Question

Each of the integrals or integral expressions in Exercise represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.

$2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }$

Short answer

The integral's value is$2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=\frac{\mathrm{\pi }}{4}$

Step-by-step solution

given information 

Let consider the given integral is$2{\int }_0^{\pi /2} {\int }_0^{\sin 3\theta } rdrd\theta$

Finding establish the expression

The goal of this issue is to sketch the region and assess the expression using polar coordinates $2{\int }_0^{\pi /2} {\int }_0^{\sin 3\theta } rdrd\theta$

Using, $\mathrm{r}=0,\mathrm{r}=\sin \mathrm{\theta }\text{and}\mathrm{\theta }=0,\mathrm{\theta }=\mathrm{\pi }/2$

$\theta$	$r=\sin 3\theta$
0	0
$\mathrm{\pi }/6$	$1.0$
$\mathrm{\pi }/4$	$0.7071$
$\mathrm{\pi }/3$	$0$
$\mathrm{\pi }/2$	$-1.0$

Calculations 

$\begin{array}{r}2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2{\int }_0^{\mathrm{\pi }/2} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sin 3\mathrm{\theta }}\mathrm{d\theta }\\ =2{\int }_0^{\mathrm{\pi }/2} \left[\frac{{\sin }^{2}3\mathrm{\theta }-0}{2}\right]\mathrm{\theta }\\ =2{\int }_0^{\mathrm{\pi }/2} \left[\frac{(1-\cos 6\mathrm{\theta })}{4}\right]\mathrm{d\theta }\end{array}$

Integrate in relation to $\mathrm{\theta }$,

$2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2{\left[\frac{\left\{{\displaystyle \frac{\left(\mathrm{\theta }-(\sin 6\mathrm{\theta }\right)}{6}}\right\}}{4}\right]}_0^{\mathrm{\pi }/2}\left[\int \cos \mathrm{xdx}=\sin \mathrm{x}\right]$

Pointing the limits,

$$\begin{gathered} 2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2\left[\frac{\{{\displaystyle \frac{\mathrm{\pi }/2-(\sin 3\mathrm{\pi })}{6}}-0\}}{4}\right] \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

As a result, the integral value is$2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=\frac{\mathrm{\pi }}{4}$