Q. 22 - 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 value of integer 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 function is  $2{\int }_0^{\frac{\mathrm{\pi }}{2}} {\int }_0^{\sin \mathrm{\theta }} \mathrm{rdrd\theta }$

Step 2: Calculation of integrals

The goal of this issue is to sketch the region and assess the expression using polar coordinates $2{\int }_0^{\mathrm{\pi }} {\int }_0^{1+\cos \mathrm{\theta }} \mathrm{rdrd\theta }$

using the values , $\mathrm{r}=0,\mathrm{r}=1+\cos \mathrm{\theta }\text{ and }\mathrm{\theta }=0,\mathrm{\theta }=\mathrm{\pi }$

Use the table above to draw the region $$\begin{gathered} 2{\int }_0^{\mathrm{\pi }} {\int }_0^{1+\cos \mathrm{\theta }} \mathrm{rdrd\theta }=2{\int }_0^{\mathrm{\pi }} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{1+\cos \mathrm{\theta }}\mathrm{d\theta } \\ =2{\int }_0^{\mathrm{\pi }} \left[\frac{(1+\cos \mathrm{\theta }{)}^2-0}{2}\right]\mathrm{d\theta } \\ =2{\int }_0^{\mathrm{\pi }} \left[\frac{\left(1+2\cos \mathrm{\theta }+{\cos }^2\mathrm{\theta }\right)}{2}\right]\mathrm{d\theta }\left[(\mathrm{a}+\mathrm{b}{)}^2={\mathrm{a}}^2+2\mathrm{ab}+{\mathrm{b}}^2\right] \\ =2{\int }_0^{\mathrm{\pi }} \left[\frac{\left\{1+2\cos \mathrm{\theta }+\frac{(1+\cos 2\mathrm{\theta })}{2}\right\}}{2}\right]\mathrm{d\theta } \\ =2{\int }_0^{\mathrm{x}} \left[\frac{3+4\cos \mathrm{\theta }+\cos 2\mathrm{\theta }}{4}\right]\mathrm{d\theta } \end{gathered}$$

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

$2{\int }_0^{\mathrm{\pi }} {\int }_0^{1+\cos \mathrm{\theta }} \mathrm{rdrd\theta }=2{\left[\frac{3\mathrm{\theta }+4\sin \mathrm{\theta }+\frac{(\sin 2\mathrm{\theta })}{2}}{4}\right]}_0^{\mathrm{\pi }}$

$\left[\int \sin \mathrm{xdx}=-\cos \mathrm{x},\int \cos \mathrm{xdx}=\sin \mathrm{x}\right]$

Substituting the limits of derivatives, 

$2{\int }_0^{\mathrm{x}} {\int }_0^{1\cos \mathrm{\theta }} \mathrm{rdrd\theta }=2\left[\frac{3\mathrm{\pi }+4\sin \mathrm{\pi }+\frac{(\sin 2\mathrm{\pi })}{2}-0}{4}\right]$

As a result, the integral value is ${\int }_0^{2\mathrm{\pi }} {\int }_0^{1+\sin \mathrm{\theta }} \mathrm{rdrd\theta }=\frac{3\mathrm{\pi }}{2}$