# double integrals and changing order of integration

• Nov 28th 2010, 11:19 AM
chewitard
double integrals and changing order of integration
$\int_{-3}^1\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\: x^2 \: dxdy$

I think the domain for this is the shaded green:
http://i.imgur.com/k5tOI.png

When I try to integrate as is, I wind up making it even more complicated so another method to solve this would be to change the order of integration, but if I were to do that I think I would have to split the domain up into parts. The problem is I'm not quite sure how to split this up, or if the above domain is even correct.

Any help would be greatly appreciated.
• Nov 28th 2010, 07:34 PM
Sudharaka
Quote:

Originally Posted by chewitard
$\int_{-3}^1\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\: x^2 \: dxdy$

I think the domain for this is the shaded green:
http://i.imgur.com/k5tOI.png

When I try to integrate as is, I wind up making it even more complicated so another method to solve this would be to change the order of integration, but if I were to do that I think I would have to split the domain up into parts. The problem is I'm not quite sure how to split this up, or if the above domain is even correct.

Any help would be greatly appreciated.

Dear chewitard,

$\displaystyle\int_{-3}^1\displaystyle\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\: x^2 \: dxdy=\displaystyle\int_{-3}^1\left[\frac{x^3}{3}\right]_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}dy$

$\displaystyle\int_{-3}^1\displaystyle\int_{-\sqrt{9-y^2}}^{\sqrt{9-y^2}}\: x^2 \: dxdy=\displaystyle\int_{-3}^{1}{\frac{2}{3}\sqrt{9-y^2}}dy$

Substitute $y=3\sin{\theta}$ and you will be able to find the answer.