Double integrals

Volume under a surface

A function of two variables $f(x, y)$ gives a height for each point of the plane, and its graph is a surface (lesson 4.1.1). Over a region $R$ of the $xy$-plane, the solid between $R$ and the surface has a volume. The double integral $\iint_R f(x, y)\,dA$ is that volume.

Its units are the height's units times area units. If $f$ is a depth of water in metres over a pond measured in metres, the double integral is a volume in cubic metres. If $f$ is a density in kilograms per square metre, it is a mass in kilograms.

A constant height is the simplest case. Water 0.5 m deep over a pool 4 m by 3 m has volume $0.5 \times 12 = 6$ cubic metres: height times area. When the height varies, cut the region into small cells, as lesson 2.5.1 cut an interval into strips.

Boxes on a grid

Take $f(x, y) = x^2 + 2y^2$, a bowl, over the rectangle $R = [0, 2] \times [0, 1]$: the points with $0 \le x \le 2$ and $0 \le y \le 1$. Cut $[0, 2]$ into $n = 4$ strips of width $\Delta x = 0.5$, and $[0, 1]$ into $m = 2$ strips of width $\Delta y = 0.5$. That makes 8 cells, each of area $\Delta A = \Delta x\,\Delta y = 0.25$.

Over each cell stand a box whose height is $f$ at the cell's midpoint. The heights are:

The heights add up to 15.5, so the 8 boxes have volume $15.5 \times 0.25 = 3.875$. The exact volume, found below, is 4.

Finer grids close in on the volume. The midpoint error shrinks with the square of the cell width, as in one variable (lesson 2.5.1). Cells 10 times narrower give an error 100 times smaller.

Slices: the iterated integral

The exact volume comes from slicing. Hold $x$ fixed and the solid's cross-section is the region under the curve $y \mapsto f(x, y)$, a slice as in lesson 4.1.1. Its area is an ordinary integral in $y$, with $x$ treated as a constant (lesson 2.5.3):

The slice at $x = 1$, shaded in the figure, has area $A(1) = \frac{5}{3} \approx 1.667$. At $x = 0$ the slice has area 0.667, and at $x = 2$ it has 4.667.

A thin slab of thickness $dx$ has volume about $A(x)\,dx$. Adding the slabs is a second integral, this time in $x$:

The other order

Slice the other way: hold $y$ fixed and integrate across $x$ first. The slice at a fixed $y$ has area $\int_0^2 (x^2 + 2y^2)\,dx = \frac{8}{3} + 4y^2$, and $\int_0^1 \left(\frac{8}{3} + 4y^2\right) dy = \frac{8}{3} + \frac{4}{3} = 4$ again.

The condition matters. $\frac{x - y}{(x + y)^3}$ blows up at the corner $(0, 0)$ of the unit square, and its two orders give $\frac{1}{2}$ and $-\frac{1}{2}$.

A function that blows up needs care, as in lesson 2.6.3. For a function that is never negative, such as the bell of lesson 4.7.3, the two orders always agree, even over the whole plane.

A product splits

When $f(x, y) = g(x)\,h(y)$, the inner integral in $y$ is $g(x)$ times the number $\int_c^d h\,dy$. So the double integral is a product of two single integrals:

Over $[0, 2] \times [0, 3]$, $\iint xy\,dA = 2 \times \frac{9}{2} = 9$. Lesson 4.7.3 uses this to find the Gaussian integral.

Signed volume and area

Where $f$ is negative, a box hangs below the $xy$-plane, and its volume counts as negative, as signed area did in lesson 2.5.1. Take $f(x, y) = x + y - 1$ over $[0, 1] \times [0, 2]$. It is negative on the triangle where $x + y < 1$.

The inner integral is $\int_0^2 (x + y - 1)\,dy = \left[xy + \frac{y^2}{2} - y\right]_0^2 = 2x$, and $\int_0^1 2x\,dx = 1$. That is the net volume: $\frac{7}{6}$ above the plane minus $\frac{1}{6}$ below it. The total volume, the integral of $\lvert f \rvert$, is $\frac{7}{6} + \frac{1}{6} = \frac{4}{3}$.

With $f = 1$ every box has height 1, so the volume equals the base area. That turns a double integral into a way to find areas.

Regions between two curves

Most regions are not rectangles. Take the region $R$ between the parabola $y = x^2$ and the line $y = x$; they meet at $(0, 0)$ and $(1, 1)$, and the line is on top. A vertical slice at $x$ runs from $y = x^2$ up to $y = x$, so the inner limits depend on $x$.

Its area is $\iint_R 1\,dA = \int_0^1 (x - x^2)\,dx = \frac{1}{6}$. For $f(x, y) = xy$, the inner integral holds $x$ fixed:

Horizontal slices describe the same region differently. At a fixed $y$, $x$ runs from the line, $x = y$, to the parabola, $x = \sqrt{y}$, for $y$ from 0 to 1. Then $\int_0^1 \int_y^{\sqrt{y}} xy\,dx\,dy = \int_0^1 \frac{y^2 - y^3}{2}\,dy = \frac{1}{24}$ as well.

Swapping the limits alone gives $\int_0^1 \int_{y^2}^{y}$, the region's mirror image in the line $y = x$. For $f = xy$ the mistake hides, because $xy$ is the same in the mirror. For $f = x$ it shows: the right answer is $\frac{1}{12}$, but the swapped limits give $\frac{1}{15}$.

Choosing the order

The order can decide whether an integral can be done at all. Take the triangle $T$ where $0 \le y \le x \le 1$, and

The inner integral needs an antiderivative of $e^{x^2}$, and there is none made of familiar functions (lesson 2.6.1). Slice the other way: for each $x$ from 0 to 1, $y$ runs from 0 to $x$. The inner integral is now easy, since $e^{x^2}$ does not depend on $y$:

That is about 0.859141, by the substitution $u = x^2$ of lesson 2.6.1.

In Python: dblquad

scipy.integrate.dblquad(func, a, b, gfun, hfun) computes $\int_a^b \int_{g(x)}^{h(x)} \text{func}\,dy\,dx$ numerically. It returns the value and an estimate of its error, like quad in lesson 2.5.3. The limits gfun and hfun can be numbers or functions of $x$.

The trap is the order of the arguments: func must take the inner variable first, func(y, x). Written as lambda x, y:, the letters swap roles and dblquad integrates a different function.

Lesson 4.7.2 changes variables in a double integral, and case study 4.7.5 integrates a probability density over a region to get a probability.

Common mistakes

1. A box is height times area. Each term is $f(x_i^*, y_j^*)\,\Delta x\,\Delta y$; adding the heights alone leaves out $\Delta A$.
2. The inner integral treats the outer variable as a constant. In $\int_0^2 (x^2 + 2y^2)\,dy$, $x^2$ integrates to $x^2 y$.
3. The outer limits are numbers. Only the inner limits may depend on a variable, and only on the outer one.
4. Redraw the region to change the order. $\int_0^1 \int_{x^2}^{x} \ldots\,dy\,dx$ becomes $\int_0^1 \int_y^{\sqrt{y}} \ldots\,dx\,dy$, not $\int_0^1 \int_{y^2}^{y}$.
5. dblquad wants func(y, x), the inner variable first.
6. Volume below the plane counts as negative. For the total volume, integrate $\lvert f \rvert$.
7. The product rule needs a rectangle. Over a region between curves the inner limits depend on $x$, and the integral does not split.