Question

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]`?

Answer 1

0.455

Explanation:

Area under the polar curve for any function `f(theta)` is given by `int_a^b 1/2 r^2 d(theta)`

The function `f(theta)` represents the r or radius as the curve moves from `pi/8` to `pi/6`.

So all we have to do is plugin whole `f(theta)` for r.

That would give us,
`int_(pi/8)^(pi/6) (3(theta)^2 +theta*sin(4theta-((5pi)/12))+cos(2theta-pi/3))^2d(theta)`

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

The answer would be 0.455