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Calculus / Calculus: Early Transcendentals 1 / Chapter 13.4 / Problem 14E

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

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Solution: Integrals over boxes Evaluate the following

Chapter 12, Problem 14E

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QUESTION:

Integrals over boxes Evaluate the following integrals. A sketch of the region of integration may be useful.

\(\iiint_{D} x y z e^{-x^{2}-y^{2}} \ d V\) ; \(D=\{(x, y, z): 0 \leq x \leq \sqrt{\ln 2}, 0 \leq y \leq \sqrt{\ln 4}, 0 \leq z \leq 1\}\)

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QUESTION:

Integrals over boxes Evaluate the following integrals. A sketch of the region of integration may be useful.

\(\iiint_{D} x y z e^{-x^{2}-y^{2}} \ d V\) ; \(D=\{(x, y, z): 0 \leq x \leq \sqrt{\ln 2}, 0 \leq y \leq \sqrt{\ln 4}, 0 \leq z \leq 1\}\)

ANSWER:

Solution 14E

Step 1 of 3:

In this problem we need to evaluate the integral $\int\limits_{\ }^{\ }\int\limits_{\ }^{\ }\int\limits_{\ }^{\ }xyz{e}^{-{x}^{2}-{y}^{2}}dV$

Given : Region $D\ =\ \{\ (x\ ,\ y,\ z)\ :\ 0\ \leq{}x\leq{}\sqrt{ln(2)}\ ,\ 0\leq{}y\leq{}\sqrt{ln(4)}\ ,\ 0\leq{}z\leq{}1\}$

We know that dV = dx dy dz (or) dz dy dx

Consider , $I\ =\ \int\limits_{0}^{1}\int\limits_{0}^{\sqrt{ln(4)}}\int\limits_{0}^{\sqrt{ln(2)}}xyz{e}^{-{x}^{2}-{y}^{2}}dx\ dy\ dz$

$=\ \int\limits_{0}^{1}\int\limits_{0}^{\sqrt{ln(4)}}(\int\limits_{0}^{\sqrt{ln(2)}}xyz{e}^{-{x}^{2}-{y}^{2}}dx)\ dy\ dz$ …………(1)

First we will evaluate the first (inner) integral with respect to x , and the result is a function of y and z , then evaluate the second (middle) integral with respect to y , and the result is a function of z , and then evaluate the last (outer) integral with respect to z .

The inner integral is : $\int\limits_{0}^{\sqrt{ln(2)}}xyz{e}^{-{x}^{2}-{y}^{2}}dx$

Take the constants out , we get :

$\int\limits_{0}^{\sqrt{ln(2)}}xyz{e}^{-{x}^{2}-{y}^{2}}dx\ =\ yz\ \int\limits_{0}^{\sqrt{ln(2)}}x{e}^{-{x}^{2}-{y}^{2}}dx$ …………(2)

Consider , $\int\limits_{\ }^{\ }x{e}^{-({x}^{2}+{y}^{2})}dx$

Put ${x}^{2}+{y}^{2}\ =\ r$ , then 2x dx = dr

$\Rightarrow{}xdx\ =\frac{dr}{2}$

Then the integral becomes :

$\int\limits_{\ }^{\ }x{e}^{-({x}^{2}+{y}^{2})}dx\ =\int\limits_{\ }^{\ }{e}^{-r}\frac{dr}{2}$

$=\ (\frac{1}{2})(-{e}^{-r})$ , since $\int\limits_{\ }^{\ }{e}^{x}dx\ =\ {e}^{x}$

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