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

##### 1 Answer

Oct 2, 2017

The area is about 19.1849 by computational integration.

#### Explanation:

The area

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:

This integration is so hard…

According to www.wolframalpha.com

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