says different people can make different choices.
A yellow point inside the red rectangle is labeled .
Because a subrectangle is a whole region rather than a single input, the video explains that one must select a specific point inside that subrectangle and plug it into the function to obtain the height of the box above the subrectangle. The choice of point is not unique.
There is a function defined on the domain.
A subrectangle has been chosen.
A point inside that subrectangle is selected as the sample point.
The lecturer says that in the limit, where exactly you choose the , inside of the rectangle is not going to actually matter.
At 00:00:00-00:00:16 a yellow point labeled (,) is shown inside a red rectangle with side labels and .
For the construction being introduced, the exact location of the sample point (,) inside each subrectangle is arbitrary at the finite-sum stage, but the lecturer states that in the limit this choice no longer affects the result.
Applies in the limiting process as the partition becomes infinitely fine.
Each (,) is chosen inside the corresponding subrectangle.
At 00:01:02-00:01:42 the first step reads: Partition the region [a,b] × [c,d] into little rectangles .
The lecturer says the region has to be rectangular and that he will break it up into a bunch of small little rectangles.
The first formal step is to divide the rectangular domain [a,b] × [c,d] into smaller rectangles. Each small rectangle has side lengths and , and its area is denoted .
The overall region must be rectangular.
The region is subdivided into finitely many little rectangles indexed by k.
At 00:01:43-00:01:52 the second step reads: Choose a point (,) in each rectangle.
The lecturer says that in each rectangle he is going to choose some particular point.
After partitioning the region, the next step is to select one point (,) inside each subrectangle. This point determines the height used for the corresponding box in the Riemann sum.
One point is chosen for each subrectangle.
The point lies inside its associated rectangle.
At 00:00:17-00:01:01 a 3D surface is shown with a vertical box rising from a base rectangle; the height is labeled .
The lecturer says the volume of this box is base times height, so it is the product of and in the base multiplied by the height .
Each subrectangle in the domain supports a rectangular box whose height is the function value at the chosen sample point. The volume of that single box is the base area times the height.
Uses one chosen sample point (,) in the k-th subrectangle.
The base is the k-th subrectangle with area .
At 00:01:53-00:02:33 the third step reads: Volume ≈ .
The lecturer says this is not exactly the answer, but if he partitions the region into n small rectangles then he will take the sum from one up to n of all the volumes of these little boxes.
Adding the volumes of all representative boxes gives a finite sum that approximates the total volume under the surface over the rectangular region.
There are n subrectangles in the partition.
Each term uses the sample point (,) chosen in the k-th subrectangle.
At 00:02:34-00:03:00 the fourth step reads: _{||P||→0} .
The lecturer says he is going to define the volume under this region by talking about a particular limit as the length of P goes to zero.
The clip ends before the lecturer completes the definition of ||P||.
The exact volume under the surface is defined not by any one finite sum, but by the limit of the Riemann sums as the partition norm tends to zero.
The partition P is refined so that ||P|| → 0.
The limit is taken of the finite Riemann sums over the partition.
The speaker says the multivariable calculus problem is completely analogous to a single variable problem answered back in first year calculus, namely the area underneath of a curve.
The multivariable problem of finding the quantity under a surface is presented as completely analogous to the single-variable problem of finding the area under a curve.
One is working in single-variable calculus for the curve case.
One is working in multivariable calculus for the surface case.
For the examples shown in the video.
The speaker says that when the box sticks above the surface you are adding too much, and when it is beneath the surface you are adding too little, but as you increase the number of them and the sizes get smaller, these errors are going to get smaller and smaller and smaller.
From about 01:21 onward, a single highlighted box is shown protruding above and sitting below parts of the curved surface.
For a box standing over a subrectangle, parts of the box may lie above the surface and parts may lie below it, creating local overestimation and underestimation; the video claims that increasing the number of subdivisions and shrinking the boxes makes these errors smaller.
The surface is curved over the subrectangle.
Boxes are used to approximate the volume under the surface.
The subdivision is refined so the boxes become smaller.
For the highlighted box and the refining family of boxes shown in the video.
The lecturer states: "where exactly you choose the xk yk inside of the rectangle is not going to actually matter" in the limit.
No proof is given in this clip; the statement is presented as part of the conceptual setup.
As the partition is refined toward infinitesimal subrectangles, the precise location of (,) inside each subrectangle does not affect the limiting volume.
The construction is approaching the limit of increasingly fine partitions.
Each (,) is chosen inside its corresponding subrectangle.
For each subrectangle, any admissible choice of sample point leads to the same limiting result.
The lecturer says, "for this video the region has to be a rectangular region, a region where it's just a change in x and a change in y."
At 00:01:02-00:01:42 the displayed domain is [a,b] × [c,d].
This restriction is stated for the present development, not proved here.
In this presentation, the region being partitioned is required to be a rectangular region of the form [a,b] × [c,d].
The discussion is the one in this clip.
The region is described as a product of an x-interval and a y-interval.
For the region considered in this construction.
The lecturer says the volume of the box is base times height, namely multiplied by .
At 00:00:31-00:01:01 the height of the representative box is labeled above the base rectangle.
The volume of the box over the k-th subrectangle is the product of its base area and its height .
A subrectangle with side lengths and has been chosen.
A sample point (,) inside that subrectangle has been selected.
For each individual subrectangle k.
The lecturer says, "this is not exactly the answer, but if I'm going to do an approximation..." before introducing the sum.
At 00:01:53-00:02:33 the formula uses ≈ rather than =.
The sum gives an approximation to the volume under the surface, not the exact value.
The partition has only finitely many subrectangles.
The sum is formed before taking any limit.
For any fixed finite partition.
The lecturer says, "I am going to define the volume under this region by talking about a particular limit."
At 00:02:34-00:03:00 the formula changes from ≈ to = with lim_{||P||→0}.
The clip does not prove existence of the limit; it only defines volume by that limiting process.
The volume under the surface over the rectangular region is defined to be the limit of the Riemann sums as ||P|| → 0.
A sequence or refinement of partitions P is considered.
The Riemann sums are formed from sample points in the subrectangles.
In the limit as the partition norm tends to zero.
The speaker says that when the length of P goes to zero, the largest rectangle goes to zero area, all the other rectangles go to zero as well, and the number of rectangles goes to infinity.
This verbal explanation is attached to the displayed limit .
The statement is given informally rather than as a formal theorem with hypotheses.
As \|P\|, the largest subrectangle area tends to 0, hence all subrectangle areas tend to 0, while the number of subrectangles tends to infinity.
A rectangular region is partitioned into subrectangles.
\|P\| denotes the size of the largest rectangle in the partition.
For a sequence of refinements of the partition with mesh tending to 0.
The speaker says he could have chosen the top right corner or any corner, but for this approximation he chooses the bottom-left corner.
For a finite Riemann-sum approximation, the sample point in each subrectangle may be chosen in different ways; the bottom-left corner is only one valid convention.
A partition into subrectangles is already fixed.
One is constructing an approximate sum rather than proving a unique choice.
For each subrectangle in the example partition.
The speaker says: So this is an approximation to the area under this particular surface.
The displayed line begins with Volume ≈.
The spoken phrase says “area under this particular surface,” while the displayed quantity is labeled Volume; the mathematical meaning in context is the volume approximation.
Using only four sample terms produces an approximation to the volume under the surface, not the exact value.
The surface is sampled at finitely many points.
Each contribution uses only one representative height per subrectangle.
For the displayed 4-term sum, the result is approximate.
The slide shows = [] .
The speaker says plugging those into the function gives multiplied by 2 squared, giving 80.
For on [-2,2] × [-2,2] with lower-left sample points, the displayed four-term Riemann sum equals 80.
Use the four sample points (-2,-2), (0,-2), (-2,0), (0,0).
Each subrectangle has area .
For this specific example and partition.
The speaker moves from breaking a region under a curve into rectangles, to adding up little boxes in multivariable calculus, to refining the subdivision, to discussing the local error of a single box relative to the surface.
The visuals progress from smooth graphs to tiled rectangles and boxes, then to finer tilings, then to a zoomed single box under the surface.
The video does not write a summation formula or limit notation in this clip.
Start with the single-variable problem of area under a curve.
The speaker explicitly frames the multivariable discussion as analogous to the first-year calculus problem of area under a curve.
Break the region under the curve into many rectangles and add their areas to approximate the total area.
This is stated directly in the audio and illustrated by red rectangles filling the area under the blue curve.
Transfer the same strategy to a surface by replacing rectangles with boxes standing over subrectangles of the domain.
The speaker says the multivariable situation is exactly the same except that instead of little rectangles he is adding up little boxes.
Refine the partition by using more and smaller subdivisions to improve the approximation.
The speaker says a larger and larger number of individual subdivisions gives a better and better approximation, and the animation shows the tiling becoming finer.
Inspect one box to see why the approximation is imperfect: the box may protrude above the surface in some places and fall below it in others.
The speaker explicitly describes this local mismatch and the animation highlights one box under the curved surface.
Conclude that as the boxes get smaller, these local errors shrink.
The speaker states that as the number of boxes increases and their sizes decrease, the errors get smaller and smaller.
The clip develops an intuitive derivation of double integration via Riemann sums: approximate volume under a surface by summing box volumes, and expect the approximation to improve as the partition is refined.
The speaker says, "Let's see how we can define this a bit more formally," then isolates one rectangle in the domain, names its sides and , and chooses a point inside it to determine the box height.
The labels , , and appear sequentially on the isolated rectangle.
The clip stops before the full summation expression is written.
Focus on the domain rather than the surface itself.
The speaker says he is going to focus in just on the domain and specify some rectangle in the domain.
Name the width of the chosen subrectangle.
The speaker defines the rectangle's width as delta x and then indexes it by .
Name the height change of the chosen subrectangle in the -direction.
The speaker defines the rectangle's height change as delta y and then indexes it by .
Choose a specific sample point inside the subrectangle.
The speaker says he needs to figure out the height of the box above the rectangle and therefore must plug a specific point into the function.
Use that sample point to determine the height of the box above the subrectangle.
The speaker says the pair is some point anywhere inside the rectangle and that different choices are possible.
One term in the multivariable approximation is built from a subrectangle with side lengths and and a chosen sample point that determines the box height.
The lecturer moves from describing one box, to summing over all boxes, to taking a limit.
Steps 3 and 4 appear sequentially: first Volume ≈ ..., then _{||P||→0} ....
The derivation is conceptual and definitional rather than a rigorous proof of convergence.
Start with a rectangular region and partition it into small subrectangles; the area of the k-th subrectangle is the product of its side lengths.
Displayed in step 1 of the on-screen procedure.
Over each subrectangle, choose a sample point (,) and build a box whose height is the function value there; its volume is base area times height.
Stated verbally and illustrated by the 3D box labeled with height .
Add the volumes of all n representative boxes to obtain a finite Riemann-sum approximation to the total volume.
Displayed in step 3 and explained as summing all the little box volumes.
Replace the finite approximation by the exact definition: take the limit of the sums as the partition becomes arbitrarily fine.
Displayed in step 4 and introduced verbally as the real magic of Riemann integration.
The volume under the surface is built from representative box volumes, summed over a partition, and then defined exactly as the limit of those sums as ||P|| → 0.
The lecturer says that as one goes toward infinitely many infinitesimally small subdivisions, the exact choice of , inside the rectangle is not going to matter.
This is an intuitive claim in the clip, not a proved theorem within this excerpt.
Each subrectangle admits many possible interior sample points.
Shown by the yellow point inside the red rectangle in the 2D diagram.
When the subdivision becomes infinitely fine, the lecturer states that the particular interior choice no longer changes the resulting limiting volume.
Direct verbal assertion at the beginning of the clip.
Within this introductory explanation, the limiting volume is presented as independent of the specific interior sample-point choice.
The board first shows Volume , then Volume , then = [].
The speaker explains that there are four boxes, each side change is 2, and then he evaluates at the four chosen points.
The final numerical evaluation is not reached within the clip.
Start from the general Riemann-sum volume approximation and specialize to four subrectangles.
Direct substitution of into the displayed definition.
In this example each small square has side length 2 in both coordinate directions.
Stated verbally by the speaker from the four equal regions over [-2,2]×[-2,2].
Replace each area factor by the common value .
Algebraic substitution using .
Write out the four terms using the chosen bottom-left sample points.
By the stated rule that (,) is the bottom-left point of the k-th square.
The example reduces the volume approximation to a common factor times the sum of f at the four bottom-left corners; the clip stops before numerical evaluation.
The speaker interprets \|P\| as the largest rectangle going to zero area, forcing all rectangles to shrink, and the number of rectangles to go to infinity.
This explanation is tied to the displayed limit in step 4.
This is an intuitive explanation rather than a formal proof.
Begin with the displayed limit condition on the partition norm.
Shown in step 4 of the board.
Interpret the norm as the size of the largest subrectangle.
Explicit verbal explanation by the speaker.
If the largest one shrinks to zero, every smaller one must also shrink to zero.
Order relation among rectangle sizes in the same partition.
To keep covering the fixed region while each piece shrinks, the number of pieces must increase without bound.
Speaker's verbal description of finer and finer partition.
The limit \|P\| corresponds intuitively to refining the partition until all subrectangle areas vanish and their number grows without bound.
The slide progresses from to [] to [].
The speaker verbally lists the four points and then states the substituted values 1, 5, 5, 9 and final result 80.
Start from the general finite double Riemann sum shown on the slide.
Displayed definition of the approximation.
Because the rectangle is divided into four equal squares, each subrectangle has side lengths 2 and 2.
Slide text “Use four equal regions” and the simplified factor .
Expand the sum using the chosen lower-left sample points for the four subregions.
Slide explicitly lists these four points; audio repeats them.
Substitute the four points into to obtain the displayed heights.
Audio says “plug those into the function”; the resulting numbers are shown on screen.
Add the four heights and multiply by the common area 4.
Arithmetic from the displayed line [].
The displayed four-term lower-left Riemann sum approximates the volume as 80.
The slide reads Ex: Volume under above [-2,2] × [-2,2] and Use four equal regions.
A 3D plot shows the surface and four red boxes; a 2D plot labels ,,, and marks bottom-left sample points.
The board later shows Volume .
The clip ends before the function values are computed and before a final numeric answer is given.
Approximate the volume under above the square [-2,2]×[-2,2] using four equal subregions and bottom-left sample points.
Base region [-2,2]
Use four equal regions ,,,
Choose (,) to be the bottom-left point of the k-th square
Set up the four-term Riemann-sum approximation for the volume.
The domain is partitioned into four congruent subsquares.
Explicit instruction on the slide: Use four equal regions.
The sample points are the bottom-left corners of the four subsquares.
Stated on the slide and reinforced by the yellow dots in the 2D diagram.
Apply the general Riemann-sum volume formula to this four-rectangle partition.
Direct use of the definition introduced earlier in the clip.
Each subsquare has side length 2, so each area factor is 4.
Spoken explanation plus the displayed simplification to .
Substitute the four chosen sample points into the sum.
Algebraic expansion of the four-term sum.
The clip sets the approximation up as [], but does not reach a final numerical value within the provided duration.
No numerical verification is shown in the clip; the setup can be checked against the displayed partition, sample-point rule, and common area factor .
Title line: Ex: Volume under above [-2,2] × [-2,2].
Upper-right 3D plot shows the surface and four red boxes; lower-right plot shows the partition and yellow sample points.
The speaker computes the four sampled values and concludes the result is an approximation.
Approximate the volume under the surface above the rectangle [-2,2] × [-2,2] using four equal regions and lower-left sample points.
Domain: [-2,2] × [-2,2]
Four equal subregions
Sample rule: (,) is the bottom-left point of each square
Compute the displayed four-term Riemann-sum approximation.
Split the interval [-2,2] in each direction into two equal parts, producing four squares of side length 2.
Slide text “Use four equal regions” and the factor in the formula.
Choose the lower-left corner of each square as the sample point.
Explicit slide instruction and matching yellow dots in the diagram.
Evaluate the surface height at each selected sample point.
Shown on the slide after substitution and stated in the audio.
Sum the four heights and multiply by the common subrectangle area.
Directly displayed algebraic line on the slide.
Carry out the arithmetic to get the numerical approximation.
Final displayed result and spoken conclusion.
80
The slide itself marks the result with ≈, and the speaker explicitly calls it an approximation rather than the exact volume.
The opening title appears word by word on a chalkboard background: "INTEGRATING", then "MULTIVARIABLE", then "FUNCTIONS".
Chalkboard background
Title text "INTEGRATING MULTIVARIABLE FUNCTIONS"
The title builds sequentially from one word to the full phrase.
The background remains a dark chalkboard texture.
This identifies the topic as integration of multivariable functions.
Two panels appear with headings "Single Variable Calculus" and "Multivariable Calculus"; the left shows a blue curve in a 2D plot and the right shows a colored 3D surface.
The speaker says the multivariable problem is completely analogous to the single-variable problem of area under a curve.
Presenter
Left 2D graph with blue curve
Right 3D surface plot
Panel titles
The presenter gestures toward the left panel and then the right panel while explaining the analogy.
The left panel remains a 2D curve plot and the right panel remains a 3D surface plot during this interval.
The visuals establish the conceptual parallel between area under a curve and volume under a surface.
At about 00:30, red rectangles fill the area under the curve on the left, and many red boxes appear under the surface on the right.
The speaker describes breaking the region into rectangles and, in multivariable calculus, adding up little boxes.
Blue curve
Red rectangles under the curve
3D surface
Red boxes under the surface
Empty regions under the curve and surface become tiled by rectangles and boxes respectively.
The underlying curve and surface remain visible above the approximating shapes.
This visualizes the basic approximation idea: sum simple geometric pieces to estimate the target area or volume.
From about 01:03 to 01:06, the number of rectangles on the left and boxes on the right increases and their sizes decrease.
The speaker says that taking a larger and larger number of subdivisions gives a better and better approximation.
Left rectangle tiling
Right box tiling
Curve
Surface
Coarse rectangles and boxes are replaced by finer, more numerous ones.
Both panels continue to represent the same underlying curve and surface.
The animation illustrates convergence intuition: refining the partition improves the approximation.
From about 01:21, the view isolates one translucent red box under the surface, showing portions above and below the curved surface.
The speaker says that at some points the box is higher than the graph of the function and at other points the graph of the function is higher, producing error.
3D surface
One highlighted translucent red box
Coordinate axes
The global tiling disappears and attention shifts to a single representative box.
The surface shape remains the same while the box is held fixed for inspection.
This visualizes why finite boxes only approximate a curved surface: the box can overestimate in some places and underestimate in others.
From about 02:02, the display changes to a 2D coordinate plane with a single red rectangle; labels and appear, followed by a yellow point labeled inside the rectangle.
The speaker says he will focus on the domain, define the rectangle's width and height change, index them by , and choose a point inside the rectangle to plug into the function.
2D coordinate axes
Single red rectangle
Labels and
Yellow point
The presentation moves from 3D geometry to a 2D domain diagram, then adds side-length labels and finally a sample point.
The red rectangle remains the same subrectangle throughout this interval.
This is the formal setup for one term of a double Riemann sum: a subrectangle in the domain plus a chosen sample point.
A 2D coordinate plot shows a red rectangle with side labels and and a yellow point labeled (,).
The lecturer is discussing where exactly to choose , inside the rectangle.
Cartesian axes
red subrectangle
yellow point labeled (,)
labels and
The lecturer gestures while the static 2D diagram remains on screen.
The rectangle stays fixed in the plane.
The sample point remains inside the rectangle.
This visual isolates one partition cell and shows the data needed to form one term of the Riemann sum: base dimensions and a chosen sample point.
The display switches to a 3D surface plot with a vertical box rising from a base rectangle under the surface.
At 00:00:31-00:01:01 the box height is labeled .
The lecturer explains that f of the chosen point gives the height of the box and that volume is base times height.
colored surface
base rectangle in the xy-plane
vertical rectangular box
height label
The view changes from 2D domain to 3D graph.
A vertical box is highlighted beneath the surface.
The height label is added.
The box stands over the same kind of subrectangle introduced in 2D.
The surface remains fixed above the region.
The animation connects the planar partition data (, , (,)) to a geometric volume element whose height is determined by the function value.
At 00:01:02-00:01:42 step 1 appears with .
At 00:01:43-00:01:52 step 2 appears with Choose a point (,) in each rectangle.
At 00:01:53-00:02:33 step 3 appears with Volume ≈ .
At 00:02:34-00:03:00 step 4 appears with _{||P||→0} .
numbered text steps
summation formula
limit formula
Step 1 is introduced first.
Step 2 is added below step 1.
Step 3 replaces the geometric description with a finite sum.
Step 4 upgrades the approximation to a limit definition.
The same symbols , , (,), and persist across steps.
The overall topic remains volume under a surface over a rectangular region.
The visual progression mirrors the mathematical logic: partition, sample, sum, then limit.
A chalkboard-style slide lists four numbered steps for defining volume by Riemann sums.
Around 4 seconds, green text appears near step 4 reading Largest Rectangle going to zero area with an arrow toward \|P\|.
Four numbered steps on a dark board
Formula
Summation formulas for Volume
Green annotation Largest Rectangle going to zero area
The green annotation appears after the main four-step list is already visible.
The lecturer gestures while explaining the meaning of \|P\|.
The four-step structure remains on screen throughout this interval.
The base region [a,b]×[c,d] and the summation notation remain unchanged.
The visual layout presents the definition in procedural order: partition, choose sample points, approximate, then pass to the limit. The green note clarifies that the limit refers to refinement of the partition, not to a single rectangle index.
The slide changes to a 3D plot of over [-2,2]×[-2,2] with four red boxes beneath the surface.
The speaker says the graphic shows four different boxes, each of different heights, and the approximate volume is the sum of their volumes.
Colored surface
Four red rectangular boxes
Coordinate axes labeled x, y, z
The board transitions from the abstract definition to a concrete graph.
The boxes are shown as standing on the four subsquares below the curved surface.
The surface itself stays fixed.
The base region remains [-2,2]×[-2,2].
The picture translates the algebraic Riemann sum into geometry: each summand is the volume of one box whose height is determined by the surface value at the chosen sample point.
A 2D plot appears in the lower right showing four red squares labeled ,,,.
Yellow dots appear at the bottom-left corners of the squares as the speaker discusses the sample-point choice.
Four red squares labeled ,,,
Yellow dots at lower-left corners
Axes spanning -2 to 2 in x and y
The lower-right graphic is added after the 3D plot is already present.
Yellow sample-point markers appear one by one or together to indicate the chosen corners.
The four squares partition the whole square domain without overlap.
The labeling order ,,, remains fixed.
This view isolates the base-region bookkeeping needed for the sum: it shows where the rectangles are and which corner is being used to determine each box height.
The speaker says that because the subrectangle is a whole little region, he needs to choose a specific point to plug into the function, and that different people can make different choices as to where that point is going to be.
One might think the height of the box above a subrectangle is automatically determined just by naming the subrectangle.
The video explains that the subrectangle is an entire region, so one must choose a specific sample point inside it to evaluate the function and determine the box height.
The speaker says that at some points in the box the box is higher than the graph of the function and at other points the graph of the function is higher, so there is a little error appearing.
A single highlighted box visibly protrudes above and falls below the curved surface.
One might think a box standing over a subrectangle exactly equals the volume under the surface over that subrectangle.
The video shows and states that a box can stick above the surface in some places and sit below it in others, so each finite box contributes local over- and under-estimation error.
The lecturer explicitly says the finite sum is not exactly the answer, but an approximation.
Step 3 uses ≈ rather than =.
One might think the sum over finitely many boxes already equals the volume under the surface.
The clip distinguishes the finite sum as an approximation and reserves equality for the limiting process as ||P|| → 0.
The lecturer says that although choices might matter at intermediate stages, in the limit the exact choice of , inside the rectangle is not going to matter.
The clip asserts this without proof.
Because each box height depends on the chosen point (,), one may expect the final volume to depend on those arbitrary choices.
In the limiting construction described here, the exact interior choice is said not to affect the final limiting volume.
The speaker explicitly clarifies that \|P\| means the largest rectangle in the partition goes to zero area, not just some individual rectangle.
The green annotation Largest Rectangle going to zero area points to the limit notation.
One might think the limit refers to one particular rectangle or to the index k going to zero.
The video explains that \|P\| refers to the whole partition's mesh: the largest rectangle's area tends to 0, which forces every rectangle in the partition to shrink.
The speaker says he could have chosen the top right corner or any corner, but for this approximation he chooses the bottom-left corner.
A learner may believe there is only one correct point to choose inside each subrectangle.
For a finite approximation, many choices are allowed; the bottom-left rule is just the convention used in this example.
The speaker says four rectangles are used because they can actually compute it in the video, unlike eighteen or another number.
The displayed result is only an approximation symbol , not an equality to the true volume.
This misconception is inferred from the contrast between the approximate four-box sum and the earlier limiting definition; the video does not state it as a named error.
One might treat the four-box sum as the exact volume under the surface.
The video distinguishes the finite approximation from the limiting definition of volume; four boxes are only a computable first approximation.
The speaker says “approximation to the area under this particular surface.”
The displayed quantity is labeled “Volume ≈”.
A learner may take the phrase “area under this surface” literally and think the computation is about planar area.
In this context the displayed mathematics is a volume approximation under a surface over a rectangle; the spoken wording is imprecise relative to the on-screen label.
The slide uses ≈ in front of the finite sum.
The speaker explicitly calls the result an approximation.
One might think that evaluating four sample rectangles already gives the exact volume under the surface.
Editorial clarification: the displayed quantity is a finite Riemann-sum approximation; exact volume would require passing to the limit defining the double integral.
The speaker says the multivariable problem is completely analogous to the single-variable problem and then says the multivariable situation is exactly the same except instead of little rectangles he is adding up little boxes.
The two-panel layout directly compares a curve with rectangles to a surface with boxes.
The video contrasts the one-dimensional rectangle sum with the two-dimensional box sum to motivate double integration by analogy.
The speaker says that in both stories, taking a larger and larger number of subdivisions gives a better and better approximation.
Both tilings become finer at the same time.
The refinement principle is applied to the box-sum method to explain why the approximation improves.
The speaker first names the subrectangle dimensions and , then says he needs to choose a specific point inside the rectangle to determine the box height.
The labels appear in that order on the same isolated rectangle.
The indexed subrectangle notation provides the domain piece on which the sample point is chosen.
The speaker describes local error from one box and then says that as the number of boxes increases and their sizes decrease, the errors get smaller.
The intuitive claim that local box-surface mismatch shrinks under refinement is what supports the statement that finer partitions give better approximations.
Step 1 introduces the partition; step 2 then says to choose a point (,) in each rectangle.
One must first partition the rectangular region into subrectangles before choosing a sample point inside each one.
The lecturer uses the chosen point to determine the height of the box.
The 3D box height is labeled by above the base rectangle.
The selected sample point is applied to the function to produce the height of the representative box.
The lecturer says he will compute that volume for one box and then add it up over all of them.
Step 3 sums terms of the form .
The finite Riemann sum is formed by collecting the volumes of all individual representative boxes.
Step 3 shows Volume ≈ ... and step 4 shows _{||P||→0} ....
The lecturer says the real step is to define the volume by taking a particular limit of the sum.
The exact volume definition generalizes the finite approximation by passing to the limit as the partition norm goes to zero.
The opening statement says the exact sample-point choice will not matter in the limit.
The later limit formula is written with arbitrary sample points (,) inside the sum.
The independence claim is asserted earlier than the formal limit definition and is not proved in the clip.
The claim that the limiting volume is independent of the particular sample points supports the legitimacy of defining volume through the limit of such sums.
After defining volume by a limit of sums, the speaker says Let's see an example and applies the same procedure to on [-2,2]×[-2,2].
The worked example directly instantiates the general Riemann-sum definition with a specific function, region, and four-rectangle partition.
Step 1 defines , and steps 3–4 use inside the sum.
The volume sum depends on the rectangle-area formula to interpret each summand as height times base area.
Step 2 introduces (,), and steps 3–4 use in the summand.
Choosing sample points is a required step before the Riemann sum can be formed.
The speaker asks how to find the area underneath of a surface and compares it to area underneath of a curve.
The speaker says that in multivariable calculus, instead of little rectangles, he is adding up little boxes.
The speaker says more subdivisions give a better approximation and explains that smaller boxes make the local errors smaller.
The speaker defines the rectangle's width and height change and indexes them by .
The labels and are shown on the rectangle.
The speaker says he needs to choose a specific point to plug into the function because the rectangle is a whole region.
The point is marked inside the rectangle.
Step 1 displays .
Step 2 says to choose a point (,) in each rectangle.
The lecturer explains that this point is used to get the height of the box.
The lecturer says the volume of the box is base times height.
The height label is shown on the 3D box.
Step 3 uses Volume ≈ .
The lecturer says this is not exactly the answer but an approximation.
Step 4 shows lim_{||P||→0}.
The lecturer calls P the partition and refers to the length of P going to zero.
The clip does not complete the formal definition of ||P||.
The lecturer says the exact choice of , inside the rectangle is not going to matter in the limit.
This is stated intuitively, not proved in the excerpt.
The four-step board gives the full definition.
The speaker says this limit is my definition of volume.
Covered · Opening title card identifying the topic as integrating multivariable functions.
Covered · Introduction of the problem of finding the quantity under a surface and analogy to area under a curve.
Covered · Explanation of approximating area under a curve by summing rectangle areas.
Covered · Translation of the rectangle idea to boxes in multivariable calculus.
Covered · Statement and animation showing that finer subdivisions improve the approximation.
Covered · Zoom into one box to explain local overestimation and underestimation error.
Covered · Formal setup begins by isolating one rectangle in the domain and labeling and .
Covered · Choice of a sample point inside the subrectangle to determine box height.
Covered · Final instant contains no additional distinct mathematical content beyond the ongoing explanation already covered.
Covered · Opening explanation that the exact sample-point choice inside each rectangle does not matter in the limit.
Covered · Transition from the 2D subrectangle to the 3D surface picture, explaining how one box volume is formed from base area and height .
Covered · Formal step 1: partition the rectangular region [a,b] × [c,d] into little rectangles with .
Covered · Formal step 2: choose a point (,) in each rectangle.
Covered · Formal step 3: approximate volume by the finite sum .
Covered · Formal step 4: define exact volume as the limit as ||P|| → 0 of the same sum; the clip ends before fully defining ||P||.
Covered · General definition of volume by double Riemann sums and verbal interpretation of the partition limit.
Covered · Transition to the worked example with on [-2,2]×[-2,2] and four equal regions.
Covered · Display of the four subsquares ,,, and selection of bottom-left sample points.
Covered · Construction of the four-term sum, simplification using common area , and expansion to the four explicit function evaluations; no final numeric evaluation occurs before the clip ends.
Covered · Mathematical worked example, formulas, diagrams, and spoken explanation are all present in this interval.
Covered · Presenter outro and subscription prompt; no additional mathematical content beyond the preceding worked example.
Reviewed subject paths
A point must be chosen because the subrectangle is a two-dimensional region, not a single input value. To determine the height of the representative box standing over that subrectangle, one must evaluate the function at a specific location.
Conditions: The domain is partitioned into subrectangles.; A function defines the surface height.
The finite Riemann sum is written with because it represents an approximation of the volume, not the exact value. A finite number of boxes cannot perfectly match a curved surface; some boxes will protrude above the surface (overestimating) and others will fall below it (underestimating).
Conditions: The sum involves a finite number of subrectangles.; The surface is curved (not a flat plane).
The notation means that the size of the largest subrectangle in the partition approaches zero. As the largest rectangle shrinks to zero area, all other rectangles in the partition must also shrink to zero.
Conditions: Applies to the limit step in the definition of volume by double Riemann sums.; represents the partition of the rectangular region.
The volume is defined by a four-step limiting process. First, partition the rectangular region into small subrectangles with area .
Conditions: The base region must be a rectangle .; The function defines the height of the surface.; The limit is taken as the largest subrectangle area approaches zero.
To approximate the volume, partition the square into four equal subsquares, each with side length 2 and area . Choose a sample point in each subsquare (e.g., the bottom-left corner).
Conditions: The base region is .; The partition consists of four equal subsquares.; Sample points are chosen according to a specific rule (e.g., bottom-left corners).
The sample point serves as the specific input location where the surface function is evaluated to determine the vertical height of the representative rectangular box. Geometrically, it anchors the top of the box to the surface directly above that point, allowing the box's volume to be calculated as base area times this specific height.
Conditions: A subrectangle has been defined in the domain.; A point is chosen inside that subrectangle.
In the limiting process where the partition becomes infinitely fine (), the exact location of the sample point inside each subrectangle does not matter. While different choices of will yield different finite Riemann sums (approximations), the limit of these sums as the subrectangles shrink to zero size is independent of the specific interior point chosen.
Conditions: Applies in the limit as the partition norm .; Each must be chosen inside its corresponding subrectangle.
The infinitesimal base area represents the footprint or base of a single representative rectangular box in the Riemann sum. It is the area of the -th subrectangle in the domain partition.
Conditions: The domain is partitioned into subrectangles.; and are the side lengths of the -th subrectangle.
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