Q. 22 - Calculus

Question

Find a definite integral expression that represents the area of the given region in polar plane and then find the exact value of the expression

The region inside both of the cardioids $r=3+3\sin \theta and r=1+\sin \theta$

Step-by-Step Solution

Answer

The area can be given as ${\int }_0^{\frac{\pi }{6}}{(3-3\sin \theta )}^2-{(1+\sin \theta )}^{2}d\theta$

And the value of the integral is 1.879.

Step 1: Given inofrmation

We are given two cardioids $r=3-3\sin \theta and r=1+\sin \theta$

Step 2: Find the area

The area can be given as

${\int }_0^{\frac{\pi }{6}}{(3-3\sin \theta )}^2-{(1+\sin \theta )}^{2}d\theta$

On evaluating we get,

$$\begin{gathered} {\int }_0^{\frac{\pi }{6}}(9-18\sin \theta +9{\sin }^2\theta -1-2\sin \theta +{\sin }^2\theta )d\theta \\ ={\int }_0^{\frac{\pi }{6}}(8-20\sin \theta +10{\sin }^2\theta )d\theta \\ ={{[8\theta +20\cos \theta +\frac{\theta }{2}+\frac{\sin 2\theta }{4}]}^{\frac{\pi }{6}}}_0 \\ =21.879-20 \\ =1.879 \end{gathered}$$