Q. 41 - 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.

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

Step-by-Step Solution

Answer

The value of integral is ${\int }_0^{2r}{\int }_0^2\left(4r-r^3\right)drd\theta =8\pi$

Step 1: Given information

The expression is $I={\int }_0^{2\pi }{\int }_0^2\left(4r-r^3\right)drd\theta$

Step 2: Calculation

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

Integrate with respect to $r$first,

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

Put the limits

$$\begin{gathered} I={\int }_0^{2\pi }\left[\frac{4(2{)}^2}{2}-\frac{(2{)}^4}{4}\right]d\theta \\ I={\int }_0^{2\pi }4d\theta \end{gathered}$$

Now integrate with respect to$\theta$

$I=4[\theta {]}_0^{2\pi } I=8\pi$

Thus, the value of integral is

${\int }_0^{2r}{\int }_0^2\left(4r-r^3\right)drd\theta =8\pi$