   Chapter 15, Problem 7P

Chapter
Section
Textbook Problem

(a) Show that ∫ 0 1 ∫ 0 1 ∫ 0 1 1 1 − x y z d x   d y   d z = ∑ n − 1 ∞ 1 n 3 (Nobody has ever been to find the exact value of the sum of this series.)(b) Show that ∫ 0 1 ∫ 0 1 ∫ 0 1 1 1 + x y z d x   d y   d z = ∑ n − 1 ∞ ( − 1 ) n − 1 n 3 Use this equation to evaluate the triple integral correct to two decimal places.

(a)

To determine

To show: 01010111xyzdxdydz=n=11n3.

Explanation

Given:

The triple integral 01010111xydxdydz is an improper integral and can be defined as the limit of double integrals over the rectangle [0,t]×[0,t]×[0,t] as t1.

Calculation:

As t1, it is observed that the value of |xyz|<1. So, the sum of the geometric series is given by, 11xyz=n=0(xyz)n except at the point (1,1,1).

01010111xyzdxdydz=010101(n=0(xyz)n)dxdydz=n=0(010101(xyz)ndxdydz)=n=0(01yndy01xndx01zndz)

Integrate it and apply the limit

(b)

To determine

To show: 01010111+xyzdxdydz=n=1(1)n1n3.

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