I have to evaluate the integral between 0 and 2a of the integral between 0 and sqrt(2ax-x^2) of x^2 dxdy
I am supposed to draw a diagram and draw a radial slice and I have to convert to polar co-ordinates.
The only thing I know how to do is x becomes rcos(theta) and dxdy becomes r dr dtheta
Any help would be great thanks.
You are going from a botton curve to a top curve . For the latter, squaring both sides gives
(Just the top half).
In polar coords this becomes
or just .
So we have a 1/2 circle centered at with radius . In order to find limits we ask, "for what values of and will sweep out this half circle. Here will go from 0 to and r from to the outside curve . That's how I got the limits.
I evaluated the integral that you wrote down after changing to polar co-ords and got:
Int between Pi/2 & 0 of (cos^2(theta)*(2acos(4theta)))^4 /4 d theata
which is: Int between Pi/2 & 0 of (16a^4(cos^6(theta))) /4 d theta
= 4a^4(sin^6(theta)) , subs in values of theat
which is: 4a^4 - 0 = 4a^4 i think ?????