Question

find the area enclosed by the curve r^2=4cos theta

Answer 1

Let's graph the curve $r^2=4cos \theta.$

The graph of the curve is symmetrical to the origin and the area consists of 4 equal parts.

So we can find the area of one part ($\theta$ ranges from 0 to $\frac{\pi}{2}$) and multiply it by 4:

$A=4\cdot\frac{1}{2}∫_{0}^{\frac{\pi}{2}}[r(\theta)]^2dθ=$

$=2∫_{0}^{\frac{\pi}{2}}4cos \theta dθ=8∫_{0}^{\frac{\pi}{2}}cos \theta dθ=$

$=8 sin \theta |_0^{\frac{\pi}{2}}=8(sin\frac{\pi}{2}-sin0)=8(1-0)=8.$

Answer: 8 square units.