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Finding volume between surface and x-y plane

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data

    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)

    2. Relevant equations



    3. 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]
     
    Last edited: Apr 15, 2009
  2. jcsd
  3. Apr 15, 2009 #2
    Your setup looks good. But what a messy integral. Have you learned change of variables in multiple integrals yet?
     
  4. Apr 15, 2009 #3
    change of variables... as in changing to polar coordinates rsp. cylinder coordinates?
     
  5. Apr 15, 2009 #4
    More general than that, where you use the Jacobian. Have you learned it yet?
     
  6. Apr 15, 2009 #5
    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
     
  7. Apr 15, 2009 #6
    I was thinking if you transform the unit disc S [tex]u^2+v^2\le1[/tex] into the ellipse R [tex]\left(\frac{x+1}{4}\right)^2+\left(\frac{y-2}{\sqrt{8}}\right)^2\le1[/tex] the integral might be easier.
     
  8. Apr 16, 2009 #7
    Hmm, ok we havnt done that yet, but ill check it out, thanks :)
     
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