How to Set Up Double Integrals

1. Formula
   Observation

   The displayed formula is $\int_a^b f(x)\,dx$.

2. Audio
   Observation

   The narrator says that in single-variable calculus, integrals can be used to calculate the area under a curve.

$\int_a^b f(x)\,dx$

Single-variable definite integral representing the area under the graph of $f(x)$ from $x=a$ to $x=b$.

$x$ ranges over $[a,b]$; $f$ is a one-variable function.

1. Formula
   Observation

   The displayed formula is $\int_a^b \int_c^d f(x,y)\,dy\,dx$.

2. Audio
   Observation

   The narrator says a double integral can be used to find the three-dimensional volume under a surface and is evaluated by doing two integrals one after another.

$\int_a^b \int_c^d f(x,y)\,dy\,dx$

Iterated double integral for the volume under a surface over a rectangular region.

Outer variable $x$ ranges over $[a,b]$; inner variable $y$ ranges over $[c,d]$.

1. Formula
   Observation

   The slice is labeled $\text{Area} = A(x)$.

2. Audio
   Observation

   The narrator explains that one first finds a formula computing the area of any particular slice located at any given point.

$A(x)$

Cross-sectional area of a solid slice as a function of position along the slicing axis.

Function of the coordinate labeling the slice position; in this segment the visual example uses $x$.

1. Formula
   Observation

   The displayed formula is $\int_a^b A(x)\,dx$.

2. Audio
   Observation

   The narrator says the volume is computed by integrating the area function along the axis where the slices are located.

$\int_a^b A(x)\,dx$

Volume obtained by integrating cross-sectional area along an axis.

$x$ ranges over $[a,b]$; $A(x)$ is the slice area at position $x$.

1. Formula
   Observation

   The purple slice is labeled $\int_c^d g(y)\,dy$.

2. Audio
   Observation

   The narrator says that for a solid whose slices look like parabolas, the area is found by taking an integral of the function describing the parabola.

$\int_c^d g(y)\,dy$

Area of one parabolic cross-section, computed as a single-variable integral in $y$.

Inner variable $y$ ranges over $[c,d]$; $g$ describes the parabola within the slice.

1. Formula
   Observation

   The displayed formula becomes $\text{Volume} = \int_a^b \int_c^d g(y)\,dy\,dx$.

2. Audio
   Observation

   The narrator says that to use volume by cross-sections here, the area formula for a slice is itself an integral, and in this case it will be an integral in $y$ because the slices are parallel to the $y$-axis.

$\int_a^b \int_c^d g(y)\,dy\,dx$

Double integral expressing volume by cross-sections when each slice area is itself computed by an integral.

Outer variable $x$ ranges over $[a,b]$; inner variable $y$ ranges over $[c,d]$.

1. Formula
   Observation

   The displayed formula reads $\text{Volume} = \int_a^b \int_c^d g(x,y)$\,dy\,dx.

Volume

The volume of the solid represented by the iterated integral.

A real number when the integral is evaluated.

1. Formula
   Observation

   The integrand in the first displayed formula is $g(x,y)$.

2. Audio
   Observation

   The narration says that changing x changes the y-function that must be integrated to get the slice's cross-sectional area.

$g(x,y)$

A two-variable height function whose inner y-integral gives the cross-sectional area of a slice at fixed x.

Defined for the variables x and y in the illustrated solid.

1. Formula
   Observation

   From about 34 seconds onward the displayed formula changes to $\int_a^b \int_c^d f(x,y)$\,dy\,dx.

$f(x,y)$

The general two-variable integrand used in the later visual explanation of double integrals.

Defined for the variables x and y in the illustrated surface and solid.

1. Diagram
   Observation

   The coordinate plane is labeled with an x-axis.

2. Audio
   Observation

   The narration repeatedly refers to sweeping along the x-direction and to outer x-integral bounds.

x

One horizontal coordinate variable; in the displayed iterated integral it is the outer integration variable.

The narration gives the example interval from $x=-2$ to $x=2$.

1. Diagram
   Observation

   The coordinate plane is labeled with a y-axis.

2. Audio
   Observation

   The narration repeatedly refers to sweeping along the y-direction and to inner y-integral bounds.

y

One horizontal coordinate variable; in the displayed iterated integral it is the inner integration variable.

The narration gives the example interval from $y=-3$ to $y=2$.

1. Formula
   Observation

   The outer integral is written $\int_a^b$ ... dx.

2. Audio
   Observation

   The bounds of each integral reflect the extents of the sweeping motions in each direction.

a,b

Lower and upper bounds of the outer integral with respect to x.

In the worked example, these correspond to $x=-2$ and $x=2$.

1. Audio
   Observation

   The narrator says, 'Back in single variable calculus, probably one of the very first things you learned about integrals is that they can be used to calculate the area under a curve.'

2. Diagram
   Observation

   A 2D graph shows a blue curve above the x-axis with the region between the curve and the x-axis shaded from $a$ to $b$.

3. Formula
   Observation

   The shaded region is labeled $\int_a^b f(x)\,dx$.

The video introduces the definite integral $\int_a^b f(x)\,dx$ as the standard single-variable way to compute the area under a curve between $x=a$ and $x=b$.

1. One-variable setting.

2. Integration bounds are $a$ and $b$ on the horizontal axis.

3. The visual interpretation is area under the graph of $f(x)$.

1. Audio
   Observation

   The narrator says, 'In a very similar fashion, a double integral can be used to find the three-dimensional volume under a surface.'

2. Diagram
   Observation

   A 3D plot shows a purple surface above a rectangular base in the $xy$-plane, with a blue solid volume beneath it.

3. Formula
   Observation

   The solid is labeled $\int_a^b \int_c^d f(x,y)\,dy\,dx$.

The video presents the double integral $\int_a^b \int_c^d f(x,y)\,dy\,dx$ as the two-variable analogue of the single integral: instead of area under a curve, it gives the three-dimensional volume under a surface.

1. Two-variable function $f(x,y)$.

2. Rectangular integration region with $x\in[a,b]$ and $y\in[c,d]$.

3. Visual interpretation is volume under a surface.

1. Single integral as area under a curve

1. Audio
   Observation

   The narrator says, 'And the usual way we evaluate a double integral is by doing two integrals one after another, in a construction called an iterated integral, where you take an integral of an integral.'

2. Formula
   Observation

   The displayed notation nests one integral inside another: $\int_a^b \int_c^d f(x,y)\,dy\,dx$.

The video defines the usual evaluation method for a double integral as an iterated integral: perform one integral, then integrate the result again. This is described verbally as 'an integral of an integral.'

1. Applies to the displayed double-integral setup.

2. The order shown is inner $dy$ and outer $dx$.

1. Double integral as volume under a surface

1. Caption evidence
   Observation

   The title card reads 'Volume by Cross-Sections'.

2. Audio
   Observation

   The narrator explains that if a solid's slices along a particular axis all have a regular shape, one can compute the volume by first finding a formula for the area of any slice and then integrating that area function along the slicing axis.

3. Diagram
   Observation

   A red solid is sliced perpendicular to the $x$-axis; one slice is labeled $\text{Area} = A(x)$.

4. Formula
   Observation

   The final displayed formula is $\int_a^b A(x)\,dx$.

The video reviews the single-integral method for computing volume: label each slice by its position along an axis, write its area as a function such as $A(x)$, and integrate that area function over the interval of slice positions.

1. The solid is described by slices perpendicular to a chosen axis.

2. Each slice has an area expressible as a function of the slicing coordinate.

3. The integration runs along the axis that labels the slices.

1. Single integral as area under a curve

1. Audio
   Observation

   The narrator says, 'But what if I gave you a solid whose slices look like parabolas? There's no quick and dirty formula to compute the area under a parabola, but we still know how to do it. You take an integral of the function describing the parabola.'

2. Diagram
   Observation

   A purple solid is shown with a highlighted parabolic slice labeled $\int_c^d g(y)\,dy$.

3. Formula
   Observation

   The overall volume is rewritten as $\text{Volume} = \int_a^b \int_c^d g(y)\,dy\,dx$.

The video motivates double integrals by showing that if cross-sectional shapes are not elementary polygons or semicircles, the area of a slice may itself require an integral. Substituting that inner integral into the cross-section formula produces a double integral.

1. The solid is being computed by cross-sections.

2. The slice area is not available as a simple closed-form geometric formula.

3. In the example, slices are parallel to the $y$-axis, so the inner integral is in $y$.

1. Volume by cross-sections
2. Iterated integral construction

1. Formula
   Observation

   $\text{Volume} = \int_a^b \int_c^d g(x,y)$\,dy\,dx is shown above the 3D solid.

2. Audio
   Observation

   The narrator says this is the nested double integral expression for the volume.

The video presents a nested double integral as the formula for the volume of a solid. The inner integral computes the area of an arbitrary slice, and the outer integral sweeps those slices to build the full solid.

1. The displayed setup uses an inner y-integral and an outer x-integral.

2. The narration explains the case where the slice shape depends on x.

1. Audio
   Observation

   The narrator says we first have to compute the inner integral to find the proper area formula of an arbitrary slice.

2. Animation
   Observation

   A purple slice moves along the x-axis while the formula remains on screen.

In the displayed order dy dx, the inner integral is responsible for finding the cross-sectional area of a slice at a fixed x-value. Changing x changes the y-function that must be integrated.

1. Applies to the inner integral in the displayed iterated integral.

2. The slice is taken along the y-direction for fixed x.

1. Double integral as a volume formula

1. Audio
   Observation

   The narrator says we take an outer integral in order to have that slice sweep out the corresponding volume of the solid.

2. Animation
   Observation

   The moving slice accumulates into a larger solid under the surface.

After the inner integral produces a slice area, the outer integral moves that slice along the other axis so that the collection of slices carves out the whole solid.

1. The outer variable is x in the displayed formula.

2. The inner integral has already produced the slice area.

1. Inner integral gives slice area

1. Formula
   Observation

   The displayed formula becomes $\int_a^b \int_c^d f(x,y)$\,dy\,dx.

2. Audio
   Observation

   The narrator says this is the visual to keep in mind when thinking about what a double integral is doing.

3. Animation
   Observation

   A blue slice sweeps along one axis and then the resulting slab sweeps along the other axis.

The video defines the meaning of a double integral geometrically: the inner integral sweeps out an arbitrary slice of the solid along one axis, and the outer integral sweeps that slice along the other axis to carve out the full solid.

1. The explanation is presented for the displayed order dy dx.

2. The solid is shown above a rectangular base in the xy-plane.

1. Double integral as a volume formula

1. Audio
   Observation

   If the inner integral is a y-integral, you sweep along the y-direction. If the inner integral is an x-integral, you sweep along the x-direction.

The differential in the inner integral tells which coordinate direction is being swept first. A y-integral sweeps along y; an x-integral sweeps along x.

1. Applies to iterated integrals written with explicit differentials such as dy dx or dx dy.

1. Visual interpretation of a double integral

1. Audio
   Observation

   Also note that the bounds of each integral reflect the extents of the sweeping motions in each direction.

2. Animation
   Observation

   The formula bounds are highlighted while the solid remains on screen.

Each pair of bounds records how far the corresponding sweep extends along its axis. The inner bounds set the extent of the slice, and the outer bounds set the extent over which the slice is moved.

1. Discussed for the displayed iterated integral with separate inner and outer bounds.

1. Visual interpretation of a double integral

1. Audio
   Observation

   Using only constant bounds, a double integral can only sweep out solids whose base is a rectangular region of the xy-plane.

2. Animation
   Observation

   The base is shown as a rectangle with labels $y=-3$, $y=2$, $x=-2$, $x=2$.

When both inner and outer bounds are constants, every slice has the same extent in the inner variable, so the projection of the solid onto the xy-plane is a rectangle. The video describes these solids as box-like shapes with a curvy top.

1. All four bounds a,b,c,d are constants.

2. The base region is therefore rectangular in the xy-plane.

1. Integral bounds encode sweep extents

1. Audio
   Observation

   The narrator states that the usual way to evaluate a double integral is by doing two integrals one after another, in a construction called an iterated integral, where you take an integral of an integral.

2. Formula
   Observation

   The displayed notation $\int_a^b \int_c^d f(x,y)\,dy\,dx$ visually nests one integral inside another.

A double integral is evaluated by performing two integrals successively, i.e. as an iterated integral.

1. The object under discussion is a double integral of the form shown in the video.

For the displayed setup $\int_a^b \int_c^d f(x,y)\,dy\,dx$, the evaluation procedure is sequential.

1. Audio
   Observation

   The narrator says, 'So that's volume by cross sections in a nutshell, and we only needed one integral to do it.'

2. Formula
   Observation

   The preceding displayed formula is $\int_a^b A(x)\,dx$.

If the cross-sectional area function is already known, the volume can be computed with a single integral.

1. The solid is described by slices along an axis.

2. An explicit area function such as $A(x)$ is available.

For the cross-section method as presented, one integral suffices once $A(x)$ is known.

1. Audio
   Observation

   The narrator says there is no quick and dirty formula to compute the area under a parabola, but we still know how to do it by taking an integral of the function describing the parabola.

2. Formula
   Observation

   The slice is labeled $\int_c^d g(y)\,dy$.

For a slice shaped like a parabola, its area is obtained by integrating the function that describes the parabola.

1. The slice boundary is described by a function $g(y)$.

2. The slice extends over $y\in[c,d]$.

For the parabolic-slice example shown, the slice area is $\int_c^d g(y)\,dy$.

1. Audio
   Observation

   The narrator says, 'In this case, it will be an integral in y because our slices are parallel to the y-axis.'

2. Formula
   Observation

   The displayed expression is $\text{Volume} = \int_a^b \int_c^d g(y)\,dy\,dx$.

3. Diagram
   Observation

   The highlighted slice lies in a plane parallel to the $y$-axis direction within the 3D plot.

In the displayed example, the inner integral is with respect to $y$ because the slices are parallel to the $y$-axis.

1. The solid is being decomposed into slices.

2. The slice orientation is parallel to the $y$-axis.

For the specific parabolic-slice construction shown, the inner differential is $dy$.

1. Audio
   Observation

   This is because the shape of a slice also depends on where along the x-axis the slice is located.

2. Audio
   Observation

   Change the x value, and you change the y-function that needs to be integrated to get its cross-sectional area.

For the displayed solid, the cross-sectional shape at a fixed x depends on the value of x, so the inner y-integrand changes when x changes.

1. The solid is represented by $\text{Volume} = \int_a^b \int_c^d g(x,y)$\,dy\,dx.

2. The slice is taken along the y-direction at fixed x.

For the illustrated family of slices indexed by x.

1. Audio
   Observation

   Using only constant bounds, a double integral can only sweep out solids whose base is a rectangular region of the xy-plane.

2. Audio
   Observation

   Restricting us to solids which look like a box with a curvy top.

If a double integral uses only constant bounds, the swept solid has a rectangular base in the xy-plane.

1. The integral is an iterated double integral with constant limits.

2. The solid is generated by the described sweep process.

For the class of solids produced by constant-bound iterated integrals shown in the video.

1. Audio
   Observation

   We first need to find a formula to compute the area of an arbitrary slice of the solid across some axis.

2. Audio
   Observation

   This is the job of our inner integral.

When setting up a double integral by cross sections, the first task is to find a formula for the area of an arbitrary slice, and that task is performed by the inner integral.

1. The solid's base may be non-rectangular.

2. The setup follows the volume-by-cross-sections method described in the video.

For the general setup procedure introduced at the end of the clip.

1. Audio
   Observation

   At 00:00-00:11 the speaker says the region looks easiest to slice along the y direction because the two bounding curves seem to naturally describe a floor and ceiling, as opposed to two walls.

2. Diagram
   Observation

   The shaded region is bounded above by a red curve and below by a green line.

For the displayed region, slicing along the y-direction is presented as the easier choice because the two bounding curves naturally serve as lower and upper bounds for each slice.

1. The region is bounded by two curves.

2. One curve lies above the other over the relevant x-range.

For the region shown in this example.

1. Audio
   Observation

   At 00:18-00:23 the speaker says the bounds of the integral will depend on which slice we're looking at.

2. Audio
   Observation

   At 00:46-00:57 the speaker says the bounds of the inner integral will themselves be functions of the outer variable, in this case x, since the width of each slice depends on its x location.

3. Diagram
   Observation

   Different highlighted slices show different y-values at their lower and upper ends.

When the vertical extent of each slice changes with x, the lower and upper limits of the inner integral must be written as functions of x rather than constants.

1. The integration order is dy dx.

2. The slice width depends on x.

For each slice position x in the region.

1. Audio
   Observation

   At 00:12-00:17 the speaker says, 'To compute the area of one of these slices, we use a single integral, the inner integral.'

2. Formula
   Observation

   At 01:53-02:00 the substituted inner integral is boxed and labeled 'Area of the xth slice'.

In the displayed setup, the inner integral with variable y-limits gives the area of the slice of the solid located at a given x-position.

1. The inner integral is taken with respect to y.

2. The limits are the lower and upper y-values of the slice at that x.

For an arbitrary fixed x within the outer bounds.

1. Audio
   Observation

   At 02:00-02:17 the speaker says to set up the bounds of the outer integral by plugging in the leftmost x position of the region and the rightmost x position of the region as the lower and upper bounds.

2. Formula
   Observation

   The labels $x = a$ and $x = b$ are replaced by $x = -2$ and $x = 2$.

For this example, the outer integral bounds are obtained directly from the minimum and maximum x-values of the shaded region.

1. The outer variable is x.

2. The region has identifiable leftmost and rightmost x-positions.

For the whole region shown.

1. Audio
   Observation

   At 02:33-02:38 the speaker says, 'This double integral computes the volume of the original solid we started with.'

2. Formula
   Observation

   The final displayed integral is $\text{Volume} = \int_{-2}^{2} \int_{-1/2 x - 1}^{-1/2 x^2 - 1/2 x + 1} g(x,y)$\,dy\,dx.

1. The video does not explicitly define $g(x,y)$ as a height function, though the visual context presents the integral as a volume computation.

After substituting the explicit inner and outer bounds, the displayed double integral computes the volume of the original solid.

1. The bounds are correctly read from the region.

2. The integrand $g(x,y)$ is the function being integrated for volume.

For the solid depicted in the animation.

1. Audio
   Observation

   The narrator first explains volume by cross-sections, then asks why double integrals are needed, then gives the parabolic-slice example and substitutes the slice-area integral into the outer volume integral.

2. Formula
   Observation

   The sequence of displayed formulas is $\int_a^b A(x)\,dx$, then $\int_c^d g(y)\,dy$, then $\text{Volume} = \int_a^b \int_c^d g(y)\,dy\,dx$.

3. Diagram
   Observation

   The visuals move from a generic sliced solid to a highlighted parabolic slice and then to the combined nested integral.

1. Expression

   $$V = \int_a^b A(x)\,dx$$
   Explanation

   Start with the cross-section method: integrate the slice area function along the slicing axis.

   Justification

   This is the formula explicitly shown and verbally explained for volume by cross-sections.

   Shown in the video
2. Expression

   $$A(x) = \int_c^d g(y)\,dy$$
   Explanation

   For the parabolic-slice example, the area of one slice is itself computed by integrating the function describing the parabola.

   Justification

   The narrator states that there is no quick formula for the area under a parabola, so one takes an integral of the describing function; the slice is labeled with this integral.

   Shown in the video
3. Expression

   $$V = \int_a^b \int_c^d g(y)\,dy\,dx$$
   Explanation

   Substitute the slice-area integral into the outer volume integral to obtain a nested double integral.

   Justification

   The video visually replaces $A(x)$ by $\int_c^d g(y)\,dy$ and displays the combined formula.

   Shown in the video

A double integral arises naturally when the cross-sectional area itself must be computed by integration.

1. Audio
   Observation

   The narrator explains that changing x changes the y-function to be integrated, then says we end up with a nested double integral expression for the volume.

2. Formula
   Observation

   $\text{Volume} = \int_a^b \int_c^d g(x,y)$\,dy\,dx is displayed throughout this derivation.

3. Animation
   Observation

   A slice moves along x while the solid is built under the surface.

1. Expression

   $$\int_c^d g(x,y)\,dy$$
   Explanation

   For a fixed x, integrate in y to obtain the area of the corresponding slice.

   Justification

   The narration states that the inner integral finds the proper area formula of an arbitrary slice.

   Shown in the video
2. Expression

   $$\int_a^b \left(\int_c^d g(x,y)\,dy\right) dx$$
   Explanation

   Then integrate that slice-area expression over x so the slice sweeps through the whole solid.

   Justification

   The narration states that the outer integral makes the slice sweep out the corresponding volume.

   Shown in the video

The volume is represented by the nested integral $\text{Volume} = \int_a^b \int_c^d g(x,y)$\,dy\,dx.

1. Audio
   Observation

   The narrator gives the example bounds $y=-3$ to $y=2$ for the inner y-integral and $x=-2$ to $x=2$ for the outer x-integral.

2. Diagram
   Observation

   The top-down view labels the rectangle with $y=2$, $y=-3$, $x=-2$, and $x=2$.

3. Animation
   Observation

   A vertical slice is shown inside the rectangle and then the rectangle is lifted back into 3D.

1. Expression

   $$y=-3 \text{ to } y=2$$
   Explanation

   The inner y-integral sweeps a slice from the lower horizontal boundary to the upper horizontal boundary of the rectangle.

   Justification

   The narration explicitly says any given y-slice should be swept starting from $y=-3$ and ending at $y=2$.

   Shown in the video
2. Expression

   $$x=-2 \text{ to } x=2$$
   Explanation

   The outer x-integral then moves that slice from the left vertical boundary to the right vertical boundary.

   Justification

   The narration explicitly says the slice should be swept along the x-direction starting from $x=-2$ and ending at $x=+2$.

   Shown in the video

For the displayed rectangular example, the inner bounds are $y\in[-3,2]$ and the outer bounds are $x\in[-2,2]$.

1. Audio
   Observation

   The narration proceeds from choosing slice direction, to inner integral bounds, to outer integral bounds, to the final statement that the double integral computes volume.

2. Formula
   Observation

   The formula evolves from $\int_a^b \int_c^d g(x,y) dy dx$ to $\int_a^b \int_{c(x)}^{d(x)} g(x,y) dy dx$ and then to $\int_{-2}^{2} \int_{-1/2 x - 1}^{-1/2 x^2 - 1/2 x + 1} g(x,y) dy dx$.

1. Expression

   $$Volume = \int_a^b \int_c^d g(x,y)\,dy\,dx$$
   Explanation

   Start from a generic iterated integral for volume with outer variable x and inner variable y.

   Justification

   This is the initial formula displayed on screen.

   Shown in the video
2. Expression

   $$\text{Choose slicing along } y$$
   Explanation

   Select the y-direction for the inner slice because the bounding curves act like floor and ceiling.

   Justification

   Stated in the audio at 00:00-00:11.

   Shown in the video
3. Expression

   $$c, d \to c(x), d(x)$$
   Explanation

   Replace the constant inner bounds with functions of x because different slices have different vertical extents.

   Justification

   Stated in the audio at 00:18-00:57 and shown by changing slice labels.

   Shown in the video
4. Expression

   $$c(x) = -\frac{1}{2}x - 1, \quad d(x) = -\frac{1}{2}x^2 - \frac{1}{2}x + 1$$
   Explanation

   Read the lower and upper bounding curves as functions of x from the given line and parabola equations.

   Justification

   Given explicitly in the audio and formulas at 01:34-01:48.

   Shown in the video
5. Expression

   $$Volume = \int_a^b \int_{-\frac{1}{2}x - 1}^{-\frac{1}{2}x^2 - \frac{1}{2}x + 1} g(x,y)\,dy\,dx$$
   Explanation

   Substitute the curve equations into the inner limits, yielding the area of the xth slice inside the outer integral.

   Justification

   Shown on screen at 01:50-02:00 and described in the audio.

   Shown in the video
6. Expression

   $$a = -2, \quad b = 2$$
   Explanation

   Use the leftmost and rightmost x-values of the region as the outer bounds.

   Justification

   Stated in the audio at 02:00-02:27 and shown by labels $x=-2$ and $x=2$.

   Shown in the video
7. Expression

   $$Volume = \int_{-2}^{2} \int_{-\frac{1}{2}x - 1}^{-\frac{1}{2}x^2 - \frac{1}{2}x + 1} g(x,y)\,dy\,dx$$
   Explanation

   Substitute the outer bounds to obtain the fully explicit iterated integral for the solid's volume.

   Justification

   Displayed as the final formula and affirmed in the audio at 02:33-02:38.

   Shown in the video

The video derives the explicit double integral for the volume by first determining variable inner bounds from the bounding curves and then fixing the outer bounds from the region's extreme x-values.

1. Audio
   Observation

   The narrator asks, 'But what if I gave you a solid whose slices look like parabolas?' and then explains that the slice area is found by integrating the function describing the parabola.

2. Diagram
   Observation

   A purple solid is shown with a highlighted parabolic slice labeled $\int_c^d g(y)\,dy$.

3. Formula
   Observation

   The final displayed formula is $\text{Volume} = \int_a^b \int_c^d g(y)\,dy\,dx$.

Compute the volume of a solid whose slices parallel to the $y$-axis are parabolic in shape.

1. The solid is decomposed into slices along the $x$-direction.

2. Each slice is bounded by a function $g(y)$.

3. The inner variable range is $c\le y\le d$.

4. The outer slicing coordinate runs from $a$ to $b$.

Express the volume as an iterated integral.

1. Expression

   $$V = \int_a^b A(x)\,dx$$
   Explanation

   Use the general volume-by-cross-sections formula.

   Justification

   The video has just reviewed this method and displays the formula.

   Shown in the video
2. Expression

   $$A(x) = \int_c^d g(y)\,dy$$
   Explanation

   Replace the unknown slice-area formula by an integral over the parabolic boundary function.

   Justification

   The narrator states that the area under a parabola is found by integrating the function describing it, and the slice is labeled with this expression.

   Shown in the video
3. Expression

   $$V = \int_a^b \int_c^d g(y)\,dy\,dx$$
   Explanation

   Combine the outer and inner integrals into a double integral.

   Justification

   This is the final displayed formula in the clip.

   Shown in the video

$\displaystyle V=\int_a^b\int_c^d g(y)\,dy\,dx$

The answer matches the on-screen formula and the narrator's explanation that the inner integral is in $y$ because the slices are parallel to the $y$-axis.

1. Diagram
   Observation

   A top-down rectangle in the xy-plane is labeled $y=2$, $y=-3$, $x=-2$, $x=2$.

2. Audio
   Observation

   The narrator walks through the inner y-bounds and then the outer x-bounds for this example.

3. Animation
   Observation

   The 2D rectangle is converted back into the 3D solid with a wavy top.

Interpret the bounds of $\int_a^b \int_c^d f(x,y)$\,dy\,dx for a solid whose base is the rectangle shown in the xy-plane.

1. The displayed integral order is dy dx.

2. The rectangle is bounded by $y=-3$ and $y=2$.

3. The rectangle is bounded by $x=-2$ and $x=2$.

Identify which bounds belong to the inner integral and which belong to the outer integral.

1. Expression

   $$\int_{-3}^{2} f(x,y)\,dy$$
   Explanation

   Because the inner integral is with respect to y, its bounds come from the horizontal edges of the rectangle.

   Justification

   The narration says the inner y-integral indicates what portion of the y-axis should be swept to produce a slice.

   Derived from the video
2. Expression

   $$\int_{-2}^{2} \left(\int_{-3}^{2} f(x,y)\,dy\right) dx$$
   Explanation

   The outer integral is with respect to x, so its bounds come from the vertical edges of the rectangle.

   Justification

   The narration says the outer x-integral indicates the portion of the x-axis along which the slice is swept.

   Derived from the video

Inner bounds: y from -3 to 2. Outer bounds: x from -2 to 2.

The top-down diagram explicitly labels the rectangle with $y=2$, $y=-3$, $x=-2$, and $x=2$, matching the stated bounds.

1. Audio
   Observation

   Take, for example, this solid. Its base in the xy-plane is the region bounded by a line segment on the bottom and a parabola on top.

2. Animation
   Observation

   The view changes to a solid whose base is a curved region between a straight lower boundary and a parabolic upper boundary.

3. Audio
   Observation

   How do we set up a double integral here? Think back again to how volume by cross sections worked.

1. The clip ends before the actual bounds or final integral setup are written out.

Set up a double integral for a solid whose base is bounded below by a line segment and above by a parabola.

1. The base is not a rectangle.

2. The lower boundary is a line segment.

3. The upper boundary is a parabola.

4. The method to use is volume by cross sections.

Begin the setup by identifying the role of the inner integral.

1. Expression

   $$\text{Find the area of an arbitrary slice.}$$
   Explanation

   Return to the cross-section method: first determine a formula for the area of a slice across some axis.

   Justification

   The narration explicitly says we first need to find a formula to compute the area of an arbitrary slice of the solid.

   Shown in the video
2. Expression

   $$\text{Use the inner integral for that area formula.}$$
   Explanation

   That slice-area computation is assigned to the inner integral.

   Justification

   The narration says this is the job of our inner integral.

   Shown in the video

The clip introduces the setup strategy but does not reach a completed explicit integral within the provided duration.

The final spoken line in the clip is that finding the slice-area formula is the job of the inner integral, and no completed bound expression is shown before the clip ends.

1. Diagram
   Observation

   A purple shaded region in the xy-plane is bounded below by a green line and above by a red downward-opening parabola.

2. Formula
   Observation

   The clip supplies the equations $y = -1/2 x - 1$ and $y = -1/2 x^2 - 1/2 x + 1$, and later the outer bounds $x = -2$ and $x = 2$.

3. Audio
   Observation

   The narration uses this region to explain how to set up the double integral for volume.

1. The exact algebraic derivation of the intersection points $x=-2$ and $x=2$ is not shown; the values are given directly.

Given a solid whose base region in the xy-plane is bounded below by $y = -\frac{1}{2}x - 1$ and above by $y = -\frac{1}{2}x^2 - \frac{1}{2}x + 1$, set up a double integral that computes its volume.

1. Lower bounding curve: $y = -\frac{1}{2}x - 1$.

2. Upper bounding curve: $y = -\frac{1}{2}x^2 - \frac{1}{2}{x} + 1$.

3. Leftmost x-value of the region: $x = -2$.

4. Rightmost x-value of the region: $x = 2$.

5. Integrand shown symbolically as $g(x,y)$.

Write the iterated double integral for the volume using slicing along the y-direction.

1. Expression

   $$\text{Slice along } y$$
   Explanation

   Choose vertical slices because the two curves naturally give a lower and upper y-bound for each x.

   Justification

   Stated in the audio at 00:00-00:11.

   Shown in the video
2. Expression

   $$\int_{c(x)}^{d(x)} g(x,y)\,dy$$
   Explanation

   Use an inner integral in y to compute the area of one slice at fixed x.

   Justification

   Stated in the audio at 00:12-00:17 and reinforced by the label Area of the xth slice.

   Shown in the video
3. Expression

   $$c(x) = -\frac{1}{2}x - 1$$
   Explanation

   Take the lower curve as the lower inner bound.

   Justification

   Given explicitly at 01:34-01:41.

   Shown in the video
4. Expression

   $$d(x) = -\frac{1}{2}x^2 - \frac{1}{2}x + 1$$
   Explanation

   Take the upper curve as the upper inner bound.

   Justification

   Given explicitly at 01:42-01:48.

   Shown in the video
5. Expression

   $$\int_{-2}^{2}$$
   Explanation

   Use the extreme x-values of the region as the outer bounds.

   Justification

   Given explicitly at 02:19-02:27.

   Shown in the video
6. Expression

   $$Volume = \int_{-2}^{2} \int_{-\frac{1}{2}x - 1}^{-\frac{1}{2}x^2 - \frac{1}{2}x + 1} g(x,y)\,dy\,dx$$
   Explanation

   Combine the inner and outer bounds into the final iterated integral.

   Justification

   Displayed as the final formula and stated to compute the volume at 02:33-02:38.

   Shown in the video

$\text{Volume} = \int_{-2}^{2} \int_{-\frac{1}{2}x - 1}^{-\frac{1}{2}x^2 - \frac{1}{2}x + 1} g(x,y)$\,dy\,dx

The answer matches the final on-screen formula and the narrator's statement that this double integral computes the volume of the original solid.

1. Diagram
   Observation

   A 2D coordinate system appears with tick marks labeled $a$ and $b$; a blue curve is drawn above the x-axis and the region under it is shaded blue.

2. Formula
   Observation

   The shaded region is labeled $\int_a^b f(x)\,dx$.

1. Cartesian axes

2. Blue curve

3. Shaded region between curve and x-axis

4. Labels $a$, $b$

5. Formula $\int_a^b f(x)\,dx$

1. Axes appear first.

2. Tick marks $a$ and $b$ are added.

3. The curve is drawn.

4. The region under the curve fills with color.

5. The integral label appears.

1. The setting remains two-dimensional.

2. The horizontal axis represents the integration variable $x$.

The animation visually identifies a single definite integral with the area under a curve from $a$ to $b$.

1. Diagram
   Observation

   A 3D plot shows a purple surface above a rectangular grid in the $xy$-plane; a blue solid volume forms beneath the surface.

2. Formula
   Observation

   The solid is labeled $\int_a^b \int_c^d f(x,y)\,dy\,dx$.

1. 3D axes labeled $x$ and $y$

2. Purple surface

3. Rectangular base region

4. Blue solid volume

5. Formula $\int_a^b \int_c^d f(x,y)\,dy\,dx$

1. The scene shifts from 2D to 3D.

2. A surface is introduced above a rectangular domain.

3. A blue volume appears beneath the surface.

4. The double-integral formula is overlaid on the solid.

1. The base remains a rectangle in the $xy$-plane.

2. The volume is bounded above by the surface and below by the base region.

The animation presents the double integral as the three-dimensional analogue of area under a curve: volume under a surface.

1. Diagram
   Observation

   The screen splits vertically, showing the 2D area example on the left and the 3D volume example on the right.

2. Formula
   Observation

   Left side shows $\int_a^b f(x)\,dx$; right side shows $\int_a^b \int_c^d f(x,y)\,dy\,dx$.

3. Animation
   Observation

   Boxes highlight parts of the right-hand double-integral formula.