Q. 37 - Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: 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 $A=\frac{3}{4} \mathrm{units}$