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.

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

Short answer

The value of integral is${\int }_0^z{\int }_0^{\pi }\left(R^{2}r-r^3\right)drd\theta =\frac{\pi R^4}{4}$

Step-by-step solution

Given information

The expression is $I={\int }_0^{\pi }{\int }_0^R\left(R^{2}r-r^3\right)drd\theta$

Calculation

The objective of this problem is to use polar coordinates to describe the solid and evaluate the integral expression.

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

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

Integrate with respect to r first,

$I={\int }_0^{\pi }{\left[\frac{R^2{r}^2}{2}-\frac{r^4}{4}\right]}_0^{R}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

Substitute the limits

$I={\int }_0^{\pi }\left[\frac{R^4}{2}-\frac{R^4}{4}\right]d\theta$

$I=in{t}_0^{\pi }\left[\frac{R^4}{4}\right]d\theta$

Now integrate with respect to$\theta$

$I=\frac{R^4}{4}[\theta {]}_0^{\pi }I=\frac{\pi R^4}{4}$

Thus, the value of integral is

${\int }_0^z{\int }_0^{\pi }\left(R^{2}r-r^3\right)drd\theta =\frac{\pi R^4}{4}$