Q. 19 - 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. 

${\int }_0^{\mathrm{\pi }/2}{\int }_0^{2\sin \theta }{r}^3 drd\theta$

Step-by-Step Solution

Answer

${\int }_0^{\mathrm{\pi }/2}{\int }_0^{2\sin \theta }{r}^3 drd\theta =\frac{3}{4}\mathrm{\pi }$

Step 1: Draw the region

The region determined by the limits is shown below,  

Step 2: Evaluate the integral

$$\begin{gathered} I={\int }_0^{\mathrm{\pi }/2}{\int }_0^{2\sin \theta }{r}^3 drd\theta \\ I={\int }_0^{\mathrm{\pi }/2}{\left[\frac{r^4}{4}\right]}_0^{2 \sin \theta } d\theta \\ I=\frac{16}{4}{\int }_0^{\mathrm{\pi }/2}{\sin }^4\theta d\theta \\ I=4{\int }_0^{\mathrm{\pi }/2}{\sin }^4\theta d\theta \\ I=4\left[\frac{3}{4}·\frac{1}{2}·\frac{\mathrm{\pi }}{2}\right] \\ I=\frac{3}{4}\mathrm{\pi } \end{gathered}$$