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

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

    http://img.photobucket.com/albums/v87/hbombblack/3dgraphcopy.jpg [Broken]

    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.
    Last edited by a moderator: May 2, 2017
  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


    the second


    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
    [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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook