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Math Help - volume of sphere using disk method.

  1. #1
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    volume of sphere using disk method.

    Say I have a sphere and it has a radius of 3. How could I find it's volume using the disk method and integration?
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    Member Mentia's Avatar
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    By disk method I assume you mean using cylindrical coordinates. In this case you need to think about your limits on z. At what z is the top and bottom of the circle? At \pm 3.

    So what is the equation of our circle? Its:

    x^2+y^2+z^2 = 9

    We know from polar coordinates that r^2 = x^2 + y^2. Then the equation of our circle is now:

    r^2+z^2 = 9 or r = \pm \sqrt[]{9-z^2}
    But in polar coordinates, a negative r doesn't make sense so we will say:
    r = \sqrt[]{9-z^2}.

    So now what is the area of any disk in our circle? It is:

     \int_{0 }^{2\pi }  \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta
    But we want to add up all our disks, so we multiply them by a little bit of height, dz, and then integrate over z also:

    \int_{-3}^{3} \int_{0 }^{2\pi }  \int_{0 }^{\sqrt[]{9-z^2} } rdrd\theta dz

    Does it work? Lets evaluate it, we know we should get \frac{4}{3}\pi(3^3) = 36 \pi

    Indeed, if you went through and solved the triple integral you would get 36 \pi.
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  3. #3
    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by gammaman View Post
    Say I have a sphere and it has a radius of 3. How could I find it's volume using the disk method and integration?
    Start by drawing the top half of a circle. Since I assume you were not given a specific location of the sphere on the graph, I would just center it at the origin. The top half of (since radius is 3) will go from (-3,0) to (3,0). Your disk will rotate about the x axis.

    Since you are only using 1/2 of a circle, use the equation:

    y=\sqrt{r^2 - x^2}

    Volume = 2\pi \int\limits^{3}_{-3} (\sqrt{9 - x^2})^2

    Can you take it from here?
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