So these are two paraboloids. My guess is I would want to find the intersection of these two which would be $2x^2 + 3y^2 = 10$ and construct a triple integral based on its projection. No idea how to do this but the answer comes out to be $\dfrac{50\pi}{\sqrt{6}}$.
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1I say this to everyone: whenever you see $x^2+y^2$ in the boundary equations for a domain, consider switching to polar coordinates. I also say this: whenever you do a double or triple integral by hand (a barbaric practice, but a necessary evil in our day and age), draw or plot the boundaries. This will give you a better idea of the geometry of the problem, which will later help with the algebra. – Gyu Eun Lee May 20 '13 at 04:16
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$f(x, y) = x^2 + y^2$ and $g(x, y) = 10- x^2 -2y^2$. Now as you said you need to finds out where they intersect. That means we want $x^2 + y^2 = 10- x^2 -2y^2$ which means $2x^2+3y^2 = 10$. If you graph this in 2D you will see that it is an ellipse. Following the analogy with the 2D case, we have to compute
$\int \int g(x, y) − f(x, y)dxdy = \int \int_D (10 - 2x^2 -3 y^2) dxdy$, where $D$ is the interior of the ellipse. Now all that you need to find out is the limits of integration. Can you do it?
Here is the intersection of $f$ and $g$
Srijan
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If you've learned about change of variables in double integrals, you can easily turn this into an integral over a disk, i.e., the inside of a circle, where polar coordinates makes it a snap. – Ted Shifrin May 20 '13 at 12:34