   Chapter 15.2, Problem 16E

Chapter
Section
Textbook Problem

Set up iterated integrals for both orders of integration. Then evaluate the double integral using the easier order and explain why it’s easier.16. ∬ D y 2 e x y   d A , D is bounded by y = x, y = 4, x = 0

To determine

To setup: The iterated integrals for both the orders of integration and then evaluate the double integral by easier method.

Explanation

Given:

The function is f(x,y)=y2exy .

The region D is bounded by the curves x=0,y=x,y=4 .

Definition used:

Region of type 1:

A plane region D is said to be of type 1 if it lies between two continuous functions of x.

That is, D={(x,y)|axb,g1(x)yg2(x)} , where g1(x) and g2(x) are the continuous functions of x.

Region of type 2:

A plane region D is said to be of type 2 if it lies between two continuous functions of y.

That is, D={(x,y)|ayb,h1(y)xh2(y)} , where h1(y) and h2(y) are the continuous functions of y.

If the region D is of type 1, then the region that has to be evaluated is given below in the Figure 1.

Therefore, from Figure 1, it is observed that when x varies from 0 to 4, y varies from x to 4.

Similarly, if the region D is of type 2, then the region that has to be evaluated is given below in the Figure 2.

Therefore, from Figure 2, it is observed that x varies from 0 to y and y varies from 0 to 4.

So, if the region D is of type 1, then the integration by parts has to be used, but if the region D is of type 2, then it is easy to evaluate the integration when compared to the previous case. Hence, this method is considered as the easier method.

So, the value of double integral is obtained as follows.

Df(x,y)dA=040yy2exydxdy=04[y2exyy]0ydy=04[ye(y)yye(0)y]dy=04yey2dy04ydy

Integrate the first term with respect to y

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