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Found in: Page 714

### Essential Calculus: Early Transcendentals

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

## Use polar coordinates to find the volume of the given solid.

Under the cone $${\rm{z = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} {\rm{ }}$$and above the disk$${{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ }} \le {\rm{ 4}}$$

The volume of the given solid is $$\frac{{16\pi }}{3}$$.

See the step by step solution

## Step 1: Convert into polar coordinates.

In polar coordinates$$x = r\cos \theta$$,$$y = r\sin \theta$$.

$$\Rightarrow {x^2} + {y^2} = {r^2}$$

So, the given cone $$z = \sqrt {{x^2} + {y^2}}$$will be,

$$\begin{array}{l}z = \sqrt {{r^2}} \\ \Rightarrow z = r\end{array}$$

And the disk $${x^2} + {y^2} \le 4$$will be,

$$\begin{array}{l}0 \le {r^2} \le 4\\ \Rightarrow 0 \le r \le 2\end{array}$$

The region of integration is a polar rectangle described as $$0 \le \theta \le 2\pi$$and$$0 \le r \le 2$$.

## Step 2: Compute the volume.

The volume (V) will be

\begin{aligned}V &= \iint\limits_D {zdA} \\ &= \int\limits_0^{2\pi } {\int\limits_0^2 {r \cdot rdrd\theta } } \\ &= \int\limits_0^{2\pi } {\int\limits_0^2 {{r^2}drd\theta } } \\ \end{aligned}

## Step 3: Evaluate the integral.

$$I = \int\limits_0^{2\pi } {\int\limits_0^2 {{r^2}drd\theta } }$$

Integrating on $$r$$ we get,

$$\begin{array}{c}I = \int\limits_0^{2\pi } {\left( {\frac{1}{3}{r^3}} \right)} _0^2d\theta \\ = \int\limits_0^{2\pi } {\left( {\frac{1}{3}\left( {{2^3} - {2^0}} \right)} \right)} d\theta \\ = \int\limits_0^{2\pi } {\frac{8}{3}} d\theta \end{array}$$

Integrating on $$\theta$$ we get,

$$\begin{array}{c}I = \frac{8}{3}\left( {\left( \theta \right)} \right)_0^{2\pi }\\ = \frac{8}{3}\left( {2\pi - 0} \right)\\ = \frac{{16\pi }}{3}\end{array}$$

Therefore, volume of the given solid is $$\frac{{16\pi }}{3}$$.