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Multivariable calculus, Integral using spherical coordinates

  1. Aug 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Using spherical coordinates, set up but DO NOT EVALUATE the triple integral of f(x,y,z) = x(x^2+y^2+z^2)^(-3/2) over the ball x^2 + y^2 + z^2 ≤ 16 where 2 ≤ z.


    2. Relevant equations
    x = ρ sin ϕ cos θ
    y = ρ sin ϕ sin θ
    z = ρ cos ϕ
    ρ^2 = x^2 + y^2 + z^2

    ∫∫∫w f(x,y,z) dxdydz
    = ∫from θ1 to θ2 ∫from ϕ1 to ϕ2 ∫from ρ1 to ρ2 f(ρ*sinϕcosθ,ρ*sinϕsinθ,ρ*cosϕ) (ρ^2*sinϕ dρ dϕ dθ)


    3. The attempt at a solution
    First, I graphed xy, xz, yz plane, and from there I tried to find θ, ϕ, and ρ. For ρ, I got 0≤ρ≤4, I am not too sure about the bottom bound (0). For ϕ, I got 0 ≤ ϕ≤ pi/4, I was told that the top bound (pi/4) is wrong. For θ, I got 0 ≤ θ ≤ pi, I think both the bounds are right for this one.

    Then I plugged them into the triple intergal and changed x(x^2+y^2 + z^2)^(-3/2) into spherical coordinate to get "ρ^-2*sinϕcosθ"

    ∫from 0 to pi ∫from 0 to pi/4 ∫from 0 to 4 ρ^-2*sinϕcosθ (ρ^2*sinϕ dρ dϕ dθ)
    = ∫from 0 to pi ∫from 0 to pi/4 ∫from 0 to 4 (sinϕ)^2 cosθ dρ dϕ dθ

    Can you please help me find the correct bounds for θ, ϕ, and ρ and I believe the function in spherical coodinates is right. Thanks in advance.
     
  2. jcsd
  3. Aug 8, 2011 #2

    tiny-tim

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    hi wildleaf! :smile:

    (have a a pi: π and try using the X2 icon just above the Reply box :wink:)
    yes, in this as in many integrals over a sphere, you can have either 0 ≤ θ ≤ π or 0 ≤ θ ≤ 2π … which one you use for θ will affect the limits you use for ϕ :smile:

    before we go any further, how would you describe the region (in words)? :wink:
     
  4. Aug 8, 2011 #3
    ρ doesn't go from 0 because you have that plane at z=2 and figure out where that intersects the sphere it will help with the angle. I dont want to help to much.
     
  5. Aug 9, 2011 #4
    In the xy plane, the region is a circle with radius of 4. The yz and xz plane looks similar, it also a circle with radius 4 but there is a line z = 2, and we want the top of the circle. HELP!!! ME!!!
     
  6. Aug 9, 2011 #5
    In the xyz space, it would want the top half of the circle.
     
  7. Aug 9, 2011 #6
    on the bounds of rho , we have z=2 and [itex] z=\rho cos(\phi) [/itex]
    so [itex] \rho [/itex] will go from [itex] 2sec(\phi) [/itex] to 4 and that will make sure we are in the top half of the sphere .
     
  8. Aug 9, 2011 #7
    Ohhh...
    This is if I choose 0 ≤ θ ≤ π ?
     
  9. Aug 9, 2011 #8
    θ will go from 0 to 2π . ϕ will go from 0 to where the plane intersects the sphere. which would be. and rho will go to what i said above.
     
  10. Aug 9, 2011 #9
    0 ≤ θ ≤ 2n, 2sec(ϕ) ≤ ρ ≤ 4, 0 ≤ ϕ ≤ pi/4?
     
  11. Aug 10, 2011 #10
    everything looks good except that top bound for ϕ , cos(ϕ)=1/2
    what angle has a cosine of 1/2 , i got the 1/2 from 2/4
     
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