Q 54. - Calculus

Question

Prove that the area enclosed by one petal of the polar rose $r = \cos 3\theta$ is the same as the area enclosed by one petal of the polar rose $r = \sin 3\theta$.

Step-by-Step Solution

Answer

It is proved that the area enclosed by one petal of the polar rose r = cos 3θ is the same as the area enclosed by one petal of the polar rose r = sin 3θ.

Step 1. Given information

Two petals of the polar rose are r = cos 3θ and r = sin 3θ.

Step 2. Explanation

Consider the petal $r=\cos 3\theta$

Area $=2\left(\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{6}}{\left(\cos 3\theta \right)}^2 d\theta \right)$

          $$\begin{gathered} =\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{6}}1+\cos 6\theta d\theta \\ =\frac{1}{2}\left(\frac{\mathrm{\pi }}{6}+\frac{\sin 6 \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}{6}-0\right) \\ =\frac{\mathrm{\pi }}{12} \end{gathered}$$

Consider the petal width="67" style="max-width: none; vertical-align: -4px;" $r=\sin 3\theta$

Area $=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}{\left(\sin 3\theta \right)}^2 d\theta$

         $$\begin{gathered} =\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{1-\cos 6\theta }{2} d\theta \\ =\frac{1}{4}{\left[\theta -\frac{\sin 6\theta }{6}\right]}_0^{\frac{\mathrm{\pi }}{3}} \\ =\frac{\mathrm{\pi }}{12} \end{gathered}$$

So, the area enclosed by one petal of the polar rose r = cos 3θ is the same as the area enclosed by one petal of the polar rose r = sin 3θ.