Double integrals in Cartesian coordinates

Integrating with respect to y first corresponds to summing values along the y-direction for a fixed x. Geometrically, this is equivalent to calculating the area of a slice cut parallel to the yz-plane at that fixed x.

Conditions: The integration order is dy dx (y inner, x outer).; The slices are taken perpendicular to the x-axis (parallel to the yz-plane).

The inner integral represents the area of a vertical cross-section of the solid taken parallel to the yz-plane at a fixed x-value. Specifically, for a chosen point x₀, the slice extends in the y-direction from the lower boundary curve φ₁(x₀) to the upper boundary curve φ₂(x₀), and the height of the solid at any point (x₀, y) is given by $z=f$(x₀,y).

Conditions: The slice is taken at a fixed $x = x$₀ within [a,b].; The boundaries in the y-direction are defined by curves $y=φ$₁(x) and $y=φ$₂(x).

Changing the order of integration requires re-describing the same planar domain D using the other variable as the outer limit. If the original order was dy dx (y inner, x outer), the domain was described by x ranging from constants a to b, and y ranging between functions of x.

Conditions: The domain D is the same in both cases.; The function $f(x,y)$ is integrable over D.

The volume interpretation strictly holds when the function $f(x,y)$ is non-negative ($f \ge 0$) over the domain D, representing the physical volume of the solid above the xy-plane. However, the iterated-integral identity also holds for signed functions (where f can be negative) under suitable integrability conditions.

Conditions: For volume interpretation: $f(x,y) \ge 0$ on D.; For signed functions: f is integrable (e.g., continuous on a compact region).; The domain D is bounded by continuous curves $x=a$, $x=b$, $y=φ$₁(x), $y=φ$₂(x).

The base region D is a closed planar domain in the xy-plane. It is bounded by two vertical lines $x=a$ and $x=b$, and two curves $y=φ$₁(x) and $y=φ$₂(x).

Conditions: The region is closed and bounded.; x ranges from a to b.; y is bounded by functions φ₁(x) and φ₂(x) with φ₁(x) ≤ φ₂(x).

The functions φ₁(x) and φ₂(x) define the lower and upper boundaries of the integration domain D in the y-direction for any given x. In the iterated integral $\int _a^b$ ($\int$_{φ₁(x)}^{φ₂(x)} $f(x,y) dy$) dx, they serve as the variable limits of the inner integral.

Conditions: φ₁(x) represents the lower boundary curve $y = φ$₁(x).; φ₂(x) represents the upper boundary curve $y = φ$₂(x).; φ₁(x) ≤ φ₂(x) for all x in [a,b].

The solid region is constructed above the xy-plane with the top surface defined by $z=f(x,y)$ and the base defined by the domain D. Evaluating the double integral $\iint _D f(x,y) dx dy$ corresponds to calculating the volume of this solid (if $f \ge 0$) or the net signed volume (if f changes sign).

Conditions: The solid lies above the xy-plane.; Top surface: $z = f(x,y)$.; Base: Domain D in the xy-plane.; Volume interpretation assumes $f \ge 0$; otherwise, it is signed accumulation.

The double integral is computed as an iterated integral by slicing the solid parallel to the yz-plane. First, for a fixed x, the inner integral calculates the area of the cross-section by integrating the function $f(x,y)$ with respect to y from the lower boundary curve φ₁(x) to the upper boundary curve φ₂(x).

Conditions: The base region D is bounded by vertical lines $x=a$, $x=b$ and curves $y=φ$₁(x), $y=φ$₂(x) with φ₁(x) ≤ φ₂(x).; The integration order is y first, then x.