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Use the cylindrical coordinates to find the volume

  1. Apr 29, 2005 #1
    I'm having trouble figuring out this volume.

    Use the cylindrical coordinates to find the volume of the solid S bounded by z=x^2 + y^2 and
    z=12 - 2x^2 - 2y^2

    I've included a pic of what I think the regions and the solid look like

    3d Graph

    I think this is what the graph looks like. I would appreciate it if some one could show the steps of getting the volume, cause I'm kinda confused. I have an idea of what needs to be done, I just want to see if I'm on the right track before I get started.
     
  2. jcsd
  3. Apr 29, 2005 #2
    The integral you need to use will be of the form

    [tex]\int \int \int r \ dz \ dr \ d \theta [/tex]

    Choosing your upper and lower limits should not be too difficult.

    Here's a hint: which function z(r, theta) is above the other?
     
  4. Apr 29, 2005 #3
    I know all about cylindrical coordinates and how I should integrate. My question is when I integrate r. There's a constant radius where the two functions meet. So my question is, do I make two seperate integrals? One for the top of the volume in question and another for the bottom. I solved the two system of equations and I came up with x^2 + y^2=4, which means that the radius is 2 where these two functions intersect. So for the first volume for dr I would do from x^2 + y^2 to 2. And for the second volume for dr I would do from 2 to 12 - 2x^2 - 2y^2. Is this correct? Or can I just go from x^2 + y^2 to 12 - 2x^2 - 2y^2?
     
  5. Apr 29, 2005 #4
    I'm having a hard time understanding your problem, sorry. :frown:

    However, I think you'll find it simpler to disregard x and y completely. That way, you can deal with z as a function of r and theta.

    The first such function is

    [tex]z=r^2[/tex]

    the second

    [tex]z=12-2{r^2}[/tex]

    Envision the r, z plane. In that plane, those two functions should intersect at (2, 4).

    In the integral I showed you, the first integration is over z, not r. Integrate from the lower z to the upper z.

    After that, integrate over r to receive the area in the r, z plane, which can be further integrated over theta.

    Am I making any sense?
     
  6. Apr 29, 2005 #5
    Yes, it does make sense. What should my limits be for r though? That's the problem I'm having. The radius changes except for where the two functions intersect.
     
  7. Apr 29, 2005 #6
    Integrate r from 0 to 2.

    It's a measure of distance from the z axis, so the lowest value it assumes is 0. The highest, in this case, is the point where the surfaces closes together, 2.
     
  8. Apr 30, 2005 #7
    It's
    [tex]V=\int_0^{2\pi}d\phi \int_0^{2}rdr \int_{r^{2}}^{12-2r^{2}}dz=...=24\pi[/tex]
    Because the z=x^2+y^2 and z=12-2x^2-2y^2 intersect at r=2, and r increases from 0 to the maximum of 2.
     
    Last edited: Apr 30, 2005
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