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Math Help - Find volume of graph through one revolution on an axis

  1. #1
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    Find volume of graph through one revolution on an axis

    I have two questions here, the first i need someone to show me my mistake somewhere, the second I have no idea how to do.
    1. The area defined by the inequalities
    0\leq y\leq x^2, 0\leq x\leq 2
    is rotated about the y-axis. Find the volume generated.

    I take the raduis of the figure generated from the y-axis outward. so then its value would be x_{further}-x_{nearer}=2-\sqrt{y}
    V=\int^4_0 \pi x^2 dy
    V=\pi \int^4_0(2-\sqrt{y})^2 dy
    V=\pi \int^4_0(4-4\sqrt{y}+y) dy
    V=\pi [4y-\frac{8}{3}y^{\frac{3}{2}}+\frac{1}{2}y^2]^4_0
    V=\frac{8\pi}{3}
    answer is 8\pi

    2. Find the volume generated when the area in the first quadrant enclosed by y=|x^2-1|, and the line y=1 is rotated about the line y=1.
    I only know that the lines intersect at x=0 and \sqrt{2}. Also the radii of the figure would be 1-|x^2-1|.

    Thanks!
    Last edited by arze; February 9th 2010 at 12:14 AM.
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  2. #2
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    Quote Originally Posted by arze View Post
    I have two questions here, the first i need someone to show me my mistake somewhere, the second I have no idea how to do.
    1. The area defined by the inequalities
    0\geq y\geq x^2, 0\geq x\geq 2
    is rotated about the y-axis. Find the volume generated.

    I take the raduis of the figure generated from the y-axis outward. so then its value would be x_{further}-x_{nearer}=2-\sqrt{y}
    This rotates the region under the curve, you should be rotating the region above it.

    V=\int^4_0 \pi x^2 dy
    V=\pi \int^4_0(2-\sqrt{y})^2 dy
    V=\pi \int^4_0(4-4\sqrt{y}+y) dy
    V=\pi [4y-\frac{8}{3}y^{\frac{3}{2}}+\frac{1}{2}y^2]^4_0
    V=\frac{8\pi}{3}
    answer is 8\pi

    2. Find the volume generated when the area in the first quadrant enclosed by y=|x^2-1|, and the line y=1 is rotated about the line y=1.
    I only know that the lines intersect at x=0 and \sqrt{2}. Also the radii of the figure would be 1-|x^2-1|.

    Thanks!
    2
    For x between 0 and 1, 1-|x^2-1|=1-\left(1-x^2\right)
    For x between 1 and \sqrt{2},\ 1-|x^2-1|=2-x^2

    You can rotate the 2 regions seperately about y=1.
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  3. #3
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    Oops I made a mistake in the equations, the signs are the other way around
    its
    0\leq y\leq x^2 and 0\leq x\leq 2
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  4. #4
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    It was wrong for the x,
    are you sure the y isn't x^2\le{y}\le{4} ?
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  5. #5
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    No its right now its 0\leq y\leq x^2
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  6. #6
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    Sorry arze,
    your radius is incorrect,

    you have to subtract the volume calculated using r=\sqrt{y}
    from the volume calculated using r=2.

    8\pi is correct.

    Your volume of rotation using washers is the difference between 2 volumes of rotation.
    Last edited by Archie Meade; February 9th 2010 at 12:38 AM.
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  7. #7
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    here's the graph
    Attached Thumbnails Attached Thumbnails Find volume of graph through one revolution on an axis-washers.jpg  
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  8. #8
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    sorry i don't get it.
    I thought i already calculated the radii, by the difference between the two distances from the y-axis?
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  9. #9
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    No, the length you calculated is the width of the washer at any height y.
    That's the distance between the pink circles.
    As the shape is being revolved around the y axis,
    you have 2 radii, one outer radius=2 and one inner radius = \sqrt{y}

    Hence you must calculate

    \pi\int_{0}^4{2^2}dy-\pi\int_{0}^4{\left(\sqrt{y}\right)^2}dy
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