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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.

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

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

    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.

  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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