Finding volume between surface and x-y plane

[foges]

Homework Statement

Find the volume between the x-y plane (z=0) and the function z = x^2+2y^2, given that (x+1)^2+2(y-2)^2=4^2 (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 V = \int^{ x^2+2y^2}_{0}<area\_of\_ellipse> dz, I get some function dependent upon x and y, so i should first integrate with respect to z, so i get \int^a_b \int^c_d x^2+2y^2 dx dy, but what should i use as a,b,c,d?

ThanksEDIT: would it be the following: 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

 

[Billy Bob]

Your setup looks good. But what a messy integral. Have you learned change of variables in multiple integrals yet?

 

[foges]

change of variables... as in changing to polar coordinates rsp. cylinder coordinates?

 

[Billy Bob]

change of variables... as in changing to polar coordinates rsp. cylinder coordinates?

More general than that, where you use the Jacobian. Have you learned it yet?

 

[foges]

umm, our teacher has showed us an example, how when chaning to polar coordinates you need to multiply by the jacobian determinant, nothing in depth though, ill read up on it online though ... if you could give me a tip in what direction i should be searching

 

[Billy Bob]

I was thinking if you transform the unit disc S u^2+v^2\le1 into the ellipse R \left(\frac{x+1}{4}\right)^2+\left(\frac{y-2}{\sqrt{8}}\right)^2\le1 the integral might be easier.

 

[foges]

Hmm, ok we havnt done that yet, but ill check it out, thanks :)