   Chapter 15.6, Problem 8E

Chapter
Section
Textbook Problem

Evaluate the iterated integral.8. ∫ 0 1 ∫ 0 1 ∫ 0 2 − x 2 − y 2   x y e z   d z   d y   d x

To determine

To evaluate: The iterated integral.

Explanation

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Given:

The function is f(x,y,z)=xyez .

The region is B={(x,y,z)|0x(yz),0yz2,0z2} .

Calculation:

The given integral is, 010102x2y2xyezdzdydx .

Integrate the given integral with respect to z and apply the limit of it.

010102x2y2xyezdzdydx=0101xy[ez]02x2y2dydx=0101xy[e(2x2y2)e0]dydx=0101xy[e(2x2y2)1]dydx=0101[xye(2x2y2)xy]dydx

Use (1) to separate the integrals,

010102x2y2xyezdzdydx=0101[xye2ex2ey2xy]dydx=e201xex2dx01yey2dy01xdx01ydy

Here, 01xex2dx and 01yey2dy both are same and 01xdx and 01ydy are same. So, take this as (01xey2dx)2 and (01xdx)2

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