Question

What is the area enclosed by `r=theta^2-2sintheta` for `theta in [pi/4,pi]`?

Answer 1

Area `=21.86658""" """"square units"`

Explanation:

`r=theta^2-2*sin theta`

Area `=1/2int_(pi/4)^pi r^2 d theta`

Area `=1/2int_(pi/4)^pi(theta^2-2*sin theta)^2 d theta`

Area `=1/2int_(pi/4)^pi(theta^4-4*theta^2*sin theta+4sin^2theta) d theta`

Area `=1/2*[theta^5/5+4*theta^2*cos theta-8*theta*sin theta-8*cos theta+2*theta-sin 2theta]_(pi/4)^pi`evaluate from `pi/4` to `pi`

Area `=21.86658""" """"square units"`

I hope the explanation is useful....God bless...