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Homework Help: 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

    LCKurtz

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    No. It is exactly right.
     
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