   Chapter 14.2, Problem 8E

Chapter
Section
Textbook Problem

Evaluating a Double Integral In Exercises 7-12, sketch the region R and evaluate the iterated integral ∫ R ∫ f ( x , y ) d A . ∫ 0 π ∫ 0 π □ 2 sin 2 x cos 2 y   d y   d x

To determine

To Graph: The iterated integral 0π0π2(sin2xsin2y)dydx and evaluate its value.

Explanation

Given:

The integral, 0π0π2(sin2xsin2y)dydx

Graph:

The limit of y goes from 0 to π2 and x goes from 0 to π.

The region R can be sketched with the following values:

Formula used:

The trigonometric identity cos2y=1sin2y2.

The integral of, cos2xdx=sin2xx.

Calculation:

Now, evaluate the iterated integral,

0π0π2(sin2xsin2y)dydx=0π0π2(1cos2y2sin2x)dydx<

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