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

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Answer 1 · 2016-04-22T21:50:58

The answer is 142.523.

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.