1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Triple integral (spherical coordinate)

  1. Jul 8, 2011 #1
    find the volume of the solid D that lies above the cone z = (x^2 + y^2)^1/2
    and below the sphere z = (x^2 + y^2 + z^2)

    i've done the integration until i need to substitute cos phi = u..
    however.. i dont know to change the range..

    http://imageshack.us/photo/my-images/839/spherical.jpg/"
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Jul 8, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    [Edit] Never mind the below paragraph. Your work is correct; I misread the equation at first. All you have to do is put in the correct u limit in your last equation. Since u = cos(φ) and φ goes from 0 to pi/4, your upper u limit is 1/sqrt(2). Be careful with your lower limit for u, it isn't 0. Just finish your integration.


    I don't think your integral is set up correctly. I see a sphere centered at (0,0,1/2) with radius 1/2 intersecting the cone at z = 1/2. So you apparently want the volume under the top half of the sphere but above the cone. Are you working this in two parts, the hemisphere and the cone with a flat base? If so, for the cone, ρ goes from 0 to z=1/2 which is ρcos(φ)=1/2 or ρ = (1/2)sec(φ). And you have to add the hemisphere.
     
    Last edited: Jul 8, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Triple integral (spherical coordinate)
Loading...