Question

What is the area under the polar curve `f(theta) = theta^2sin((5theta)/2 )-cos((2theta)/3+pi/2)` over `[pi/6,(3pi)/2]`?

Answer 1

The answer is 142.523.

Explanation:

The formula for finding the area of a polar curve goes like this...

`A=1/2int_a^b(f(theta))^2d theta`

With this formula in mind, plug in your f graph and your endpoints into the above formula, like so...

`A=1/2int_(pi/6)^((3pi)/2)(theta^2sin((5theta)/2)-cos((2theta)/3+pi/2))^2d theta`

Yuck. This looks like a job for a calculator.

I'm getting... 142.523...sounds good to me.