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Math Help - Shell Method

  1. #1
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    Shell Method

    I have a problem that I have been struggling with and need help:

    Use the shell method to determine the volume of a 24-cm diameter sphere with a 10-cm diameter hole drilled through the centerline.

    If someone could explain this to me so I can stop pulling my hair out I would be very happy.
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  2. #2
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    Quote Originally Posted by jigger1 View Post
    I have a problem that I have been struggling with and need help:

    Use the shell method to determine the volume of a 24-cm diameter sphere with a 10-cm diameter hole drilled through the centerline.

    If someone could explain this to me so I can stop pulling my hair out I would be very happy.
    For the Shell method around the y-axis the equation is...

    V=2 \pi \int_a^bxf(x)dx

    imagine that the part being drilled out is centered on the y-axis. It will have a radius of 5cm. The radius of the sphere will be 12cm.

    So the equation of the circle will be

    x^2+y^2=144

    Solving for y and using the positive root we get an equation of the upper half of the circle.

    y=\sqrt{144-x^2}

    Then using all of this we can set up the integral

    V= 2\pi \int_5^{12}x\sqrt{144-x^2}dx

    Since we used the upper half of the circle we will only get half of the volume so we will need to multiply by 2. Doing so we get..

    V= 4\pi \int_5^{12}x\sqrt{144-x^2}dx

    setting u=144-x^2 then du=-2xdx

    subbing in we get...

    -2\pi\int_{119}^0u^{1/2}du=2\pi \int_0^{119}u^{1/2}=\frac{4\pi}{3}u^{3/2}|_0^{119}=\frac{4\pi}{3}(119)^{3/2}=\frac{476\pi}{3}\sqrt{119}
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  3. #3
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    Thanks Empty Set, I follow you all the way till you sub in at the end. Where are you getting the 119 from? Can you explain more when you are subbing in?
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    Quote Originally Posted by jigger1 View Post
    Thanks Empty Set, I follow you all the way till you sub in at the end. Where are you getting the 119 from? Can you explain more when you are subbing in?
    u=144-x^2

    So when you change your limits of integration we evaluate the u expression.

    So for the upper limit x=12

    u=144-12^2=0

    so the new upper limit is u=0.

    for the lower limit x=5

    u=144-5^2=129

    so the new lower limit is 129.

    Then we flip over the integral using the property

    \int_a^bf(x)dx=-\int_b^af(x)dx

    I hope this clears it up...

    P.s you could just back substitue after integrating and use your original limits of integration. I just think that it is more work than is needed.
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