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.

$h{\int }_0^{2\pi }{\int }_0^R\left(r-\frac{r^2}{R}\right)drd\theta$

Short answer

The value of integral is

$h{\int }_0^{2\pi }{\int }_0^R\left(r-\frac{r^2}{R}\right)drd\theta =\frac{\pi h{R}^2}{3}$

Step-by-step solution

Given information

The expression is$I=h{\int }_0^{2\pi }{\int }_0^R\left(r-\frac{r^2}{R}\right)drd\theta$

Calculation

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

Integrate with respect to r first,

$I=h{\int }_0^{2\pi }{\left[\frac{r^2}{2}-\frac{r^3}{3R}\right]}_0^{R}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

Put the limits

$I=h{\int }_0^{2\pi }\left[\frac{R^2}{2}-\frac{R^3}{3R}\right]d\theta$

localid="1651390798747" $I=h{\int }_0^{2\mathrm{\pi }}\frac{R^2}{6}d\theta$

Now integrate with respect to $\theta$

$I=\frac{h{R}^2}{6}[\theta {]}_0^{2\pi }I=\frac{h{R}^2}{6}[2\pi ]$

$I=\frac{\pi h{R}^2}{3}$

Thus, the value of integral is

localid="1651390833501" $h{\int }_0^{2\mathrm{\pi }}{\int }_0^R\left(r-\frac{r^2}{R}\right)drd\theta =\frac{\pi h{R}^2}{3}$