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Math Help - Volume of solid revolution

  1. #1
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    Volume of solid revolution

    Have a problems with the following question:

    Consider the solid of revolution obtained when the region surrounded by x=3 ,x=6 ,y=0 and y= x+5 is rotated about the line y=2

    Enter an expression for the cross-sectional area of this solid at a point

    I can find the value..but how to get the expression?

    I reckon the ans is suppose to be

    ((x+3)^2)*pi

    but its wrong
    Last edited by anthony162; March 23rd 2011 at 01:33 AM.
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  2. #2
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    Quote Originally Posted by anthony162 View Post
    Have a problems with the following question:

    Consider the solid of revolution obtained when the region surrounded by x=3 ,x=6 ,y=0 and y= x+5 is rotated about the line y=2

    Enter an expression for the cross-sectional area of this solid at a point

    I can find the value..but how to get the expression?

    I reckon the ans is suppose to be

    ((x+3)^2)*pi

    but its wrong
    1. I assume that you want to calculate the volume of the solid (?)

    2. A trapezium rotates about the line y = 2 which yields a frustrum of a cone.

    3. Use thin cylinders whose radius is y-2=x+3, whose thickness is dx. Use the formula to calculate the volume of a cylinder and add up all cylinders between x = 3 and x = 6.

    4. V=\int_3^6\left(\pi(x+3)^2\right)dx
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  3. #3
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    Nope..thats the second part of the question..which i have already answered..
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  4. #4
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    Have you sketched it? I see a radius of x. I don't know where the 3 is coming from - probably formula-dazzle.
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