Q. 39 - Calculus

Question

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in ${\mathrm{ℝ}}^3$. Use polar coordinates to describe the solid, and evaluate the expressions.

39. $2{\int }_0^{2\pi }{\int }_0^{4}r\sqrt{16-r^2}drd\theta$

Step-by-Step Solution

Answer

The value of the integral is 

${\int }_0^{2\pi }{\int }_0^{4}r\sqrt{16-r^2}drd\theta =\frac{256\pi }{3}$

Step 1: Given information

Given expression : 

$I=2{\int }_0^{2\pi }{\int }_0^{4}r\sqrt{16-r^2}drd\theta$

Step 2: Using polar coordinates to describe the solid, and evaluating the expressions.

Here, $r=0, r=4$and $\theta =0,\theta =2\pi$

To integrate with respect to $r$ first,

Put $16-r^2=t^2$

$$\begin{gathered} -2rdr=2tdt \\ rdr=-tdt \\ r=0\Rightarrow t=4\text{ and }r=4\Rightarrow t=0 \end{gathered}$$

So integral $I$ becomes:

$$\begin{gathered} I=2{\int }_0^{2\pi }\left[{\int }_4^0\sqrt{t^2}(-tdt)\right]d\theta \left[{\int }_0^{b}f\left(x\right)dx=-{\int }_0^{a}f\left(x\right)dx\right] \\ I=2{\int }_0^{2\pi }\left[{\int }_0^4{t}^{2}dt\right]d\theta \\ I=2{\int }_0^{2\pi }{\left[\frac{t^3}{3}\right]}_0^{4}d\theta \\ I=2{\int }_0^{2\pi }\left[\frac{4^3-0}{3}\right]d\theta \\ I=2{\int }_0^{2\pi }\frac{64}{3}d\theta \\ I=\frac{128}{3}{\int }_0^{2\pi }d\theta \\ I=\frac{128}{3}[\theta {]}_0^{2\pi } \\ \frac{256\pi }{3} \end{gathered}$$

Thus, the value of integral is

$2{\int }_0^{2\pi }{\int }_0^{4}r\sqrt{16-r^2}drd\theta =\frac{256\pi }{3}$