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Math Help - Surface area question

  1. #1
    s3a
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    Surface area question

    Question:
    x = 1/3 * (y^2 + 2)^(3/2), 1 <= y <= 2

    Correct final answer:
    21/2 * pi

    My work:
    SA = 2pi * integral y ds = 2pi*integral (y * sqrt(1 + (dx/dy)^2)) dy
    x = 1/3 * (y^2 + 2)^(3/2)
    dx/dy = y*sqrt(y^2 + 2)

    Then I'm stuck. Wolfram also says my answer is wrong. I used y variables and the y limits of integration and the answer differs from that of the back of the book. (http://www.wolframalpha.com/input/?i=integrate(sqrt(1+%2B+y^4+%2B+y^2)*y)+*+2pi+from +1+to+2)

    Any help would be GREATLY appreciated!
    Thanks in advance!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by s3a View Post
    Question:
    x = 1/3 * (y^2 + 2)^(3/2), 1 <= y <= 2

    Correct final answer:
    21/2 * pi

    My work:
    SA = 2pi * integral y ds = 2pi*integral (y * sqrt(1 + (dx/dy)^2)) dy
    x = 1/3 * (y^2 + 2)^(3/2)
    dx/dy = y*sqrt(y^2 + 2)

    Then I'm stuck. Wolfram also says my answer is wrong. I used y variables and the y limits of integration and the answer differs from that of the back of the book. (http://www.wolframalpha.com/input/?i=integrate(sqrt(1+%2B+y^4+%2B+y^2)*y)+*+2pi+from +1+to+2)

    Any help would be GREATLY appreciated!
    Thanks in advance!
    Surface Area  = \int dS about the surface S.

    What is dS? Well, in a function only in the XY plain it is

    dS= \sqrt {1+ ( \frac{dx}{dy})^2 } dy

    if  x=f(y) and  a \le y \le b

    Simply integrate to find your answer.
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  3. #3
    s3a
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    But that's exactly what I did: ds = sqrt(1 + y^2 + y^4) dx

    Edit: (I forgot to mention that it is rotating about the x-axis)
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by s3a View Post
    But that's exactly what I did: ds = sqrt(1 + y^2 + y^4) dx
    Then you integrated wrong but I think your dS is incorrect as well

     x = \frac{ (y^2 + 2)^{ \frac{3}{2} } }{3}

     \frac{dx}{dy} = y \sqrt {y^2+2}

    ds= \sqrt { 1 + y^2 (y^2+2) } = \sqrt { 1 + y^4 + 2y^2 }

    I have a 2 where you don't. Perhaps I'm wrong (i didnt go over it lol) and if I am, then you messed up the integration. If not, then try again with that equation

    ...The rotating about the y-axis means we have

    S=\int 2 \pi x dS recompute with the new dS.
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  5. #5
    s3a
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    I said x-axis but that coefficient of 2 was my mistake; thanks!
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