Question

Use Theorem 9.13 to show that the area of the circle defined by the polar equation$r=2a\cos \theta$is$\pi a^2$.

Short answer

The area of the circle is$\pi a^2$.

Step-by-step solution

Step 1. Given Information

The circle is bounded by the curve $r=2a\cos \theta$.

Step 2. Use the formula for area of a region bounded by a curve 

• The region is bounded by the curve $r=2a\cos \theta$.
• The region is a circle, so the boundary arcs will be $\theta =0\mathrm{to}\mathit{}\theta \mathit{=}\mathit{2}\pi$.
• But the curve is traced twice in this region.
• So, the area of the region will be calculated as follows:

$\begin{array}{rcl}\frac{1}{2}\times \frac{1}{2}{\int }_0^{2\pi }4a^2{\cos }^2\theta d\theta & =& a^2{\int }_0^{2\pi }{\cos }^2\theta d\theta \\ & =& a^2{\int }_0^{2\pi }\frac{1+\cos \left(2\theta \right)}{2}d\theta \\ & =& a^2\times \frac{1}{2}\left({\int }_0^{2\pi }1d\theta +{\int }_0^{2\pi }\cos \left(2\theta \right)d\theta \right)\\ & =& \frac{1}{2}{a}^2\left(2\pi +0\right)\\ & =& \pi a^2\end{array}$

• Hence, the area of the curve is $\pi a^2$.