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Math Help - Volumes of revolution

  1. #1
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    Volumes of revolution

    A region is bounded by the given curves. Find the volume of the solid of revolution obtained by rotating the region about the -axis. (give your answer in terms of or approximated to the nearest tenth).
    y=3x x=1 x=2 x axis.
    Im having alot of trouble with this assignment because I just dont understand this stuff, any help would be great!

    Thanks
    Greg
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    Re: Volumes of revolution

    You don't need to use Calculus at all. The line y= 3x, from x= 0 to 1, rotated around the x-axis, gives a cone with "height" 1 and base radius 3(1)= 3. The lline y= 2x, from x= 0 to 1, rotated around the x-axis, cone with "height" 1 and base radius 2(1)= 2. Look up the formula for volume of a cone, find each of those areas, and subtract the inner cone from the outer cone.

    If you really want to use Calculus, you can use the "washer method". The a vertical line between those two lines, rotated around the x-axis will form a "washer", a disk with a hole in it. Its area is the are between the two cirles, \pi(3x)^2- \pi(2x)^2. Its thickness, dx, gives volume \pi(9- 4)x^2 dx= 5\pi x^2 dx. Integrate that from x= 0 to 1.
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    Re: Volumes of revolution

    Quote Originally Posted by HallsofIvy View Post
    You don't need to use Calculus at all. The line y= 3x, from x= 0 to 1, rotated around the x-axis, gives a cone with "height" 1 and base radius 3(1)= 3. The lline y= 2x, from x= 0 to 1, rotated around the x-axis, cone with "height" 1 and base radius 2(1)= 2. Look up the formula for volume of a cone, find each of those areas, and subtract the inner cone from the outer cone.

    If you really want to use Calculus, you can use the "washer method". The a vertical line between those two lines, rotated around the x-axis will form a "washer", a disk with a hole in it. Its area is the are between the two cirles, \pi(3x)^2- \pi(2x)^2. Its thickness, dx, gives volume \pi(9- 4)x^2 dx= 5\pi x^2 dx. Integrate that from x= 0 to 1.
    You have read the question wrong HallsOfIvy. The region is bounded by y = 3x, x = 1, x = 2 and the x axis. Not y = 3x and y = 2x. The OP needs to realise that the region will be DISCS or CYLINDERS (not washers) of radius 3x, so the volume of revolution will be \displaystyle V = \int_1^2{ \pi \left( 3x \right) ^2 \, dx }.
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