91Ó°ÊÓ

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

${∫}_0^{2n}{∫}_0^{1\sin \mathrm{θ}}rdrd\mathrm{θ}$

Here, $r=0,r=1+\sin \mathrm{θ}$and $\mathrm{θ}=0,\mathrm{θ}=2\mathrm{π}$

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

To sketch the region use the above table.

The integral can be evaluated as follows:

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}={∫}_0^{2z}{\left[\frac{r^2}{2}\right]}_0^{1+\sin \mathrm{θ}}d\mathrm{θ}$

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}={∫}_0^{2\mathrm{π}}\left[\frac{(1+\sin \mathrm{θ}{)}^2}{2}\right]d\mathrm{θ}$

${∫}_0^{2\mathrm{π}}{∫}_0^{1\sin \mathrm{θ}}rdrd\mathrm{θ}={∫}_0^{2r}\left[\frac{\left(1+2\sin \mathrm{θ}+{\sin }^2\mathrm{θ}\right)}{2}d\mathrm{θ}\left[(a+b{)}^2=a^2+2ab+b^2\right]\right.$

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}={∫}_0^{2\mathrm{π}}\left[\frac{\{1+2\sin \mathrm{θ}+(1-\cos 2\mathrm{θ})/2\}}{2}\right]d\mathrm{θ}$

${∫}_0^{2r}{∫}_0^{1\sin \mathrm{θ}}rdrd\mathrm{θ}={∫}_0^{2\mathrm{π}}\left[\frac{\{3+4\sin \mathrm{θ}-\cos 2\mathrm{θ}\}}{4}\right]d\mathrm{θ}$

Integrate with respect to $\mathrm{θ}$.

${∫}_0^{2\mathrm{π}}{∫}_0^{1+sen\mathrm{θ}}rdrd\mathrm{θ}={\left[\frac{\{3\mathrm{θ}-4\cos \mathrm{θ}-\sin 2\mathrm{θ}/2\}}{4}\right]}_0^{2\mathrm{π}}\left[∫\sin xdx=-\cos x,∫\cos xdx=\sin x\right]$

Put the limits

${∫}_0^{2\mathrm{π}}{∫}_0^{1\sin \mathrm{θ}}rdrd\mathrm{θ}=\frac{\{6\mathrm{π}-4\cos 2\mathrm{π}-\sin 4\mathrm{π}/2\}}{4}+1$

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}=\frac{\{6\mathrm{π}-4\}}{4}+1$

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}=\frac{3\mathrm{π}}{2}$

Thus the value of the integral is,

${∫}_0^{2\mathrm{π}}{∫}_0^{1+\sin \mathrm{θ}}rdrd\mathrm{θ}=\frac{3\mathrm{π}}{2}$