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Found in: Page 713

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

## Sketch the region whose area is given by the integral and evaluate the integral.

$$\int\limits_{{\rm{\pi /2}}}^{\rm{\pi }} {\int\limits_{\rm{0}}^{{\rm{2 sin\theta }}} {{\rm{r dr d\theta }}} }$$

$$\int\limits_{\pi /2}^\pi {\int\limits_0^{2\sin \theta } {r{\rm{ }}dr{\rm{ }}d\theta } } = \frac{\pi }{2}$$

See the step by step solution

## Step 1: Introduction.

The given integral is,

$$I = \int\limits_{\pi /2}^\pi {\int\limits_0^{2\sin \theta } {r{\rm{ }}dr{\rm{ }}d\theta } }$$

## Step 2: Evaluate the integral.

Integrating first on $$r$$, we get

$$\begin{array}{l}I = \int\limits_{\pi /2}^\pi {\frac{1}{2}} \left. {{r^2}} \right|_0^{2\sin \theta }d\theta \\ = \int\limits_{\pi /2}^\pi {\frac{1}{2}} \left( {{2^2}{{\sin }^2}\theta - {0^1}} \right)d\theta \\ = \int\limits_{\pi /2}^\pi {2{{\sin }^2}\theta } {\rm{ }}d\theta \\{\sin ^2}\theta = \frac{{1 - \cos \left( {2\theta } \right)}}{2}\\\int\limits_{\pi /2}^\pi {2{{\sin }^2}\theta } {\rm{ }}d\theta = \int\limits_{\pi /2}^\pi {1 - \cos \left( {2\theta } \right)} {\rm{ }}d\theta \end{array}$$

Integrating on $$\theta$$, we get

$$\begin{array}{l}I = \theta - \frac{1}{2}\sin \left. {2\theta } \right|_{\pi /2}^\pi \\ = \left( {\pi - 0} \right) - \left( {\frac{\pi }{2} - 0} \right)\\ = \pi - \frac{\pi }{2}\\ = \frac{\pi }{2}\end{array}$$

## Step 3: Sketch the region.

The region R whose area is given by $$I = \int\limits_{\pi /2}^\pi {\int\limits_0^{2\sin \theta } {r{\rm{ }}dr{\rm{ }}d\theta } }$$ is