How do you find the area of the region bounded by the polar curve r=3cos(theta) ?

1 Answer
Sep 27, 2014

The area of the region is $\frac{9}{4} \pi$.

Let us look at some details.

The region is a disk, which looks like this:

If you are allowed to use the formula for the area of a circle, then

$A = \pi {r}^{2} = \pi {\left(\frac{3}{2}\right)}^{2} = \frac{9}{4} \pi$

If you wish to use integration, then

$A = {\int}_{0}^{\pi} \setminus {\int}_{0}^{3 \cos \theta} r \mathrm{dr} d \theta$

$= {\int}_{0}^{\pi} {\left[{r}^{2} / 2\right]}_{0}^{3 \cos \theta} d \theta$

$= {\int}_{0}^{\pi} \frac{9 {\cos}^{2} \theta}{2} d \theta$

by the trig identity: ${\cos}^{2} \theta = \frac{1}{2} \left(1 + \cos 2 \theta\right)$,

$= \frac{9}{4} {\int}_{0}^{\pi} \left(1 + \cos 2 \theta\right) d \theta$

$= \frac{9}{4} {\left[\theta + \frac{\sin 2 \theta}{2}\right]}_{0}^{\pi}$

$= \frac{9}{4} \pi$

I hope that this was helpful.