Q. 23 - 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.  

$2\underset{\frac{-\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{2-\mathrm{sin\theta }}{\int }}\mathrm{r}\mathrm{dr}\mathrm{d\theta }$

Step-by-Step Solution

Answer

The Integral value is $2\underset{\frac{-\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{2-\mathrm{sin\theta }}{\int }}\mathrm{r}\mathrm{dr}\mathrm{d\theta }=\frac{9\mathrm{\pi }}{2}$

Step 1: Given information

The Given integral is $2\underset{\frac{-\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{2-\mathrm{sin\theta }}{\int }}\mathrm{r}\mathrm{dr}\mathrm{d\theta }$

Step 2: Calculation of integrals

The goal of this challenge is to sketch the region in polar coordinates and assess the expression,$2\underset{\frac{-\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\underset{0}{\overset{2-\sin \theta }{\int }}rdrd\theta$

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

$\mathrm{\theta }$	$\mathrm{r}=2-\mathrm{sin\theta }$
$\mathrm{\pi }/2$	$3$
$-\mathrm{\pi }/3$	$2.8660$
$-\mathrm{\pi }/6$	$2.5$
$0$	$2$
$\mathrm{\pi }/6$	$1.5$
$\mathrm{\pi }/3$	$1.1339$
$\mathrm{\pi }/2$	$1.0$

Use the table above to draw the region 

$\begin{array}{r}2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\int }_0^{2-\sin \mathrm{\theta }} \mathrm{rdrd\theta }=2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{2-\sin \mathrm{\theta }}\mathrm{d\theta }\\ =2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \left[\frac{(2-\sin \mathrm{\theta }{)}^2-0}{2}\mathrm{d\theta }\right.\\ =2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \left[\frac{\left(4-4\sin \mathrm{\theta }+{\sin }^2\mathrm{\theta }\right)}{2}\right]\mathrm{\theta }\left[(\mathrm{a}-\mathrm{b}{)}^2={\mathrm{a}}^2-2\mathrm{ab}+{\mathrm{b}}^2\right]\\ =2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \left[\frac{\{4-4\sin \mathrm{\theta }+(1-\cos 2\mathrm{\theta })/2\}}{2}\right]\mathrm{\theta }\\ =2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} \left[\frac{\{9-8\sin \mathrm{\theta }-\cos 2\mathrm{\theta }\}}{4}\mathrm{d\theta }\right.\end{array}$

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

$\begin{array}{r}2{\int }_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2} {\int }_0^{2-\sin \mathrm{\theta }} \mathrm{rdrd\theta }=2{\left[\frac{(9\mathrm{\theta }+8\cos \mathrm{\theta }-(\sin 2\mathrm{\theta })/2\}}{4}\right]}_{-\mathrm{\pi }/2}^{\mathrm{\pi }/2}\\ \left[\int \sin \mathrm{xdx}=-\cos \mathrm{x},\int \cos \mathrm{xdx}=\sin \mathrm{x}\right]\end{array}$

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

Substituting the limits of derivatives,  

$$\begin{gathered} 2\underset{\frac{-\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{2-\mathrm{sin\theta }}{\int }}\mathrm{r}\mathrm{dr}d\theta \\ =2\left[\frac{{\displaystyle \frac{\frac{9\mathrm{\pi }}{2}}{2}+8\cos \left(\frac{\mathrm{\pi }}{2}\right)-\frac{\left(\mathrm{sin\pi }\right)}{2}\}-\{-\frac{\frac{9\mathrm{\pi }}{2}}{2}+8\cos \left(\frac{\mathrm{\pi }}{2}\right)+\frac{\left(\mathrm{sin\pi }\right)}{2}\}}}{4}\right] \\ =2\left[\frac{\left\{\frac{9\mathrm{\pi }}{2}\right\}{\displaystyle -}{\displaystyle \left\{-\frac{{\displaystyle 9\mathrm{\pi }}}{{\displaystyle 2}}\right\}}}{4}\right] \\ =\frac{{\displaystyle 9\mathrm{\pi }}}{{\displaystyle 2}} \end{gathered}$$

As a result, the integral value is $2\underset{\frac{-\mathrm{\pi }}{2}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\underset{0}{\overset{2-\mathrm{sin\theta }}{\int }}\mathrm{r}\mathrm{dr}\mathrm{d\theta }=\frac{9\mathrm{\pi }}{2}$