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Math Help - Volume of Revolution

  1. #1
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    Volume of Revolution

    y = x^3
    y = 4x
    in the first quadrant about line y = 8


    Would this be right?...

    integrate (from 2 to 0) [((8-x^3)^2)-(8-4x)^2] dx *pi

    Can someone tell me if I am right, if not, could you tell me how it should be?
    (I am using the washer method)
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  2. #2
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    Quote Originally Posted by Morgan82 View Post
    y = x^3
    y = 4x
    in the first quadrant about line y = 8


    Would this be right?...

    integrate (from 2 to 0) [((8-x^3)^2)-(8-4x)^2] dx *pi

    Can someone tell me if I am right, if not, could you tell me how it should be?
    (I am using the washer method)
    close enough ...

    V = \pi \int_0^2 (8-x^3)^2 - (8-4x)^2 \, dx
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  3. #3
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    Yes you integral is set up right, now simplify the expression and integrate.
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  4. #4
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    Quote Originally Posted by skeeter View Post
    close enough ...

    V = \pi \int_0^2 (8-x^3)^2 - (8-4x)^2 \, dx
    I figured it out... and I get -344pi/21

    ???
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  5. #5
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    Volume should never be negative, so therefore check your integration. You may have made a mis-calculation somewhere.
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  6. #6
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    Quote Originally Posted by Morgan82 View Post
    I figured it out... and I get -344pi/21

    ???
    I get \frac{832\pi}{21} ... check your algebra/antiderivative/arithmetic.
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  7. #7
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    Same as skeeter, work it out spaced out neatly, the difference sign in between the two can often make one lose track.
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  8. #8
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    Quote Originally Posted by skeeter View Post
    I get \frac{832\pi}{21} ... check your algebra/antiderivative/arithmetic.

    Are you VERY sure about this? My friend didn't get that.. and it seems like she is always right.
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  9. #9
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    We are both positive, just because she seems like she is always right, doesn't mean she is right.
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