# What is the area under the polar curve f(theta) = 3theta^2+thetasin(4theta-(5pi)/12 ) +cos(2theta-(pi)/3) over [pi/8,pi/6]?

Jun 21, 2016

0.455

#### Explanation:

Area under the polar curve for any function $f \left(\theta\right)$ is given by ${\int}_{a}^{b} \frac{1}{2} {r}^{2} d \left(\theta\right)$

The function $f \left(\theta\right)$ represents the r or radius as the curve moves from $\frac{\pi}{8}$ to $\frac{\pi}{6}$.

So all we have to do is plugin whole $f \left(\theta\right)$ for r.

That would give us,
${\int}_{\frac{\pi}{8}}^{\frac{\pi}{6}} {\left(3 {\left(\theta\right)}^{2} + \theta \cdot \sin \left(4 \theta - \left(\frac{5 \pi}{12}\right)\right) + \cos \left(2 \theta - \frac{\pi}{3}\right)\right)}^{2} d \left(\theta\right)$

Doing this integral by hand would be little tedious so you can just plug integral in your calculator and get the answer.