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Math Help - Application of integrate- Volume 2

  1. #1
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    Application of integrate- Volume 2

    Region R is bounded by curve y=x^2 , y=1/x and the line x=1/2 . Calculate the volume of the solid formed by rotating region R through 360 degree about
    a) the x-axis
    b) the y-axis


    My problem is where is the region R in the graph?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sanikui View Post
    Region R is bounded by curve y=x^2 , y=1/x and the line x=1/2 . Calculate the volume of the solid formed by rotating region R through 360 degree about
    a) the x-axis
    b) the y-axis


    My problem is where is the region R in the graph?
    a.) Around the x-axis, we have the bounds as x=\frac{1}{2}...1 because \frac{1}{x}=x^2 \implies x=1.

    Note that \frac{1}{x}\geq x^2 here.

    So \pi\int_{1/2}^1\frac{1}{x^2}-x^4\,dx = \pi\left[-\frac{1}{x}-\frac{x^5}{5}\right]_{1/2}^1 = -\pi\frac{6}{5}-(-\pi\frac{321}{160}) = \boxed{\frac{129\pi}{160}}

    b.) I'd break it up into two integrals. Note that the functions are now x=\frac{1}{y} and x=\sqrt{y}

    x=\frac{1}{y} intersects x=\frac{1}{2} at y=2

    x=\frac{1}{y} intersects x=\sqrt{y} at y=1

    x=\sqrt{y} intersects x=\frac{1}{2} at y=\frac{1}{4}

    So we have \pi\int_{1/4}^1 (\sqrt{y})^2-(1/2)^2\,dy + \pi\int_1^2(1/y)^2-(1/2)^2\,dy = \frac{9\pi}{32}+\frac{\pi}{4} = \boxed{\frac{17\pi}{32}}
    Last edited by redsoxfan325; May 6th 2009 at 10:58 AM. Reason: I made a stupid mistake
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  3. #3
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    Part a) i can get your mean
    Part b) i can't
    In part b):
    intersects at
    Is'nt y=1/4 ?
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  4. #4
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    Thanks...i done.
    In part b), y = 1/4
    Then my answer will get 1.668971097
    Thanks.
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  5. #5
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    Quote Originally Posted by sanikui View Post
    Region R is bounded by curve y=x^2 , y=1/x and the line x=1/2 . Calculate the volume of the solid formed by rotating region R through 360 degree about
    a) the x-axis
    b) the y-axis


    My problem is where is the region R in the graph?
    see graph ...
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  6. #6
    Super Member redsoxfan325's Avatar
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    Yeah, it should be \frac{1}{4}. I fixed my above post. My only excuse is that it was late at night when I answered you, but still unexcusable. Sorry if I made you waste your time.
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  7. #7
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    No problem.
    It's ok.
    Thanks.
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