Question

Short answer

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

Step-by-step solution

Introduction.

The given integral is,

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

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

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

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