Recipe

Set up and evaluate a double integral over a general region

Use when: you're asked for \iint_R f\,dA where R's boundary is one or more curves, not a plain rectangle

Goal: a single number, reached through whichever order of integration the region's bounds make solvable

Taught in Reversing the order of integration

1. 1

   Draw R and label every curve or line that bounds it, along with the points where those curves cross.

   Example

   Evaluate \displaystyle\iint_R (x+y)\,dA where R is bounded below by y=x^2 and above by y=x. The two curves meet at (0,0) and (1,1).

   Why: the bounds on the inner integral come directly from reading this picture, not from manipulating the integrand.

2. 2

   Look at which pair of sides is straight and which pair curves, since that tells you which variable's bounds can stay constant.

   Example

   Here R sits between x=0 and x=1, with y running from the bottom curve y=x^2 up to the top curve y=x: Type I, 0\le x\le1, x^2\le y\le x.

   • R is bounded left and right by constants and top and bottom by functions of x — describe it as Type I, a\le x\le b, g_1(x)\le y\le g_2(x)
   • R is bounded top and bottom by constants and left and right by functions of y — describe it as Type II, c\le y\le d, h_1(y)\le x\le h_2(y)

   Why: the type determines which variable's bounds can be constants and which have to ride along as functions of the other variable, which in turn fixes which integral has to be done first.

3. 3

   Write \displaystyle\int_a^b\!\int_{g_1(x)}^{g_2(x)} f\,dy\,dx (or the Type II analogue) and integrate the inner variable first, treating the outer variable as a constant.

   Example

   \displaystyle\int_{x^2}^{x} (x+y)\,dy = \left[xy+\frac{y^2}{2}\right]_{x^2}^{x} = \left(x^2+\frac{x^2}{2}\right)-\left(x^3+\frac{x^4}{2}\right) = \frac{3x^2}{2}-x^3-\frac{x^4}{2}.

   Why: freezing the outer variable is exactly what turns a two-variable problem into an ordinary single-variable antiderivative, the same trick Fubini's theorem justifies for rectangles.

4. 4

   Check whether Step 3's inner integral actually has a closed-form antiderivative before moving on.

   Example

   \displaystyle\int_0^1\!\int_x^1 \frac{\sin y}{y}\,dy\,dx has no elementary antiderivative for \frac{\sin y}{y}. The region is 0\le x\le1,\ x\le y\le1; reading the same picture as Type II gives 0\le y\le1,\ 0\le x\le y. Rewritten: \displaystyle\int_0^1\!\int_0^y \frac{\sin y}{y}\,dx\,dy, whose inner integral is now trivial since \frac{\sin y}{y} doesn't depend on x: \int_0^y \frac{\sin y}{y}\,dx = y\cdot\frac{\sin y}{y}=\sin y.

   • the inner integral has a workable antiderivative in the order given — skip ahead to Step 5
   • the inner integral has no elementary antiderivative in the order given — re-sketch the SAME region, re-describe it as the other Type by solving the boundary curves for the other variable, and rewrite the iterated integral with the new bounds before integrating

   Why: reversing the order changes only which variable is integrated first, never the region or the value of the integral, so trading an unworkable order for a workable one costs nothing.

5. 5

   Integrate the remaining single-variable function over its constant bounds to get the final number.

   Example

   For the running example: \displaystyle\int_0^1\left(\frac{3x^2}{2}-x^3-\frac{x^4}{2}\right)dx = \left[\frac{x^3}{2}-\frac{x^4}{4}-\frac{x^5}{10}\right]_0^1 = \frac12-\frac14-\frac1{10} = \boxed{\frac{3}{20}}. For the reversed example: \displaystyle\int_0^1 \sin y\,dy = \big[-\cos y\big]_0^1 = \boxed{1-\cos1}.

   Why: this is the only step that produces the actual value of the double integral; every earlier step exists to make this last antiderivative solvable.