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Found in: Page 735

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Evaluate $$\iiint_{\text{E}}{{{\text{x}}^{\text{2}}}}\text{dV}$$, where $${\rm{E}}$$ is the solid that lies within the cylinder $${{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 1}}$$, above the plane $${\rm{z = 0}}$$, and below the cone $${{\rm{z}}^{\rm{2}}}{\rm{ = 4}}{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{y}}^{\rm{2}}}$$.Use cylindrical coordinates.

The evaluation of the given equation is $${\rm{I = }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{\pi }}$$.

See the step by step solution

## Step 1: Given data.

$$\iiint_{\text{E}}{{{\text{x}}^{\text{2}}}}\text{dV}$$

cylindrical coordinates

$$\left\{ {\begin{array}{*{20}{l}}{{\rm{x = rcos\theta }}}\\{{\rm{y = rsin\theta }}}\\{{\rm{z = z}}}\end{array}} \right.$$

$${\rm{dxdydz = rdzdrd\theta }}$$

## Step 3: Evaluation.

Solid in the plane $${\rm{z = 0}}$$ and the cone,

\begin{aligned}{{\rm{z}}^{\rm{2}}} \rm &= 4{{\rm{x}}^{\rm{2}}}{\rm{ + 4}}{{\rm{y}}^{\rm{2}}}\\{{\rm{z}}^{\rm{2}}} \rm &= 4{{\rm{r}}^{\rm{2}}}\\ \rm z &= 2r\end{aligned}

Limits, $${\rm{0}} \le {\rm{z}} \le {\rm{2r}}$$

In $${\rm{xy}}$$ plane the circle of radius is $${\rm{1}}$$

$${\rm{0}} \le {\rm{r}} \le {\rm{1}}$$

The region is the whole circle, the limits of $${\rm{\theta }}$$

$${\rm{0}} \le {\rm{\theta }} \le {\rm{2\pi }}$$

Integrate the equation

$${\rm{I = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {\int_{\rm{0}}^{{\rm{2r}}} {{{\rm{r}}^{\rm{3}}}} } } {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta dzdrd\theta }}$$

Integrate to $${\rm{z}}$$,

\begin{aligned} \rm I &= \left. {\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{3}}}} } {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta * z}}} \right|_{\rm{0}}^{{\rm{2r}}}{\rm{drd\theta }}\\\rm I &= \int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{3}}}} } {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta (2r - 0)drd\theta }}\\\rm I &= 2\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{1}} {{{\rm{r}}^{\rm{4}}}} } {\rm{co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta drd\theta }}\end{aligned}

Integrate to $${\rm{r}}$$,

\begin{aligned}\rm I &= \left. {{\rm{2}}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{co}}{{\rm{s}}^{\rm{2}}}} {\rm{\theta }}\frac{{{{\rm{r}}^{\rm{5}}}}}{{\rm{5}}}} \right|_{\rm{0}}^{\rm{1}}{\rm{d\theta }}\\\rm I &= 2\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{co}}{{\rm{s}}^{\rm{2}}}} {\rm{\theta }}\left( {\frac{{{{\rm{1}}^{\rm{5}}}}}{{\rm{5}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{5}}}}}{{\rm{5}}}} \right){\rm{d\theta }}\\\rm I &= \frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{co}}{{\rm{s}}^{\rm{2}}}} {\rm{\theta d\theta }}\end{aligned}

Integrals with trigonometry identity,

\begin{aligned}{\rm{co}}{{\rm{s}}^{\rm{2}}}\rm\theta &= \frac{{{\rm{1 + cos(2\theta )}}}}{{\rm{2}}}\\ \rm I &= \frac{{\rm{1}}}{{\rm{5}}}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{(1 + cos(}}} {\rm{2\theta ))d\theta }}\end{aligned}

Divide into two integrals,

$${\rm{I = }}\frac{{\rm{1}}}{{\rm{5}}}\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{d}} {\rm{\theta + }}\frac{{\rm{1}}}{{\rm{5}}}\int_{\rm{0}}^{{\rm{2\pi }}} {{\rm{cos}}} {\rm{(2\theta )d\theta }}$$

Integrate to $${\rm{\theta }}$$,

\begin{aligned}\rm I &= \left. {\frac{{\rm{1}}}{{\rm{5}}}{\rm{\theta }}} \right|_{\rm{0}}^{{\rm{2\pi }}}{\rm{ + }}\left. {\frac{{\rm{1}}}{{\rm{5}}}{\rm{ * }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{sin(2\theta )}}} \right|_{\rm{0}}^{{\rm{2\pi }}}\\ \rm I &= \frac{{\rm{1}}}{{\rm{5}}}{\rm{(2\pi - 0) + }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{(sin(4\pi ) - sin(0))}}\\ \rm I &= \frac{{\rm{2}}}{{\rm{5}}}{\rm{\pi }}\end{aligned}

Therefore, the evaluation of the given equation is $${\rm{I = }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{\pi }}$$.