1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    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!

Double integral in polar coordinates

  1. Apr 10, 2010 #1
    1. The problem statement, all variables and given/known data
    Ok so I solved the problem, I think. I would just like to check my work.
    So the problem is:
    Use polar coordinates to find the volume of the given solid bounded by the paraboloids z = 3x^2 + 3y^2 and z = 4 - x^2 - y^2.

    2. Relevant equations
    r^2 = x^2 + y^2
    x = r cos Θ
    y = r sin Θ

    3. The attempt at a solution
    Ok so I starts by setting the two equations equal to each other to find their intersection.
    3x^2 + 3y^2 = 4 - x^2 - y^2
    4x^2 + 4y^2 = 4 (add the x^2 and y^2 across)
    x^2 + y^2 = 1 (divide by four)
    So then that the paraboloids intersect in the circle x^2 + y^2 = 1.

    So from this, I figured that the area I was integrating over in polar coordinates was 0 ≤ r ≤ 1 and 0 ≤ Θ ≤ 2pi.

    Then, I converted the two equations to polar coordinates:
    z = 3x^2 + 3y^2 = 3r^2
    z = 4 - x^2 - y^2 = 4 - r^2

    Then I took the double integral of (4 - r^2 - 3r^2) r dr dΘ from 0 to 1 (with respect to r) and then from 0 to 2pi (with respect to Θ).

    I subtracted the two r^2 and mulitplied to other r through so that I was now taking the double integral of 4r - 4r^3 dr dΘ over the same area.

    I integrated that to get 2r^2 - r^4 and plugged in 1 and 0 for the values of r and got 1.

    Then I integrated 1 from 0 to 2pi to get a final answer of 2pi.

    Did I do something wrong?
    Thanks! :)
  2. jcsd
  3. Apr 10, 2010 #2
    I don't think you have done this properly. I think that you need to find the volume of the solid for a surface which is

    [tex] z=4(-x^2-y^2+1) [/tex]

    for z>0

    Which intersects the xy-plane on the unit circle centred at (0,0)
  4. Apr 10, 2010 #3
    That equation you gave is not in the problem I stated so I have no idea where you are coming from.
  5. Apr 10, 2010 #4
    Well you said that you need to find the volume between the two surfaces?

    [tex] z_1 = 3x^2 + 3y^2 [/tex]
    [tex] z_2 = -x^2 - y^2 + 4 [/tex]

    The difference in their height at every point is

    [tex] z=z_2-z_1=-4x^2-4x^2+4 [/tex]

    I think that the solution will be to find the area under the surface above which can be written as

    [tex] z=4(-x^2-y^2+1) [/tex]

    that is, the volume between this curve and the xy-plane. Also, to find a volume, polar co-ordinates are not sufficient, how can you get a volume when integrating for a 2D problem? Use cyclinderical polar coordinates or spherical polar coordinates.

    Also, I suggest the former method, adding small disks of varying radius and infinitesimal height dz.
  6. Apr 10, 2010 #5
    I can only use polar coordinates cause that is the unit we are doing in the book so they have to be sufficient.
  7. Apr 10, 2010 #6
    And you were saying that I was trying to find the area under the surface z = 4(-x^2 - y^2 +1).
    However, if you look, I did that because I integrated 4 - 4r^2 which is exactly that.
  8. Apr 10, 2010 #7
    Then I think actually it should be OK
  9. Apr 10, 2010 #8


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No. It is exactly right.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook