Q. 24 - Calculus

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.  

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

Step-by-Step Solution

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 1: Given information

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

Step 2: 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$$\mathrm{r}=\mathrm{sin\theta }$
0	0
$\pi /6$	0.5
$\pi /4$	0.7071
$\pi /3$	0.8660
$\pi /2$	1.0

Use the table above to draw the region  ,  

elaborating the expressions,

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

We can integrating in terms of  $\mathrm{\theta },$

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

Substituting the limits of derivatives,  

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

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