Double integral examples

To illustrate computing double integrals as iterated integrals, we start with the simplest example of a double integral over a rectangle and then move on an integral over a triangle.

Example 1

Compute the integral

∫∫Dxy2dA
where D is the rectangle 0 ≤ x ≤ 2, 0 ≤ y ≤ 1.

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

∫ 01(∫ ) 2 2 xy dx 0dy.
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
∫ 01( ∫ ) 2 2 xy dx 0dy = ∫ 01( | ) x2-2|x=2 2 y || x=0dy
= ∫ 01( ) 22 2 02 2 ---y - --y 2 2dy
= ∫ 012y2dy.
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

∫∫Dxy2dA = ∫ 012y2dy
= | 2y3 |1 ----|| 3 0 = 2(13) ----- 3 -2(03) ----- 3 = 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

∫∫Dxy2dA = ∫ 02( ∫ ) 1 2 xy dy 0dx
= ∫ 02( |y=1) y3|| x 3 | y=0dx
= ∫ 02x- 3dx
= 2|2 x--|| 6 |0 = 4- 6 = 2- 3.
As it must, this iterated integral gives the same answer.

Example 2

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

PIC

Use same function f(x,y) = xy2. Compute ∫∫ Dxy2dA when D is that triangle.

Solution: Note for the triangle defined by 0 ≤ x ≤ 2 and 0 ≤ y ≤ 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

∫ ∫Dxy2dA = ∫ 02( ∫ ) x∕2 2 xy dy 0dx
= ∫ 02( ) x 3||y=x∕2 -y | 3 y=0dx
= ∫ 02( ( ) ) x- x- 3 x-3 3 2 - 30dx
= ∫ 02 4 x-- 24dx
= 5 ||2 -x---| 5 ⋅ 24 |0 = -32--- 5 ⋅ 24 = -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 ≤ y ≤ 1 and 2y ≤ x ≤ 2.

PIC

Is it confusing that the limits of x are 2y ≤ x ≤ 2 rather than 0 ≤ x ≤ 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

∫∫Dxy2dA = ∫ 01( ∫ ) 2 2 xy dx 2ydy
= ∫ 01( | ) x2y2-||x=2 2 | x=2ydy
= ∫ 01( ) 2 (2y)2y2 2y - ---2----dy
= ∫ 01(2y2 - 2y4)dy
= 2[y3 y5 ] ---- --- 3 501
= 2( ) 1 1 -- --- (0 - 0) 3 5 = 2 ⋅2 --- 15 = 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.