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Math Help - Triple integral in a ball?

  1. #1
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    Triple integral in a ball?

    Solve the Integral in cylindrical coordinates
    ∫∫∫ dxdydz/(sqrt( x^2 + y^2 + (h-z)^2)
    B

    Where B is the Ball with a Radius R around (0,0,0), and the parameter h is greater than R.

    And then infer the average on that ball B with radius R of the distance opposite to the point (0,0,h)

    (the average of function f on some shape v is defined as
    [ ∫∫∫ f(x,y,z) dxdydz ] / vol V
    v
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  2. #2
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    re: Triple integral in a ball?

    Ok, I understand you're stuck in this assignment, but where exactly? What is your question to us?
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  3. #3
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    re: Triple integral in a ball?

    Is the following solution looks OK?
    Using Cylindrical Coordinates,
    ∫∫∫B dV/√(x + y + (h-z))
    = ∫(θ = 0 to 2π) ∫(z = 0 to R) ∫(r = -√(R - z) to √(R - z)) (r dr dz dθ) / √(r + (h-z))
    = 2π ∫(z = 0 to R) ∫(r = -√(R - z) to √(R - z)) r dr dz / √(r + (h-z))
    = 2 * 2π ∫(z = 0 to R) ∫(r = 0 to √(R - z)) r dr dz / √(r + (h-z)), by evenness
    = 4π ∫(z = 0 to R) √(r + (h - z)) {for r = 0 to √(R - z)} dz
    = 4π ∫(z = 0 to R) [√((R - z) + (h - z)) - √(h - z)] dz
    = 4π ∫(z = 0 to R) [√((R - z) + (h - z)) - (h - z)] dz, since h > R > r
    = 4π ∫(z = 0 to R) [√(R + h - 2hz) - h + z] dz
    = 4π [(-1/(3h)) (R + h - 2hz)^(3/2) - hz + z/2] {for z = 0 to R}
    = 4π {[(-1/(3h)) (R + h - 2hR)^(3/2) - hR + R/2] - (-1/(3h)) (R + h)^(3/2)}
    = 4π [(-1/(3h)) (h - R) - hR + R/2 + (1/(3h)) (R + h)^(3/2)].


    And now, how do I find the average value?
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  4. #4
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    Re: Triple integral in a ball?

    I have to specifically solve with Cylindrical Coordinates, but I am always confused with them, did I do an OK job with them in my solution? And is my final term is correct?
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  5. #5
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    Re: Triple integral in a ball?

    No, you have the "r" and "z" limits of integration reversed (perhaps that's a typo). The sphere of radius R, with center (0, 0, 0), is given by x^2+ y^2+ z^2= r^2+ z^2= R^2. That is, for every r and \theta, z varies from z= -\sqrt{R^2- r^2} to z= \sqrt{R^2- r^2}. It is r that varies from 0 to R and, of course, \theta varies from 0 to 2\pi
    Last edited by HallsofIvy; December 27th 2011 at 11:36 PM.
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  6. #6
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    Re: Triple integral in a ball?

    I see, thanks, but, hmm, the final answer is still correct, no?

    Or should I do all the calculations over again?
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  7. #7
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    Re: Triple integral in a ball?

    Do you think the following solution is correct?
    I got the coordinates a little different, is that still ok?


    In cylindricals the ball is covered by [θ=0,2π], [z=−R,R], [r=0,√(R−z)]

    ∫ = ∫∫∫ rdθdrdz / √{r+(z−h)}, [θ=0,2π], [z=−R,R], [r=0,√(R−z)]

    = 2π ∫∫ rdrdz / √{r+(z−h)}, [z=−R,R], [r=0,√(R−z)]

    = 2π ∫ √{r+(z−h)} dz, [z=−R,R], [r=0,√(R−z)]

    = 2π ∫ { √{R−z+(z−h)} − √{(z−h)} } dz, [z=−R,R]

    = 2π ∫ { √{R+h−2zh} − (h−z) } dz, [z=−R,R] since h > z over range of z

    = 2π { (−⅓h‾)(R+h−2zh)˙⁵ + (h−z) }, [z=−R,R]

    = 2π { (−⅓h‾)( (R+h−2Rh)˙⁵ − (R+h+2Rh)˙⁵ ) + (h−R) − (h+R) }

    = 2π { (−⅓h‾)( (h−R) − (h+R) ) − 2hR }

    = 2π { (−⅓h‾)(−6hR−2R) − 2hR } = 4πR/(3h)
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