Q. 21 - Calculus

Question

Each of the integrals or integral expressions in Exercises 21–28 represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta$

Step-by-Step Solution

Answer

The value of the integral is ${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta =\frac{3\pi }{2}$

Step 1 Given Information

The objective of this problem is to use polar coordinates to sketch the region and evaluate the expression

${\int }_0^{2n}{\int }_0^{1\sin \theta }rdrd\theta$

Step 2 Calculation

Here, $r=0,r=1+\sin \theta$ and $\theta =0,\theta =2\pi$

$\begin{array}{|ll|}\hline \theta & r=1+\sin \theta \\ 0& 1\\ \pi /6& 1.5\\ \pi /4& 1.7071\\ \pi /3& 1.8660\\ \pi /2& 2.0\\ \hline\end{array}$

Step 3 Diagram

To sketch the region use the above table.

Step 4 Calculation

The integral can be evaluated as follows:

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta ={\int }_0^{2z}{\left[\frac{r^2}{2}\right]}_0^{1+\sin \theta }d\theta$

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

${\int }_0^{2\pi }{\int }_0^{1\sin \theta }rdrd\theta ={\int }_0^{2r}\left[\frac{\left(1+2\sin \theta +{\sin }^2\theta \right)}{2}d\theta \left[(a+b{)}^2=a^2+2ab+b^2\right]\right.$

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta ={\int }_0^{2\pi }\left[\frac{\{1+2\sin \theta +(1-\cos 2\theta )/2\}}{2}\right]d\theta$

${\int }_0^{2r}{\int }_0^{1\sin \theta }rdrd\theta ={\int }_0^{2\pi }\left[\frac{\{3+4\sin \theta -\cos 2\theta \}}{4}\right]d\theta$

Integrate with respect to $\theta$.

${\int }_0^{2\pi }{\int }_0^{1+sen\theta }rdrd\theta ={\left[\frac{\{3\theta -4\cos \theta -\sin 2\theta /2\}}{4}\right]}_0^{2\pi }\left[\int \sin xdx=-\cos x,\int \cos xdx=\sin x\right]$

Put the limits

${\int }_0^{2\pi }{\int }_0^{1\sin \theta }rdrd\theta =\frac{\{6\pi -4\cos 2\pi -\sin 4\pi /2\}}{4}+1$

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta =\frac{\{6\pi -4\}}{4}+1$

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta =\frac{3\pi }{2}$

Thus the value of the integral is, 

${\int }_0^{2\pi }{\int }_0^{1+\sin \theta }rdrd\theta =\frac{3\pi }{2}$