Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Volume of region between two functions?

  1. Nov 15, 2004 #1
    The two functions are z=2x^2-4 and z=4-2y^2, and I'm supposed to find the volume between the two by integrating two ways: one with respect to z first, and the other with respect to z last (x and y don't have a set order).

    When integrating with respect to z first, I had the limits such that z ranged from 2x^2-4 to 4-2y^2, and since x and y only depend on z and z has been covered already, I thought that x and y both ranged from -2 to 2.

    When I integrated with respect to z last, I had x range from -(z/2+2)^1/2 to (z/2+2)^1/2 and y range from -(2-z/2)^1/2 to (2-z/2)^1/2, with z ranging from -4 to 4.

    The problem is that I got two different answers, although I feel more confident in my answer from integrating the second way, z last (128/3). Could anyone help me with where I went wrong?
  2. jcsd
  3. Nov 15, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper

    You're confusing me.

    I don't understand:
    You mention two one-dimensional functions, and you ask for the volume rather than area between them.
    You write [tex]z=2x^2-4[/tex] as if [tex]z[/tex] is a dependant variable, but then you claim to integrate with respect to [tex]z[/tex].

    If you're talking about tbhe volume bounded by [tex]f(x,y)=2x^2-4[/tex] and [tex]g(x,y)=4-2y^2[/tex], I'm not sure that you can use a rectangular bounding region:
    If I simplify [tex]2x^2-4=4-2y^2[/tex], I get [tex]x^2+y^2=2^2[/tex] which is a circle of radius 2.

    Perhaps you could look into using slices?
  4. Nov 15, 2004 #3
    I'll try and describe the three dimensional region more clearly.

    z1 = 2x^2 - 4 is a parabola in the x-z plane that runs all through the y-axis, if that makes sense. It is a three dimensional surface that happens to be independent of y, so it is the same parabola for all y-values. The same goes for z2 = 4 - 2y^2, except it lies in the y-z plane, meaning that the two functions are perpendicular to eachother. The 3d space between z1 and z2 is a volume, bordered on top by z1 and on bottom by z2. The functions aren't two dimensional, they are just independent of the third, making the same two dimensional graph for all values of the third variable not in the equation.

    Does that make any more sense? The problem I'm having is finding the bounds for dzdxdy and dxdydz.
    Last edited: Nov 15, 2004
  5. Nov 16, 2004 #4
    In order to find the bounds, make an imagination of how that volume looks in space.

    Integration bounds are there where both functions cut into each other, ea, share the same z value for a given x/y point.

    Take it from there.

  6. Nov 16, 2004 #5


    User Avatar
    Science Advisor
    Homework Helper

    I mentioned in my post that the 'shadow' that this region casts into the xy plane is a circle. It's also relatively easy to identify that the z range is [tex](4,-4)[/tex].
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook