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.

${\int }_0^{2\pi }{\int }_0^3\left(6r-2r^2\right)drd\theta$

Short answer

The value of integral is

${\int }_0^{2\pi }{\int }_0^3\left(6r-2r^2\right)drd\theta =18\pi$

Step-by-step solution

Given information

The expression is

$I={\int }_0^{2\pi }{\int }_0^3\left(6r-2r^2\right)drd\theta$

Calculation

Here, $r=0,r=3$and$\theta =0,\theta =2\pi$

Integrate with respect to$r$first,

$I={\int }_0^{2\pi }{\left[\frac{6r^2}{2}-\frac{2r^3}{3}\right]}_0^{3}d\theta \left[\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right]$

Put the limits

localid="1651390685586" $$\begin{gathered} I={\int }_0^{2\pi }\left[\frac{6(3{)}^2}{2}-\frac{2(3{)}^3}{3}\right]d\theta \\ I={\int }_0^{2\mathrm{\pi }}9d\theta \end{gathered}$$

Now integrate with respect to $\theta$

$I=9[\theta {]}_0^{2\pi }I=18\pi$

Thus, the value of integral is

${\int }_0^{2\pi }{\int }_0^3\left(6r-2r^2\right)drd\theta =18\pi$