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.

$2{\int }_0^{2\mathrm{\pi }} {\int }_0^{\mathrm{R}} \mathrm{r}\sqrt{{\mathrm{R}}^2-{\mathrm{r}}^2}\mathrm{drd\theta }$

Short answer

The value of integral is

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

Step-by-step solution

Given information

The expression is

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

Calculation

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

To integrate with respect to $r$ first,

Put

localid="1651390046153" role="math" $$\begin{gathered} R^2-r^2=t^2 \\ -2rdr=2tdt \\ rdr=-tdt \\ r=0\Rightarrow t=R\text{and}r=R\Rightarrow t=0 \end{gathered}$$

So integral $I$become

localid="1650634581886" $$\begin{gathered} I=2{\int }_0^{2\pi }\left[{\int }_R^0\sqrt{t^2}(-tdt)\right]d\theta \left[{\int }_0^{b}f\left(x\right)dx=-{\int }_0^{a}f\left(x\right)dx\right] \\ I=2{\int }_0^{2\pi }\left[{\int }_0^R{t}^{2}dt\right]d\theta \\ I=2{\int }_0^{2\pi }{\left[\frac{t^3}{3}\right]}_0^{R}d\theta \\ I=2{\int }_0^{2\pi }\left[\frac{R^3-0}{3}\right]d\theta \\ I=2{\int }_0^{2x}\frac{R^3}{3}d\theta \\ I=\frac{2R^3}{3}{\int }_0^{2\pi }d\theta \\ I=\frac{2R^3}{3}[\theta {]}_0^{2\pi } \\ I=\frac{4}{3}\pi R^3 \end{gathered}$$

Thus, the value of integral is

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