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Math Help - Area of solid formed by rotating in about the x axis

  1. #1
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    Area of solid formed by rotating in about the x axis

    A region is bounded by the lines y=√x,y=0 and y=x-2. Find the volume of the solid by rotating it about the x axis.

    what do i do?
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  2. #2
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    Before I explain my steps, is the correct answer known to you? If so, what is it?
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  3. #3
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    its a multiple choice

    it's either

    8pi/3
    16pi/3
    20pi/3
    24pi/3
    or
    32pi/3

    this text book doesnt have answers
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  4. #4
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    Ok, so the answer is 8pi/3.

    You must first draw the graph of the functions, y=rt.x, y=0 and y=x-2, so you know where the area bounded by the curves is above and below the x-axis.

    The formula for the volume is pi (integral sign) (f(x))^2 - (g(x))^2 dx.
    Your f(x) function is always your bigger function, so rt.x and so g(x) is x-2. Integrate that and you get: pi [(-x^3/3 + 5x^2/2 -4x).

    From your graph you know that your lower limit is 0 and your upper limit is 4.

    Sub. in x=0 into the anti-derivative from above and subtract is from the value you get when you sub in x=4 into the anti-derivative.

    The answer you get should be 8pi/3, which is the answer.

    Earlier, I mentioned to graph so you can see where the area is above and below, the area above and below cancels out here so 8pi/3 is the final answer.
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  5. #5
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    fanks i will do it now!!!!!!
    Last edited by -DQ-; April 22nd 2010 at 09:48 PM.
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