Q. 40 - Calculus

Question

Areas of regions bounded by polar functions: Find the areas of the following regions. The area inside both polar roses $r = \sin 3\theta \mathrm{and} r = \cos 3\theta$. 

Step-by-Step Solution

Answer

The area inside both polar roses is $\left(\frac{1}{6}-\frac{\sqrt{2-\sqrt{2}}}{6}\right) \mathrm{units}$

Step 1: Given information

$r = \sin 3\theta \mathrm{and} r = \cos 3\theta$

Step 2: Area

The graph of both the curves is,

The limits of the integration is,

$$\begin{gathered} \sin 3\theta =\cos 3\theta \\ \Rightarrow \tan 3\theta =1 \\ \Rightarrow 3\theta =\frac{\mathrm{\pi }}{4} \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{12} \end{gathered}$$

The area bounded by both the curves is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^2 d\theta \\ A=3\times \frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{12}}\sin 3\theta d\theta \\ A=\frac{1}{2}{\left(-\frac{\cos 3\theta }{3}\right)}_0^{\frac{\mathrm{\pi }}{8}} \\ A=\frac{1}{2}\left(-\frac{\cos 3\left(\frac{\mathrm{\pi }}{8}\right)}{3}+\frac{\cos 3\left(0\right)}{3}\right) \\ A=\left(\frac{1}{6}-\frac{\sqrt{2-\sqrt{2}}}{6}\right) \mathrm{units} \end{gathered}$$

The required area is $A=\left(\frac{1}{6}-\frac{\sqrt{2-\sqrt{2}}}{6}\right) \mathrm{units}$