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## Homework Statement

Fine the area of the region inside the polar curve r=4sin(theta) and outside the polar curve r=2.

## Homework Equations

area under polar curve = 1/2 integral (a,b) r^2 d[tex]\Theta[/tex]

## The Attempt at a Solution

I set up the integral like follows:

integral (0,pi/2) (4sin([tex]\Theta[/tex])^2 d[tex]\Theta[/tex] - integral (pi/2, pi/6) 2 d[tex]\Theta[/tex]

Then from plugging those into my calculator i got 4pi - (2/3)pi = (10/3) pi

Sorry i cant get the integral latex code to work for some reason.