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Surface Integral Problem

  1. Dec 3, 2013 #1
    If the mass per unit area of a surface is given by ρ=xy, find the mass if S is the part of the cylinder x2+z2=25 which is in the first octant and contained within the cylinder x2+y2=16.


    So here was my attempt.

    I parametrized the curve.
    x2+z2=25
    r(u, v) = <5cos(u), v, 5sin(u)>

    I then plugged into the bounds
    x2+y2=16.
    25cos2(u) + v2 = 16
    v = sqrt(16 - 25cos2(u))

    Next I took the cross product of ru X rv. Its magnitude is a constant 5.

    Now I solved the integral with bounds 0 < u < Pi/2 and 0 < v < sqrt(16 - 25cos2(u))
    ∫∫25 v cos(u) du dv

    This returns the result of -25/3 but since we are looking for a mass I submitted the answer of positive 25/3 and this is wrong. I checked the integration on my calculator and it gets the same result.
     
  2. jcsd
  3. Dec 3, 2013 #2

    LCKurtz

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    ##u## doesn't go from ##0## to ##\pi/2##. Look at a picture showing what angle ##u## represents.
     
  4. Dec 4, 2013 #3
    I redid my bounds and now I have arccos(sqrt(16-v^2)/5) < u < pi/2 and 0 < v < 4 which results in the answer of 22/15 which is still wrong. I fell like I found the wrong value for u.
     
  5. Dec 4, 2013 #4

    LCKurtz

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    I get 22/3 both that way and working the integral in reverse order so I think you have a mistake somewhere. But haven't you left out multiplying by the multiplier ##|r_u\times r_v|##? That would give 110/3 by my calculations. Is that the given answer?
     
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