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Finding volume between surface and x-y plane

  • Thread starter foges
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  • #1
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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]
 
Last edited:

Answers and Replies

  • #2
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Your setup looks good. But what a messy integral. Have you learned change of variables in multiple integrals yet?
 
  • #3
53
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change of variables... as in changing to polar coordinates rsp. cylinder coordinates?
 
  • #4
392
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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?
 
  • #5
53
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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
 
  • #6
392
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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.
 
  • #7
53
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Hmm, ok we havnt done that yet, but ill check it out, thanks :)
 

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