Question

What is the area enclosed by `r=-sin(theta+(11pi)/8) -theta/4` between `theta in [0,(pi)/2]`?

Answer 1

`A=pi/8+sqrt2/8+frac{pi^3}{768}-pi/8cos(15pi/8) + 1/4sin(15pi/8) - 1/4sin(11pi/8)`

`approx .3823`

Explanation:

`r=-sin(theta+frac{11pi}{8})-theta/4`

Area inside polar curves: `A=1/2int_(theta_1)^(theta_2)r^2 d theta`

`A=1/2int_(0)^(pi/2)(-sin(theta+(11pi)/8)-theta/4)^2 d theta`

(If this problem allows a calculator, use a graphing calculator here)
Expand what's inside the integral:
`1/2int_0^(pi/2)(sin^2(theta+(11pi)/8)+theta/2 sin(theta+(11pi)/8)+theta^2/16) d theta`

`=pi/8+sqrt2/8+frac{pi^3}{768}-pi/8cos(15pi/8) + 1/4sin(15pi/8) - 1/4sin(11pi/8)`

`approx .3823`