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Math Help - Volume (disk method)

  1. #1
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    Volume (disk method)

    find the volume of the solid genereated b revolving the triangular region bounded by the lines:

    y = 2x , y = 0 , and x = 1 about

    a. the line x = 1
    b. the line x = 2



    my problem is i understand how to use the disk method when u flip on the x or y axis.. but when u do it on x = 1 or x =2 for the flip, i dont know how to understand and approach it!
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  2. #2
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    first note that if y = 2x, then x = y/2 .

    hope you have made a sketch and know that this problem involves horizontal disks with respect to y.

    for rotation about x = 1, your radius of revolution is a horizontal segment from the line x = 1 to the line x = y/2.

    since x = 1 is to the right of x = y/2, the length of the radius is 1 - y/2.

    so, your volume found by disks is ...

    V = \pi \int_0^2 \left(1 - \frac{y}{2}\right)^2 \, dy

    for rotation about x = 2, note that this will involve the washer method w/r to y.

    your inner radius goes from x = 2 to x = 1 ... r = 2-1 = 1

    your outer radius goes from x = 2 to x = y/2 ... R = 2 - y/2
    <br />
V = \pi \int_0^2 \left(2 - \frac{y}{2}\right)^2 - (1)^2 \, dy
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  3. #3
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    Quote Originally Posted by skeeter View Post
    first note that if y = 2x, then x = y/2 .

    hope you have made a sketch and know that this problem involves horizontal disks with respect to y.

    for rotation about x = 1, your radius of revolution is a horizontal segment from the line x = 1 to the line x = y/2.

    since x = 1 is to the right of x = y/2, the length of the radius is 1 - y/2.

    so, your volume found by disks is ...

    V = \pi \int_0^2 \left(1 - \frac{y}{2}\right)^2 \, dy

    for rotation about x = 2, note that this will involve the washer method w/r to y.

    your inner radius goes from x = 2 to x = 1 ... r = 2-1 = 1

    your outer radius goes from x = 2 to x = y/2 ... R = 2 - y/2
    <br />
V = \pi \int_0^2 \left(2 - \frac{y}{2}\right)^2 - (1)^2 \, dy
    hm ok i understand part a.. but can u help clarify how for part b u got a 2 and also a -1 at the end thanks!
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    ok i m guessing the two comes from the fact your flippen it over the 2 which is "on top" with respect to y.. but the -1 ^2 i m not seeing where that is coming from
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  5. #5
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    the integral method for horizontal washers is

    V = \pi \int_c^d (R^2 - r^2) \, dy

    I told you how I determined R and r in my previous post.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    the integral method for horizontal washers is

    V = \pi \int_c^d (R^2 - r^2) \, dy

    I told you how I determined R and r in my previous post.

    i know but it is makin no sense to me , hm is there another way to explain how to set this problem up?
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