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Math Help - Volume of Revolution - Cylindrical Shell

  1. #1
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    Volume of Revolution - Cylindrical Shell

    So I'm suppose to find the volume of revolution of an area enclosed by y=2-x^{2} and y=x^2 rotated about x=1.

    So I'm using the cylindrical shell method \int_a^b 2\pi rhdx and I'm integrating with respect to x; finding the integration points for the limits of integration, I get b=1 or a=-1.

    Since the axis is x=1, then r=1-x. The height, h=(2-x^{2})-(x^2)=2-2x^2

    Therefore, the definite integral is (factoring out the 2pi from the start and 2 from the height): 4\pi \int_{-1}^{1} ((1-x)(1-x^2))dx.

    Since it's symmetrical I can change the limits to 0 and 1, multiplying the integral by a factor of 2. Then expanding yields: 8\pi \int_{0}^{1} (1-x-x^2+x^3)dx

    By FTC: =8\pi (x-\frac {x^2}{2}-\frac {x^3}{3}+\frac {x^4}{4}) evaluated from 0 to 1.

    That yields 8\pi (1-\frac {1}{2}-\frac {1}{3}+\frac {1}{4}) and simplifying gets: \frac {10\pi}{3}

    However, I'm apparently wrong as the answer key says that it should be \frac {16\pi}{3}. I can't see my mistake, anyone see where I went wrong?
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  2. #2
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    Quote Originally Posted by VectorRun View Post
    So I'm suppose to find the volume of revolution of an area enclosed by y=2-x^{2} and y=x^2 rotated about x=1.

    So I'm using the cylindrical shell method \int_a^b 2\pi rhdx and I'm integrating with respect to x; finding the integration points for the limits of integration, I get b=1 or a=-1.

    Since the axis is x=1, then r=1-x. The height, h=(2-x^{2})-(x^2)=2-2x^2

    Therefore, the definite integral is (factoring out the 2pi from the start and 2 from the height): 4\pi \int_{-1}^{1} ((1-x)(1-x^2))dx.

    Since it's symmetrical I can change the limits to 0 and 1, multiplying the integral by a factor of 2. Then expanding yields: 8\pi \int_{0}^{1} (1-x-x^2+x^3)dx

    By FTC: =8\pi (x-\frac {x^2}{2}-\frac {x^3}{3}+\frac {x^4}{4}) evaluated from 0 to 1.

    That yields 8\pi (1-\frac {1}{2}-\frac {1}{3}+\frac {1}{4}) and simplifying gets: \frac {10\pi}{3}

    However, I'm apparently wrong as the answer key says that it should be \frac {16\pi}{3}. I can't see my mistake, anyone see where I went wrong?
    Remember that

    \int_{-1}^{1}x^ndx=0 if n is odd and


    \int_{-1}^{1}x^ndx=2\int_{0}^{1} if n is even

    This gives

    \displaystyle 8\pi \left(1-\frac{1}{3}\right)=\frac{16}{3}\pi
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  3. #3
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    The cylindrical shell method divides the solid into cylinders having height y_2- y_1 and radius equal to the distance from the axis of rotation. Here, y_2- y_1= (2- x^2)- (x^2)= 2- 2x^2 and the radius is x-(-1)= x+ 1. The surface area of such a cylinder is \pi r^2 h= \pi(2- 2x^2)(x+1) and, taking "dx" to be the thickness of the cylinder, the volume is the integral \pi\int_{-1}^1 (2- 2x^2)(x+ 1)dx.

    Yes, y_2- y_1= 2- 2x^2 is symmetrical about x= 0 but the radius, x+ 1, is NOT.
    Last edited by HallsofIvy; January 29th 2011 at 06:26 PM.
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  4. #4
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    Right! I can't believe I overlooked that. Alrighty, I figured out the rest from there. Splitting up the integrals and then applying the symmetry rules. I'm definitely getting \frac{16}{3}\pi now. Thanks a lot!
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