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## Homework Statement

Find the volume between the x-y plane (z=0) and the function [tex]z = x^2+2y^2[/tex], given that [tex](x+1)^2+2(y-2)^2=4^2[/tex] (ellipse)

## Homework Equations

## The Attempt at a Solution

I know this has to do with a double/tripple integral, but if i do the following [tex]V = \int^{ x^2+2y^2}_{0}<area\_of\_ellipse> dz[/tex], I get some function dependent upon x and y, so i should first integrate with respect to z, so i get [tex]\int^a_b \int^c_d x^2+2y^2 dx dy[/tex], but what should i use as a,b,c,d?

Thanks

EDIT: would it be the following: [tex]V = \int_{-5}^3 \int^{\sqrt{\frac{4^2-(x+1)^2}{2}}+2}_{-\sqrt{\frac{4^2-(x+1)^2}{2}}+2} x^2+2y^2 dy dx[/tex]

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