Question

Use Theorem 9.14 to show that the circumference of the circle defined by the polar equation $r=2a\sin \theta$is $2\pi a$.

Short answer

The circumference is found to be$2\pi a$by using the formula of length of polar curves.

Step-by-step solution

Step 1. Given Information

The function that bounds the circle is$r=2a\sin \theta$.

Step 2. Use the formula of arc length of a polar curve 

• The region is bounded by the curve$r=2a\sin \theta$.
• The region is a circle, so the boundary arcs will be $\theta =0\mathrm{to}\theta \mathit{=}\mathit{2}\pi$.
• However, the function traces itself twice in this domain.
• So, the length of the curve will be calculated as follows:

role="math" localid="1650343133331" $\frac{1}{2}\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & 02\mathrm{\pi }\end{array}\begin{array}{rcl}& & \sqrt{{(r\text{'})}^2+r^2}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{1}{2}\end{array}\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & 02\mathrm{\pi }\end{array}\begin{array}{rcl}& & \sqrt{{\left(2a\cos \theta \right)}^2+{\left(2a\sin \theta \right)}^2}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{1}{2}\end{array}\begin{array}{rcl}& & \times \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & a\end{array}\begin{array}{rcl}& & \int \end{array}\begin{array}{rcl}& & 02\mathrm{\pi }\end{array}\begin{array}{rcl}& & \sqrt{1}\end{array}\begin{array}{rcl}& & d\end{array}\begin{array}{rcl}& & \theta \end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & a\end{array}\begin{array}{rcl}& & \times \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \mathrm{\pi }\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \pi \end{array}\begin{array}{rcl}& & a\end{array}$