# Thread: Evaluating the iterated integral by converting to polar coordinates.

1. ## Evaluating the iterated integral by converting to polar coordinates.

http://www.webassign.net/latexImages...38b1a15e78.gif

Please click to see the problem. The answer is 16 but I keep getting 3pi

f(x,y) = 9sqrt(x^2+y^2)
f(rcos, rsin) = 9sqrt(r^2) = 9r

From the limits we have 0 =< y =< sqrt(2x-x^2)

and 0 =< x =< 2

sqrt(2x-x^2) is a semi-circle in the first quadrant with center (1,0) and r = 1

so my iterated polar integral is:

int[(0,pi)int(0,1) 9r^2 drdtheta]

integrating first with respect to r gives me

3r^3](0,1) = 3

then int(0,pi) 3 dtheta = 3pi ... but the answer is 16 so i have no idea what i'm doing incorrectly.

2. Your semi-circle cannot be expressed as $0<\theta<\pi, 0 -- that would be a semi-circle centered on the origin. Starting with $y<\sqrt{2x-x^2}$, this can be expressed in polar coordinates as $r<2\cos\theta$. This is your correct upper bound for $r$.