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: Double integral of mass of circular cone

  1. Sep 27, 2007 #1
    Find the mass of a right circular cone of base radius r and height h given that the density varies directly with the distance from the vertex

    does this mean that density function = K sqrt (x^2 + y^2 + z^2) ?
    is it a triple integral problem?
  2. jcsd
  3. Sep 27, 2007 #2


    User Avatar
    Science Advisor

    Strictly speaking, no, because you are not given a coordinate system!!

    However, I understand what you mean. You have set up a "standard" coordinate system so that the vertex of the cone is at (0,0,0) and the central axis of the cone is on the positive z- axis. Yes, in this choice of coordinate system, "density varies directly with distance from the vertex" means that density= k[itex]\sqrt{x^2+y^2+z^2}[/itex].

    Yes, this is a triple integral (not the double integral your title implies).
  4. Sep 27, 2007 #3
    you can write it also as double integral
    [tex] \rho \propto \sqrt{r^2 +z^2} [/tex], and then write the volume element as [tex]dV=2\pi r dr dz [/tex]
  5. Sep 27, 2007 #4
    you would have problems with limits calculating this integral in x,y,z system.
  6. Sep 27, 2007 #5
    how do you get dV = 2pi r dr dz
    p ~ sqrt (r^2 + z^2)
    p ~ sqrt (p^2)
    so p ~ p

    limit on phi:
    z = sqrt (x^2 + y^2) (cone)
    z = sqrt (p^2 sin^2 phi)
    z = p sin phi
    p cos phi = r = p sin phi
    cos phi = sin phi
    phi = pi/4
    so limit of phi is 0 -> pi/4

    limit of theta is 0 -> 2pi

    [tex]\int^{2pi}_{0}\int^{pi/4}_{0}\int^{}_{} p dp dphi dtheta[/tex]

    is this correct so far?
  7. Sep 27, 2007 #6
    how do i go about looking for the limit of p?

    also it is suppose to say : p dp dphi dtheta in integral
  8. Sep 27, 2007 #7
    you should do it this way.
    [tex]\rho \propto \sqrt {r^2+z^2}~~~~(r^2 = x^2+y^2) [/tex]
    then write how radius r depend on height z. It's [tex] r=r_0(1-\frac{z}{h}) [/tex]
    Now you are integrating: [tex] 2\pi \int_{0}^{h}\int_{0}^{r_0(1-\frac{z}{h})} \sqrt{r^2+z^2} ~ r dr dz [/tex], first over r and then over z.
    2 pi comes from integrating over phi since problem has axial symmetry so volume element is 2 pi r dr dz.

    Hope I was clear enough
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook