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Mass of Region Bounded by y=sin(x), z=1-y, z=0, and x=0

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    On a sample midterm for my Calc 3 class the following question appears:

    Find the mass of (and sketch) the region E with density ##\rho = ky## bounded by the 'cylinder' ##y =\sin x## and the planes ##z=1-y, z=0, x=0## for ##0\le x\le\pi/2##.

    2. Relevant equations
    $$ m= \int_{E} \rho dV$$

    3. The attempt at a solution
    By projecting down to the x-y plane the region is given by ##(x,y,z): 0\le x\le\pi/2, \sin x\le y\le1, 0\le z\le1-y##.

    So the mass of the region is $$\int_{0}^{\pi/2} \int_{\sin x}^1 \int_{0}^{1-y} ky dzdydx$$.

    Performing all the iterated integrations gives ##k(\frac{2}{9} -\frac{\pi}{24})## but the stated answer is ##k(\frac{16-\pi}{72})##.
     
    Last edited by a moderator: Oct 24, 2016
  2. jcsd
  3. Oct 24, 2016 #2

    Mark44

    Staff: Mentor

    The two expressions are equal. Combine the two terms in the first expression using a common denominator of 72.
     
  4. Oct 24, 2016 #3
    Oh, well then my terrible algebra skills strike again. Thanks
     
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