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Math Help - volume of solid bounded above &below by sphere and inside cylinder; correct limits?

  1. #1
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    volume of solid bounded above &below by sphere and inside cylinder; correct limits?

    below and above by sphere x^2+y^2+z^2=9 and inside cylinder x^2 + y^2=4

    using cylindrical, are these limits correct?

    r: 0 to 2
    theta: 0 to 2pi
    z: - sqrt(9-r^2) to sqrt(9-r^2)
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    can someone please check my limits to see if they are correct?
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    no, again this has to be done in sections. You're going to have a cylindrical portion and then 2 spherical end cap portions and you're going to have to write it as one integral plus twice another (the end caps are identical)
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    can you please elaborate
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Quote Originally Posted by SNAKE View Post
    can you please elaborate
    volume of solid bounded above &below by sphere and inside cylinder; correct limits?-clipboard02.jpg

    Looking at the picture you see you need to just integrate a cylinder of radius 2 from $z=-\sqrt{5}$ to $\sqrt{5}$

    Then from $z=\sqrt{5}$ to $z=3$ you have a spherical cap that's the top portion of a sphere of radius 3.

    The cylinder is easily integrated in cylindrical coordinates.

    The spherical caps (there's an identical one on the bottom) are easily integrated in spherical coordinates.

    So just break the integral up into those 3 pieces which are really just 2 pieces since two are identical.

    The integral for the cylinder is trivial.

    Finding the upper limit for $\theta$ for the caps is similar to the other problem you had.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    I am sorry i didnt make it clear, but this has to be done in cylindrical.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Quote Originally Posted by SNAKE View Post
    I am sorry i didnt make it clear, but this has to be done in cylindrical.
    so integrate the caps in cylindrical. It's not that much harder than doing it in spherical.

    It's still 3 separate pieces.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Why can't he double integrate the height 2 \sqrt{9-r^2} over the disk of radius 2 centered at the origin?

    - Hollywood
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    I understand what you guys are saying now. Because of the spherical top, this can't all be done in one integral, so that is why it has to be broken up into two?
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    hence assimilate this kind of truck caps within cylindrical. It's not necessarily a whole lot of more challenging as compared to doing the work within sale paper.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Can I please have some more input
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    2\pi \int_0^2 2r\sqrt{9-r^2}\, dr = 4\pi\left(9-\frac{5\sqrt{5}}{3}\right), which is approximately 66.265.

    - Hollywood
    Thanks from SNAKE
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    I am sorry, but I am still unsure. So the first integral is:: r: 0 to 2 theta: 0 to 2pi z: - sqrt(5) to sqrt(5). And the second is:: r: 0 to 2 theta: 0 to 2pi z...
    Last edited by SNAKE; May 12th 2014 at 09:29 PM.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Quote Originally Posted by SNAKE View Post
    I am sorry, but I am still unsure. So the first integral is:: r: 0 to 2 theta: 0 to 2pi z: - sqrt(5) to sqrt(5).
    And the second is:: r: 0 to 2 theta: 0 to 2pi z...
    Or the second integral is r: 0 to 2 theta: 0 to 2pi z: - sqrt(9-r^2) to sqrt(9-r^2), and that is where hollywood gets 2\pi \int_0^2 2r\sqrt{9-r^2}\, dr = 4\pi\left(9-\frac{5\sqrt{5}}{3}\right), the 2\sqrt{9-r^2} is from the z.

    Are you saying this can be done in one integral hollywood?
    Last edited by SNAKE; May 13th 2014 at 09:12 AM.
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    Re: volume of solid bounded above &below by sphere and inside cylinder; correct limit

    Quote Originally Posted by SNAKE View Post

    Are you saying this can be done in one integral hollywood?
    Yes.
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