Q. 20 - Calculus

Question

Using polar coordinates to evaluate iterated integrals: Sketch the region determined by the limits of the given iterated integrals, and then evaluate the integrals. 

${\int }_0^{2\mathrm{\pi }}{\int }_0^{1+\sin \theta }r drd\theta$

Step-by-Step Solution

Answer

${\int }_0^{2\mathrm{\pi }}{\int }_0^{1+\sin \theta }r drd\theta =\frac{1}{2}\left[3\mathrm{\pi }-2\right]$

Step 1: Draw the region

The region determined by the limits is shown below,   

Step 2: Evaluate the integral

$$\begin{gathered} I={\int }_0^{2\mathrm{\pi }}{\int }_0^{1+\sin \theta }r drd\theta \\ I={\int }_0^{2\mathrm{\pi }}{\left[\frac{r^2}{2}\right]}_0^{1+\sin \theta }d\theta \\ I=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}{\left(1+\sin \theta \right)}^2 d\theta \\ I=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left(1+{\sin }^2 \theta +2\sin \theta \right) d\theta \\ I=\frac{1}{2}{\int }_0^{2\mathrm{\pi }}\left(1+\frac{1-\cos 2\theta }{2}+2\sin \theta \right) d\theta \\ I=\frac{1}{2}{\left[\theta +\frac{1}{2}\theta -\frac{1}{4}\sin 2\theta -2\cos \theta \right]}_0^{2\mathrm{\pi }} \\ I=\frac{1}{2}\left[2\mathrm{\pi }+\mathrm{\pi }-2\right] \\ I=\frac{1}{2}\left[3\mathrm{\pi }-2\right] \end{gathered}$$