   Chapter 15.2, Problem 68E

Chapter
Section
Textbook Problem

Use geometry or symmetry, or both, to evaluate the double integral.68. ∬ D ( 2 + x 2 y 3 − y 2 sin   x )   d A , D   =   { ( x ,   y )   |   | x | + | y |   ≤   1 }

To determine

To calculate: The value of given double integral over D using symmetry or geometry or both.

Explanation

Definition used:

Odd function: If f is a function and f(x)=f(x) , then f is said to be an odd function.

Formula used:

If f is an odd function, then aaf(x)dx=0 . (1)

And, D1dA=A(D) . (2)

Given

The function is, f(x,y)=2+x2y3y2sinx .

The domain D is, D={(x,y)||x|+|y|1} .

Calculation:

From the given conditions,

Here, second and third integrand are odd functions with respect to y and x, respectively, because, Dx2y3dA

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