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Math Help - Finding area of a region

  1. #1
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    Finding area of a region

    Would that give me the area for S? The formula that I took a picture of.
    Im doing (b)
    Attached Thumbnails Attached Thumbnails Finding area of a region-image.jpg   Finding area of a region-image.jpg  
    Last edited by asilvester635; February 7th 2013 at 09:16 PM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Finding area of a region

    I dont think so.

    This gives you the right soln

     \int_3^6  y dx - \int_3^4 f(x) dx

    Basically get the area under the curve of the line from where the line is tangent to f(x)(x=3) to where the line intersects the x-axis and subtract that from, the point where the line is tangent to f(x) (x=3) to where f(x) intersects the x-axis (x=4)
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  3. #3
    Newbie pssingh1001's Avatar
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    Re: Finding area of a region

    I think so..............
    u would be right ..............
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Finding area of a region



    subtrat from



    to get

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  5. #5
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    Re: Finding area of a region

    You are absolutely right. Area R would be integral from 0 to 4 of f(x).
    Volume of revolution would be 2pi integral of square of f(x) from 0 to 4.
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: Finding area of a region

    Guys, he is asking about The area of the region S, not the volume obtained from revolution.
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  7. #7
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    Re: Finding area of a region

    The tangent point is x=3, f(x) crosses the x-axis at x=4, and the line crosses the x-axis at x=6. So if you divide the area S into two regions with a vertical line at x=4, you have S=\int_3^4 (18-3x)-(4x^2-x^3)\,dx + \int_4^6 18-3x\,dx.

    And I think that gives the same answer as jakncoke.

    I think there's an extra 2 in ibdutt's answer to part (c) - it should be \int_0^4 \pi(4x^2-x^3)^2\,dx.

    - Hollywood
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