Question

Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one loop of $r{}^2=-\sin 2\theta$

Short answer

The required area is$\frac{3}{4}\mathrm{units}$

Step-by-step solution

Given information

$r{}^2=-\sin 2\theta$

Area

The graph of the given function $r{}^2=-\sin 2\theta$is,

The area bounded by one loop is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}-\sin 2\theta d\theta \\ A=\frac{1}{2}{\left(\frac{\cos 2\theta }{2}\right)}_0^{\frac{\mathrm{\pi }}{2}} \\ A=\frac{1}{2}\left(\frac{\cos 2\left(\frac{\mathrm{\pi }}{2}\right)}{2}-\cos 2\left(0\right)\right) \\ A=\frac{1}{2}\left(\frac{-1}{2}-1\right) \\ A=\frac{3}{4}\mathrm{units} \end{gathered}$$

The required area is$A=\frac{3}{4}\mathrm{units}$