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Volume between cylinders

  1. Apr 15, 2006 #1
    Hi, I'm having trouble with the following question.

    Q. Use triple integrals and cartesian coordinates to find the volume common to the intersecting cylinders x^2 + y^2 = a^2 and x^2 + z^2 = a^2.

    This question pops up in basically every introductory calculus text. I've seen it before but I simply don't know how to set up the integral to do this question.

    I know that the integral is of the form:

    [tex]
    V = \int\limits_a^b {\int\limits_{h_1 \left( x \right)}^{h_2 \left( x \right)} {\int\limits_{g_1 \left( {x,y} \right)}^{g_2 \left( {x,y} \right)} {dzdydx} } }
    [/tex]

    The order of integration is nominal, I can change the projections if needed but I just chose z then y then x so that I have something to begin with.

    In mos of the questions that I've done, the range of z has simply been an fairly simple and easy to see (eg. 0 <= z <= 8) but that isn't the case in this question. Since I can't find the range of z values I'll start with find the projection of the region onto the x-y plane.

    x^2 + y^2 = a^2 and x^2 + z^2 = a^2. When the cylinders intersect I'll obtain x^2 + y^2 = x^2 + z^2 => y^2 = z^2? I'm having trouble visualising the region and finding the terminals on the volume integral. Can someone please help me out?
     
  2. jcsd
  3. Apr 15, 2006 #2
    Imagine a cylinder running up and down the z-axis with a radius a

    Now imagine a cylinder running up and down the y-axis with a radius a. Thoses are your cylinders.

    combine the 2, and you will find an intersection from 0 to a on all sides



    Now I think the limits you are looking for are:

    for y and z: -(a-x) to (a-x)
    for x: 0 to a

    But dont trust me on those...
     
  4. Apr 16, 2006 #3
    I could see what the cylinders look like. It's really the region close to the intersection that I can't find a way to visualising. There's usually a way to combine the equations for the projections onto the 2D planes to find the intersection but I can't find one. Thanks for the help anyway.
     
  5. Apr 16, 2006 #4
    In this particular case you don't need to combine the equations. Note that the first equation is independent of z. So to integrate over z first only the second equation will determine the upper and lower limits of z. Then, the second equation being independant of y, only the first equation will determine the upper and lower limits of y.
     
  6. Apr 17, 2006 #5
    Thanks for the help.
     
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