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Homework Help: Triple Integration from Rectangular to Spherical Coordinates

  1. Apr 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Convert the integral from rectangular coordinates to spherical coordinates

    2 √(4-x^2) 4
    ∫ ∫ ∫ x dz dy dx
    -2 -√(4-x^2) x^2+y^2

    2. Relevant equations

    x=ρ sin∅ cosθ
    y=ρ sin∅ cosθ
    z=ρ cos∅

    In case the above integrals cannot be understood:
    -2 ≤ x ≤ 2
    -√(4-x^2) ≤ y ≤ √(4-x^2)
    x^2+y^2 ≤ z ≤ 4

    3. The attempt at a solution

    I figured that 0≤θ≤2∏ but and that the x converts to ρ sin∅ cosθ
    and I know you have to multiply the original converted function (ρ sin∅ cosθ) to ρ^2 sin∅
    but that's all i figured out
    Last edited: Apr 25, 2013
  2. jcsd
  3. Apr 25, 2013 #2


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    I assume that integtration means$$
    \int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4x\, dzdydx$$The first thing you should do is draw a picture of the 3D region described by the limits. You will need it to do the spherical limits. Are you sure you aren't asked to put it into cylindrical coordinates? That would be the natural choice.
  4. Apr 25, 2013 #3
    Actually the homework problem asked to convert to both cylindrical and spherical coordinates and I already finished the cylindrical.
    And yes thank you for cleaning up my equation!
  5. Apr 26, 2013 #4


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    [itex]\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4x\, dzdydx[/itex]
    First, mark vertical lines, on an xy- graph, at x= -2 and x= 2. Of course, [itex]y= -\sqrt{4- x^2}[/itex] and [itex]y= \sqrt{4- x^2}[/itex] are halves of the circle [itex]x^2+ y^2= 4[/itex] that lies between those vertical lines.

    Then [itex]z= x^2+ y^2[/itex] to z= 4 can be written, in spherical coordinates, as [itex]\rho cos(\phi)= \rho^2 sin^2(\phi)[/itex] or [itex]\rho= cot(\phi)csc(\phi)[/itex] and [itex]\rho cos(\phi)= 4[/itex] or [itex]\rho= sec(\phi)[/itex]
  6. Apr 26, 2013 #5


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    @enwarnock: So how are you coming on the spherical coordinate limits? Do you see that depending on what values ##\phi## takes that your ##\rho## is a two piece function and you are going to need two triple integrals to express the volume?
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