# What is the area enclosed by r=theta^2cos(theta+pi/4)-sin(2theta-pi/12)  for theta in [pi/12,pi]?

Oct 2, 2017

The area is about 19.1849 by computational integration.

#### Explanation:

The area $S$ enclosed by a polar curve $r = f \left(\theta\right)$ for $\theta \in \left[\alpha , \beta\right]$ is obtained as follows.

$S = \frac{1}{2} {\int}_{\alpha}^{\beta} {\left\{f \left(\theta\right)\right\}}^{2} d$$\theta$.

cited from a Japanese site, 極座標表示の場合の面積 http://www.geocities.jp/phaosmath/int1/polar.htm

Thus, the area in this question is:
$S = \frac{1}{2} {\int}_{\frac{\pi}{12}}^{\pi} {\left\{{\theta}^{2} \cos \left(\theta + \frac{\pi}{4}\right) - \sin \left(2 \theta - \frac{\pi}{12}\right)\right\}}^{2} d$$\theta$.

This integration is so hard…
According to www.wolframalpha.com $S$ is about 19.1849.
http://www.wolframalpha.com/input/?i=integrate+%7Bx%5E2cos(x%2Bpi%2F4)-sin(2x-pi%2F12)%7D%5E2%2F2+for+x%3D%7Bpi%2F12,pi%7D