# Calculating Polar Areas

## Key Questions

• The area inside a polar curve is approximately the sum of lots of skinny wedges that start at the origin and go out to the curve, as long as there are no self-intersections for your polar curve.

Each wedge or slice or sector is like a triangle with height $r$ and base $r$ dθ, so the area of each element is
dA = 1/2 b h = 1/2 r (r dθ) = 1/2 r^2 dθ.

So add them up as an integral going around from θ=0 to θ=2π, and using a double angle formula, we get:

A = 1/2 int_0 ^(2π)(2 + cos(2θ))^2 dθ

A = 1/2 int_0 ^(2π) [4 + 4 cos(2θ) + cos^2(2θ)] dθ

A = 1/2 int_0 ^(2π) [4 + 4 cos(2θ) + (1 + cos(4θ))/2] dθ.

Now do the integral(s) by subbing u = 2θ and then u = 4θ, and remember to change limits for the "new u." I'll let you evaluate those to get practice integrating! Remember our motto,

"Struggling a bit makes you stronger." \dansmath/

• Let us look at the region bounded by the polar curves, which looks like:

Red: $y = 3 + 2 \cos \theta$
Blue: $y = 3 + 2 \sin \theta$
Green: $y = x$

Using the symmetry, we will try to find the area of the region bounded by the red curve and the green line then double it.

$A = 2 {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \setminus {\int}_{0}^{3 + 2 \cos \theta} r \mathrm{dr} d \theta$

$= 2 {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} {\left[{r}^{2} / 2\right]}_{0}^{3 + 2 \cos \theta} d \theta$

$= {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \left(9 + 12 \cos \theta + 4 {\cos}^{2} \theta\right) d \theta$

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

$= {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \left(11 + 12 \cos \theta + 2 \cos 2 \theta\right) d \theta$

$= {\left[11 \theta + 12 \sin \theta + \sin 2 \theta\right]}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}$

$= \frac{55 \pi}{4} - 6 \sqrt{2} + 1 - \left(\frac{11 \pi}{4} + 6 \sqrt{2} + 1\right)$

$= 11 \pi - 12 \sqrt{2}$

Hence, the area of the region is $11 \pi - 12 \sqrt{2}$.

I hope that this was helpful.

• If the region is bounded by a polar curve $r = r \left(\theta\right)$ from $\theta = {\theta}_{1}$ to ${\theta}_{2}$, then its area $A$ can be found by the double-integral

$A = {\int}_{{\theta}_{1}}^{{\theta}_{2}} {\int}_{0}^{r \left(\theta\right)} r \mathrm{dr} d \theta$.

I hope that this was helpful.