Q. 34 - Calculus

Question

Areas of regions bounded by polar functions: Find the areas of the following regions. The area bounded by one petal of $r = \sin 3\theta$

Step-by-Step Solution

Answer

The area bounded by one petal is $\frac{\mathrm{\pi }}{6} {\mathrm{unit}}^2$

Step 1: Given information

The curve is $r = \sin 3\theta$

Step 2: Area

The area bounded by the curve is,

$$\begin{gathered} A=\frac{1}{2}{\int }_{\alpha }^{\beta }{r}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}{\left(\sin 3\theta \right)}^{2}d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}\left({\sin }^{2}3\theta \right)d\theta \\ A=\frac{1}{2}{\int }_0^{\frac{\mathrm{\pi }}{3}}\frac{\left(1-\cos 6\theta \right)}{2}d\theta \\ A=\frac{1}{2}{\left[\theta -\frac{\sin 6\theta }{6}\right]}_0^{\frac{\mathrm{\pi }}{3}} \\ A=\frac{1}{2}\left[\frac{\mathrm{\pi }}{3}-\frac{\sin 6{\displaystyle \frac{\mathrm{\pi }}{3}}}{6}-0\right] \\ A=\frac{\mathrm{\pi }}{6} {\mathrm{unit}}^2 \end{gathered}$$

The area of one petal is $A=\frac{\mathrm{\pi }}{6} {\mathrm{unit}}^2$