The Volume Under a Surface

From areas to volumes

The area under a curve y = f(x) is built from thin vertical strips: each is a height times a small width, and integrating adds them up. A surface z = f(x, y) standing over a region of the xy-plane encloses a solid, and the same idea, used twice, gives its volume.

Cut the solid with a vertical plane at a fixed y. The cut face is a flat region under a curve, so it has an area. Cut again a little further along, and the slab between the two cuts has a volume of about that area times the gap. Adding all the slabs gives the volume.

The dome $z=16\left(1-x^2-y^2\right)$z = 16(1 − x² − y²) over the disc $x^2+y^2\le 1$x² + y² ≤ 1, with the slice at y = 0.45 picked out in gold. The face under that curve has area $A\left(0.45\right)\approx 15.19$A(0.45) ≈ 15.19, and stacking every such face from y = −1 to y = 1 gives the dome’s volume, $8\pi \approx 25.13$8π ≈ 25.13.

The inner integral: one slice

Take f(x, y) = xy over the rectangle $0\le x\le 2$0 ≤ x ≤ 2, $0\le y\le 3$0 ≤ y ≤ 3. Hold y fixed. The slice is the curve z = xy, a function of x alone with y as a constant, and its area is $A\left(y\right)={\int }_0^{2}xydx={\left[\frac{x^{2}y}{2}\right]}_0^2=2y$A(y) = ∫₀² xy dx = [x²y/2]₀² = 2y. This is ordinary integration in x: y is carried along like any constant factor.

At y = 1.5 the slice is the line z = 1.5x from x = 0 to x = 2, and the area under it is A(1.5) = 3, the triangle with base 2 and height 3.

The slice of z = xy at y = 1.5: the line z = 1.5x over $0\le x\le 2$0 ≤ x ≤ 2. The shaded face has area $A\left(1.5\right)={\int }_0^{2}1.5xdx=3$A(1.5) = ∫₀² 1.5x dx = 3, which is 2y at y = 1.5.

The outer integral: stack the slices

A slab of thickness dy at position y has volume A(y) dy, so the volume is $V={\int }_0^{3}A\left(y\right)dy={\int }_0^{3}2ydy={\left[y^2\right]}_0^3=9$V = ∫₀³ A(y) dy = ∫₀³ 2y dy = [y²]₀³ = 9. Putting the inner integral inside the outer one gives the double integral $V={\int }_0^3{\int }_0^{2}xydxdy$V = ∫₀³ ∫₀² xy dx dy: an integral of integrals, read from the inside out, with dx belonging to the inner $\int$∫ and dy to the outer.

So the volume is the area under the graph of A(y). The slice areas grow from 0 at y = 0 to 6 at y = 3, and the area under that straight line is 9.

The slice area A(y) = 2y plotted against y from 0 to 3. The shaded area under it, ${\int }_0^{3}2ydy=9$∫₀³ 2y dy = 9, is the volume under z = xy over the rectangle.

Either order, and a check by sums

Slice the other way, holding x fixed: ${\int }_0^{3}xydy={\left[\frac{x{y}^2}{2}\right]}_0^3=4.5x$∫₀³ xy dy = [xy²/2]₀³ = 4.5x, and ${\int }_0^{2}4.5xdx={\left[2.25x^2\right]}_0^2=9$∫₀² 4.5x dx = [2.25x²]₀² = 9. The same solid, sliced in the other direction, has the same volume. And because xy is x times y, the integral also splits into ${\int }_0^{2}xdx\times {\int }_0^{3}ydy=2\times 4.5=9$∫₀² x dx × ∫₀³ y dy = 2 × 4.5 = 9.

A double integral is the limit of a double sum. Cut the rectangle into a 100 by 100 grid of small rectangles, each of area 0.02 × 0.03, and add up f times the area of each. Taking f at each lower-left corner gives 8.82; at each upper-right corner, 9.18; at each center, 9. The volume is squeezed between the first two.

When f is not a product

For $f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y² over $0\le x\le 1$0 ≤ x ≤ 1, $0\le y\le 2$0 ≤ y ≤ 2, the inner integral is ${\int }_0^1\left(x^2+y^2\right)dx={\left[\frac{x^3}{3}+x{y}^2\right]}_0^1=\frac{1}{3}+y^2$∫₀¹ (x² + y²) dx = [x³/3 + xy²]₀¹ = 1/3 + y². The $y^2$y² term is a constant in x, so it integrates to $y^2$y² times the width, 1. The outer integral is ${\int }_0^2\left(\frac{1}{3}+y^2\right)dy=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}$∫₀² (1/3 + y²) dy = 2/3 + 8/3 = 10/3.

The other order gives ${\int }_0^2\left(x^2+y^2\right)dy=2x^2+\frac{8}{3}$∫₀² (x² + y²) dy = 2x² + 8/3, and ${\int }_0^1\left(2x^2+\frac{8}{3}\right)dx=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}$∫₀¹ (2x² + 8/3) dx = 2/3 + 8/3 = 10/3 again. A midpoint sum on a 100 by 100 grid gives 3.3332.

A region that is not a rectangle

Over the triangle $0\le y\le x\le 2$0 ≤ y ≤ x ≤ 2, below the line y = x, the slices have different lengths, so the inner limits depend on the outer variable. Holding x fixed, y runs from 0 up to x: ${\int }_0^{x}xydy=\frac{x^3}{2}$∫₀ˣ xy dy = x³/2. Then ${\int }_0^2\frac{x^3}{2}dx={\left[\frac{x^4}{8}\right]}_0^2=2$∫₀² x³/2 dx = [x⁴/8]₀² = 2.

Holding y fixed instead, x runs from y to 2. The inner integral is $\frac{x^{2}y}{2}$x²y/2 at x = 2 minus its value at x = y, which is $2y-\frac{y^3}{2}$2y − y³/2, and ${\int }_0^2\left(2y-\frac{y^3}{2}\right)dy=4-2=2$∫₀² (2y − y³/2) dy = 4 − 2 = 2. The limits change with the order; the volume does not.

The triangle $0\le y\le x\le 2$0 ≤ y ≤ x ≤ 2, under the line y = x. For the x marked, 1.2, the inner integral runs in y from the point on the x-axis, y = 0, up to the point on the line, y = x = 1.2.

The usual mistakes

Multiplying the area of the region by the greatest height. For xy over the 2 by 3 rectangle that gives 6 × 6 = 36, the volume of the box around the solid; the surface is far lower than 6 over most of the rectangle, and the double integral gives 9.

Swapping the limits with the variables. In ${\int }_0^3{\int }_0^{2}xydxdy$∫₀³ ∫₀² xy dx dy the inner limits 0 and 2 belong to x, because dx is the inner differential.

Integrating the other variable in the inner integral. In ${\int }_0^1\left(x^2+y^2\right)dx$∫₀¹ (x² + y²) dx, $y^2$y² is a constant: it integrates to $y^{2}x$y²x, not $\frac{y^3}{3}$y³/3.

Keeping rectangle limits on a triangle. Over the triangle below y = x, the inner integral in y stops at y = x, not at 2.

A heap of gravel

In the application below, the depth of a heap of gravel is a function of position on a rectangular patch. Integrating the depth along the length gives the area of a cross-section, and integrating that across the width gives the volume, first of the whole heap and then of the part at one end.