# Thread: double integrals and changing order of integration

1. ## double integrals and changing order of integration

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

I think the domain for this is the shaded green:

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.

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

I think the domain for this is the shaded green:

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,

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

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

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