# Thread: polar form of double integral

1. ## polar form of double integral

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)

2. Originally Posted by cchyfly
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)
$
I=\int_{x=0}^3 \int_{y=0}^{\sqrt{9-x^2}} (x^2+y^2)^{3/2} ~dy~dx
$

is the integral of $r^3$ over the first quadrant of the interior of the circle $(x^2+y^2)=r^2=3$, so:

$
I=\int_{r=0}^3 \int_{\theta=0}^{\pi/2} (r^3)~r d \theta dr
$

RonL

3. so actually in the last step it is r^4 and then evaluate the integral

4. Originally Posted by cchyfly
so actually in the last step it is r^4 and then evaluate the integral
Yes, I left it in the form I did so you could more easily see where things came from.

RonL