Double integrals examples

Example 1

Compute the integral

$$\iint_{D}^{}$$

where D is the rectangle 0$\le$x$\le$2, 0$\le$y$\le$1.

Solution: We will compute the double integrals as the iterated integral

$$\int_{0}^{1} \biggl( \int_{0}^{2} \biggr)$$

We first integrate with respect to x inside the parentheses. Similar to the procedure with partial derivatives, we must treat y as a constant during this integration step. Since the integral of cx is cx²/2, we calculate

$$\int_{0}^{1} \biggl( \int_{0}^{2} \biggr) \int_{0}^{1} \biggl( {\frac{{x^2}}{{2}}} \bigg\vert _{{x=0}}^{{x=2}} \biggr)$$

$$\int_{0}^{1} \biggl( {\frac{{2^2}}{{2}}} {\frac{{0^2}}{{2}}} \biggr)$$

$$\int_{0}^{1}$$

Note that in the second line above, we wrote the limits as x = 2 and x = 0 so it is unambiguous that x is the variable we just integrated.

To finish, we need to compute the integral with respect to y, which is simple. Since x is gone, it's just a regular one-variable integral. We calculate that our double integral is

$$\iint_{D}^{} \int_{0}^{1}$$

$${\frac{{2y^3}}{{3}}} \bigg\vert _{0}^{1} {\frac{{2(1^3)}}{{3}}} {\frac{{2(0^3)}}{{3}}} {\frac{{2}}{{3}}}$$

To double check, we can compute the integral in the other direction, integrating first with respect to y and then with respect to x. The only trick is to remember that when integrating with respect to y, we must think of x as a constant. Since the integral of cy² is cy³/3, we calculate

$$\iint_{D}^{} \int_{0}^{2} \biggl( \int_{0}^{1} \biggr)$$

$$\int_{0}^{2} \biggl( {\frac{{y^3}}{{3}}} \bigg\vert _{{y=0}}^{{y=1}} \bigg)$$

$$\int_{0}^{2} {\frac{{x}}{{3}}}$$

$${\frac{{x^2}}{{6}}} \bigg\vert _{0}^{2} {\frac{{4}}{{6}}} {\frac{{2}}{{3}}}$$

As it must, this iterated integral gives the same answer.

Example 2

Rectangular regions were easy because the limits ( a$\le$x$\le$b and c$\le$y$\le$d) were fixed. Here's an example where we integrate over the triangle defined by 0$\le$x$\le$2, 0$\le$y$\le$x/2.

Use same function f (x, y) = xy². Compute $\iint_{D}^{}$xy²dA when D is that triangle.

Solution: Note for the triangle defined by 0$\le$x$\le$2 and 0$\le$y$\le$x/2, the limits on y depend on x. For a given value of x, y ranges from 0 to x/2, as illustrated above by the vertical dashed line from (x, 0) to (x, x/2).

In a double integral, the outer limits must be constant, but the inner limits can depend on the outer variable. This means, we must put y as the inner integration variables, as was done in the second way of computing Example 1. The only difference from Example 1 is that the upper limit of y is x/2. The double integral is

$$\iint_{D}^{} \int_{0}^{2} \biggl( \int_{0}^{{x/2}} \biggr)$$

$$\int_{0}^{2} \biggl( {\frac{{x}}{{3}}} \bigg\vert _{{y=0}}^{{y=x/2}} \biggr)$$

$$\int_{0}^{2} \biggl( {\frac{{x}}{{3}}} \Bigl( {\frac{{x}}{{2}}} \Bigr)^{3}_{} {\frac{{x}}{{3}}} \biggr)$$

$$\int_{0}^{2} {\frac{{x^4}}{{24}}}$$

$${\frac{{x^5}}{{5\cdot 24}}} \bigg\vert _{0}^{2} {\frac{{32}}{{5 \cdot 24}}} {\frac{{4}}{{15}}}$$

Example 2'

Now compute the integral over the same triangle D, but make y be the outer integration variable.

Solution: Now we need to give constant limits for y. As illustrated below, for point in the triangle y ranges between between 0 and 1. Then, for a given value of y, x takes on values between 2y and 2 (as shown by the horizontal dashed line between (2y, y) and (2, y)). Hence, we can describe the triangle by 0$\le$y$\le$1 and 2y$\le$x$\le$2.

Is it confusing that the limits of x are 2y$\le$x$\le$2 rather than 0$\le$x$\le$2 (which would more closely parallel the above Example 2)? If we let x range from 0 to 2y, then the triangle would be the upper-left triangle in the above picture. We want to compute the integral over the region D, which is the lower-right triangle shaded in red. In this triangle, y < x/2 (as used above in Example 2) which means that x > 2y (as we are now using in Example 2').

The double integral is similar to the first way of computing Example 1, with the only difference being that the lower limit of x is 2y. The integral is

$$\iint_{D}^{} \int_{0}^{1} \biggl( \int_{{2y}}^{2} \biggr)$$

$$\int_{0}^{1} \biggl( {\frac{{x^2y^2}}{{2}}} \bigg\vert _{{x=2y}}^{{x=2}} \biggr)$$

$$\int_{0}^{1} \biggl( {\frac{{(2y)^2 y^2}}{{2}}} \biggr)$$

$$\int_{0}^{1} \biggl( \biggr)$$

$$\biggl[ {\frac{{y^3}}{{3}}} {\frac{{y^5}}{{5}}} \biggr]_{0}^{1}$$

$$\biggl( {\frac{{1}}{{3}}} {\frac{{1}}{{5}}} \biggr) {\frac{{2}}{{15}}} {\frac{{4}}{{15}}}$$

Thankfully, this does agree with the answer we obtained in Example 2.

To go from Example 2 to Example 2', we "changed the order of integration." You can see more examples of changing the order of integration in double integrals here.