# Find area between 2 polar curves

## 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$$\Theta$$

## The Attempt at a Solution

I set up the integral like follows:

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

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.

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integral (0,pi/2) (4sin($$\Theta$$)^2 d$$\Theta$$ - integral (pi/2, pi/6) 2 d$$\Theta$$
Why do u think that the integral should be from $$0,\frac{\pi}{2}$$?
Don't you think its the intersection(s) of the 2 polar curves?:grumpy:

hmm..well i really dont know how i messed that up.

so it should be :

integral (pi/6, 5pi/6) r1^2 - r2^2 dtheta ?

That should be fine! :)