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Math Help - Rotation aorund x-axis (volume)

  1. #1
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    Rotation aorund x-axis (volume)

    Ok, this one is stumping me, I keep getting negative volume.

    find the volume of the solid generated by revolving the region enclosed by the triangle with verticies (2,2) (2,4) (8,4) about the x-axis.

    So I made a picture, found r(y)=2 and R(y)=y/3 (I did rise over run for the line, and solved in terms of y)
    so i know its the integral (pi) (R(y)^2)-(r(y)^2)

    found my bounds, 2 to 4, did the algebra, found anti deriv. and evaluated on the bounds, but I got -160pi/27 units^3
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  2. #2
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    Quote Originally Posted by BCHurricane89 View Post
    Ok, this one is stumping me, I keep getting negative volume.

    find the volume of the solid generated by revolving the region enclosed by the triangle with verticies (2,2) (2,4) (8,4) about the x-axis.

    So I made a picture, found r(y)=2 and R(y)=y/3 (I did rise over run for the line, and solved in terms of y)
    so i know its the integral (pi) (R(y)^2)-(r(y)^2)

    found my bounds, 2 to 4, did the algebra, found anti deriv. and evaluated on the bounds, but I got -160pi/27 units^3
    1. The line y = 4 (from x = 2 to x = 8) rotating around the x-axis generates a cylinder.

    2. The line y = \dfrac13x + \dfrac43 (from x = 2 to x = 8) generates a frustrum which is taken off from the cylinder.

    3. Therefore the volume is:

    V_{rot}=\int_2^8\left(\pi (4)^2 - \pi\left(\dfrac13x + \dfrac43  \right)^2  \right)dx

    I've got V=40\pi
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  3. #3
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    Thanks, I see where you got that, but I just made a huge mistake, it should be around the y-axis...dooH!
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  4. #4
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    Quote Originally Posted by BCHurricane89 View Post
    Thanks, I see where you got that, but I just made a huge mistake, it should be around the y-axis...dooH!
    Then the volume is:

    V-{rot}=\int_2^4\left(\pi\left(3y-4\right)^2-\pi(2)^2  \right)dy

    I've got V=48\pi
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