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Math Help - Check answer for volume by revolution

  1. #1
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    Check answer for volume by revolution

    Can you please tell me if i worked the following out correctly?

    y=x^2
    y=x+2

    1. Rotate this around the x-axis
    outside radius = x+2
    inside radius = x^2

     \pi\int_1^2 (x+2-x^2) = \frac{9\pi}{2}
    (thats suppose to be -1 to 2 but i cant get latex to work)
    2. Rotate around  y = 4
    outside radius = 4-x^2
    inside radius = 4-(x+2)

     \pi\int_1^2 (4-x^2)-(4-(x+2)= \frac{9\pi}{2}
    (thats suppose to be -1 to 2 but i cant get latex to work)
    3. Rotate around x = -2
    outside radius = \sqrt{y}+2
    inside radius = 2-y+2

    \pi\int_0^4 (\sqrt{y}+2-(2-y+2)= \frac{16\pi}{3}

    Thanks in advance
    Last edited by silencecloak; February 8th 2009 at 03:01 PM. Reason: typos
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  2. #2
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    Quote Originally Posted by silencecloak View Post
    Can you please tell me if i worked the following out correctly?

    y=x^2
    y=x+2

    1. Rotate this around the x-axis
    outside radius = x+2
    inside radius = x^2

     \pi\int_1^2 (x+2-x^2) = \frac{9\pi}{2}
    (thats suppose to be -1 to 2 but i cant get latex to work)

    Put the -1 in braces: \int_{-1}^2.
    The area of a circle is \pi r^2 you should have (x+2)^2- (x^2)^2.

    2. Rotate around  y = 4
    outside radius = 4-x^2
    inside radius = 4-(x+2)

     \pi\int_1^2 (4-x^2)-(4-(x+2)= \frac{9\pi}{2}
    (thats suppose to be -1 to 2 but i cant get latex to work)

    Again, the radii should be squared.

    3. Rotate around x = -2
    outside radius = \sqrt{y}+2
    inside radius = 2-y+2

    \pi\int_0^4 (\sqrt{y}+2-(2-y+2)= \frac{16\pi}{3}

    Thanks in advance
    same comment for the last one!
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  3. #3
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    So my answers are

    1. 72\pi/5

    2. 108\pi/5

    3. 24\pi

    Are these correct?
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  4. #4
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    Im not 100% confident in my set up can anyone please confirm?

    Thank you!
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  5. #5
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    I solved for the first two answers and also got
    and .
    However, I have not solved for the 3rd answer, as I need to verify my methods involving negative axes of revolution.
    I am sure of my first two answers, though.
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  6. #6
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    Ya my main concern is number 3 as well
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  7. #7
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    Anyone else have an opinion?
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  8. #8
    Member Abu-Khalil's Avatar
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    3) I couldn't understand how did you applied the disk method so I used shell method because i think it's easier in this case:

    V=2\pi\int_{-1}^2(x+2)(x+2-x^2)dx=2\pi\int_1^4t(t-(t-2)^2)dt=\frac{45\pi}{2}.
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