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Solving by using cylindrical shell method

  1. Sep 21, 2009 #1
    The problem statement, all variables and given/known data

    Evaluate using the cylindrical shell method, the volume of the solid of revolution obtained by rotating the region enclosed by the curves y=x^3, y=-x and y=1 around the line y=-1.

    The attempt at a solution

    The answer is suppose to be 127(pi)/42 but I keep getting 169(Pi)/42.

    I first changed y=x^3 to x= y^(1/3) and y=-x to x=-y
    then found radius = y+1
    then height as y^(1/3)+y

    then plugged into cylindrical formula.

    1
    S 2(pi)(y+1)(y^(1/3)+y) dy =
    0

    ,,,,,,,1
    2(pi) S y^(4/3)+y^2+y^(1/3)+y dy =
    ,,,,,,,0

    2(pi) [(y^(7/3)/7/3) + (y^3)/3 + ((y^4/3)/4/3) +(y^2)/2] 0 to 1 =

    2(pi) [3/7 + 1/3 + 3/4 + 1/2] =

    2(pi) [36/84 + 28/84 + 63/84 + 42/84] =

    2(pi)(169/84) = 338(pi)/84 = 169(pi)/42

    I'm not sure where I went wrong.. I went over and over but I couldn't find out the problem.

    your help is greatly appreciated, thank you!
     
  2. jcsd
  3. Sep 21, 2009 #2
    Both my hand calculations and computer algebra system agree that

    [tex]2\pi \int_{0}^{1} (y+1)(y^{1/3}+y)\; dy = \frac{169 \pi}{42}.[/tex]

    Assuming that the region and axis of revolution are correct as stated, I believe your answer is correct.

    Assuming that the boundary curves are correct and the axis of revolution is incorrect, I determined that in order to arrive at the official answer, the axis of revolution must be the line y = -3/5 (which I highly doubt would have been misconstrued as -1). I don't know what to tell you, other than to make sure you've copied the question correctly and if so, speak to your instructor. Solution manuals have been known to be wrong.

    --Elucidus
     
  4. Sep 22, 2009 #3
    My instructor mailed the class saying that the actual answer is 169(pi)/42.

    I appreciate you going over the question, thank you!
     
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