Question

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

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

Short answer

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

Step-by-step solution

Given information

The integral function is$2{\int }_0^{\frac{\mathrm{\pi }}{2}} {\int }_0^{\sin \mathrm{\theta }} \mathrm{rdrd\theta }$

Calculation of derivatives 

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

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

$\theta$	$\mathrm{r}=\sin \theta$localid="1652361865068" $\mathrm{r}=\mathrm{sin\theta }$
0	0
localid="1652361870625" $\pi /6$	0.5
localid="1652361873973" $\pi /4$	0.7071
localid="1652361878453" $\pi /3$	0.8660
localid="1652361882858" $\pi /2$	1.0

Use the table above to draw the region ,

elaborating the expressions,

localid="1652361900299" $\begin{array}{r}2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2{\int }_0^{\mathrm{\pi }/2} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{\sin 3\mathrm{\theta }}\mathrm{d\theta }\\ =2{\int }_0^{\mathrm{\pi }/2} \left[\frac{{\sin }^{2}3\mathrm{\theta }-0}{2}\right]\mathrm{d\theta }\\ =2{\int }_0^{\mathrm{\pi }/2} \left[\frac{(1-\cos 6\mathrm{\theta })}{4}\right]\mathrm{d\theta }\end{array}$

We can integrating in terms of localid="1652361916107" $\mathrm{\theta },$

localid="1652361908352" $2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2{\left[\frac{\{\mathrm{\theta }-(\sin 6\mathrm{\theta })/6\}}{4}\right]}_0^{\mathrm{\pi }/2}\left[\int \cos \mathrm{xdx}=\sin \mathrm{x}\right]$

Substituting the limits of derivatives,

localid="1652361934374" $$\begin{gathered} 2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=2\left[\frac{\{{\displaystyle \frac{\mathrm{\pi }/2-(\sin 3\mathrm{\pi })}{6}}-0\}}{4}\right] \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

As a result, the integral value is localid="1652361944670" $2{\int }_0^{\mathrm{\pi }/2} {\int }_0^{\sin 3\mathrm{\theta }} \mathrm{rdrd\theta }=\frac{\mathrm{\pi }}{4}$