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.
Thanks in advance!