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Math Help - Integration problem - Volume by Revolution

  1. #1
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    Integration problem - Volume by Revolution

    Hello everyone. I'm having trouble solving this problem:

    Finding the volume by revolution of y=3-2x, rotated around the y-axis. I have the integration set up as  pi* ((y-3)/-2)^2 dy from -3/2...3/2 . Did I set this up wrong since I'm dealing with a straight line and not a curve? I can't seem to get the answer that my book gives.

    Thanks in advance for any help!
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  2. #2
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    Quote Originally Posted by bobbert116 View Post
    Hello everyone. I'm having trouble solving this problem:

    Finding the volume by revolution of y=3-2x, rotated around the y-axis. I have the integration set up as  pi* ((y-3)/-2)^2 dy from -3/2...3/2 . Did I set this up wrong since I'm dealing with a straight line and not a curve? I can't seem to get the answer that my book gives.

    Thanks in advance for any help!
    Hi bobbert116,

    Your limits are on the x-axis while they should be on the y-axis.

    You are integrating horizontal discs parallel to the x-axis,
    whose radii are x.

    You are integrating from y=0 to y=3.

    The thicknesses of these discs are "dy"

    Hence your integral should be

    \int_{0}^3{{\pi}\left(\frac{3-y}{2}\right)^2}dy

    The shape generated is that of a cone, so your result should equal

    \frac{{\pi}r^2h}{3}=\frac{{\pi}\left(\frac{3}{2}\r  ight)^2(3)}{3}
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