Q. 39 - Calculus

Question

Areas of regions bounded by polar functions: Find the areas of the following regions. The area that is inside both lemniscates $r^2=\cos 2\theta \mathrm{and} r^2= \sin 2\theta$

Step-by-Step Solution

Answer

The area that is inside both lemniscates is $\left(1-\frac{1}{\sqrt{2}}\right) \mathrm{units}$

Step 1: Given information

$r^2=\cos 2\theta \mathrm{and} r^2= \sin 2\theta$

Step 2: Area

Find the limits of integration as,

$$\begin{gathered} \sin 2\theta =\cos 2\theta \\ \Rightarrow \tan 2\theta =1 \\ \Rightarrow 2\theta =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{8} \end{gathered}$$

The area between the curves is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^2 d\theta \\ A=4\times \frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{8}}\sin 2\theta d\theta \\ A=2{\left(-\frac{\cos 2\theta }{2}\right)}_0^{\frac{\mathrm{\pi }}{8}} \\ A=2\left(-\frac{\cos 2\left(\frac{\mathrm{\pi }}{8}\right)}{2}+\frac{\cos 2\left(0\right)}{2}\right) \\ A=\left(1-\frac{1}{\sqrt{2}}\right) \mathrm{units} \end{gathered}$$

The required area is $A=\left(1-\frac{1}{\sqrt{2}}\right) \mathrm{units}$