Question

Find the area of the region using double integral.

Short answer

The area of the region is \(\frac{{3\pi }}{2} - 4\).

Step-by-step solution

Formula used

If\(f\)is a polar rectangle\(R\)given by\(0 \le a \le r \le b,\alpha \le \theta \le \beta \), where\(0 \le \beta - \alpha \le 2\pi \), then,

Substitute \(x = r\cos \theta \) and \(y = r\sin \theta \) in the equation (1)

Integrate with respect to \(r\) apply the limit as shown below,

\(\int_0^{\frac{\pi }{2}} {\int_0^{1 - \cos \theta } r } drd\theta = \int_0^{\frac{\pi }{2}} {\left( {\frac{{{r^2}}}{2}} \right)_0^{1 - \cos \theta }} d\theta \)

\( = \int_0^{\frac{\pi }{2}} {\left( {\frac{{{{(1 - \cos \theta )}^2}}}{2} - \frac{{{0^2}}}{2}} \right)} d\theta \)

\( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\left( {1 + {{\cos }^2}\theta - 2\cos \theta } \right)} d\theta \)

\( = \frac{1}{2}\int_0^{\frac{\pi }{2}} {\left( {1 + \frac{1}{2}(1 + \cos 2\theta ) - 2\cos \theta } \right)} d\theta \)

Integrate with respect to \(\theta \) apply the limit for \(\theta \)

\(\begin{aligned}{l}\int_0^{\frac{\pi }{2}} {\int_0^{1 - \cos \theta } r } drd\theta = \frac{1}{2}\left( {\theta + \frac{1}{2}\left( {\theta + \frac{{\sin 2\theta }}{2}} \right) - 2\sin \theta } \right)_0^{\frac{\pi }{2}}\\ = \frac{1}{2}\left( {\left( {\frac{\pi }{2} + \frac{1}{2}\left( {\frac{\pi }{2} + \frac{{\sin \left( {2\frac{\pi }{2}} \right)}}{2}} \right) - 2\sin \frac{\pi }{2}} \right) - \left( {0 + \frac{1}{2}\left( {0 + \frac{{\sin (0)}}{2}} \right) - 2\sin (0)} \right)} \right)\\ = \frac{1}{2}\left( {\left( {\frac{\pi }{2} + \frac{\pi }{4} - 2} \right) - (0)} \right)\\ = \frac{1}{2}\left( {\frac{{3\pi }}{4} - 2} \right)\end{aligned}\)

Obtain the area

Thus, the area of the region is \(\frac{{3\pi }}{2} - 4\).