Q. 43 - 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.

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

Step-by-Step Solution

Answer

The value of integral is 

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

Step 1: Given information

The expression is 

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

Step 2: 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

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

Now integrate with respect to $\theta$

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

Thus, the value of integral is

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