I have to turn this in to polar
the double integral (x^2+y^2 )^(3/2) from o to 3 and Sqrt(9-x^2)
I THINK THE ANSWER IS (243PI /10)
$\displaystyle
I=\int_{x=0}^3 \int_{y=0}^{\sqrt{9-x^2}} (x^2+y^2)^{3/2} ~dy~dx
$
is the integral of $\displaystyle r^3$ over the first quadrant of the interior of the circle $\displaystyle (x^2+y^2)=r^2=3$, so:
$\displaystyle
I=\int_{r=0}^3 \int_{\theta=0}^{\pi/2} (r^3)~r d \theta dr
$
RonL