• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Essential Calculus: Early Transcendentals
Found in: Page 735
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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 by Step Solution

Step 1: Given data.


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 2: Cylindrical solid.

 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 \({\rm{I = }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{\pi }}\).

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.