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Math Help - Volume of a Solid

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    Volume of a Solid

    Find the volume of a solid that lies above the plane xOy and is bounded by the surfaces x^2+y^2+z^2=1 and x^2+y^2-z^2=0.

    Start of a solution:
    We can find f(x,y) = z, i.e., z = \sqrt{1-x^2-y^2} and z = \sqrt{x^2+y^2}.

    I know that this involves multiple integrals, but my book was not very helpful and I cannot remember how to set this up.
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    Re: Volume of a Solid

    Hey vidomagru.

    Whenever multiple integrals come, the first thing you want to do is check what kind of co-ordinate system comes up.

    Hint: Take a look at the standard co-ordinate systems and see which one is best suited (take a look at cartesian, cylindrical, polar, and paraboloid systems).
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    Re: Volume of a Solid

    Quote Originally Posted by chiro View Post
    Hey vidomagru.

    Whenever multiple integrals come, the first thing you want to do is check what kind of co-ordinate system comes up.

    Hint: Take a look at the standard co-ordinate systems and see which one is best suited (take a look at cartesian, cylindrical, polar, and paraboloid systems).
    Ok so I believe that the best system to use would be cylindrical. Here is what I have done so far:

    we have x^2+y^2+z^2=1 with x=rcos\theta and y=rsin\theta. So substituting these in we get
    (rcos\theta)^2 + (rsin\theta)^2 + z^2 = 1

    r^2(cos^2\theta + sin^2\theta) + z^2 = 1

    r^2 + z^2 = 1

    z = \pm \sqrt{1-r^2}, so the limits for z are (-\sqrt{1-r^2},\sqrt{1-r^2}).

    Now for x^2+y^2-z^2=0 we have:
    (rcos\theta)^2 + (rsin\theta)^2 - z^2 = 0

    r^2(cos^2\theta + sin^2\theta) = z^2

    r^2 = z^2

    r = \pm z, so the limits for r are (-z,z).

    And the limits for \theta are (0,2\pi).

    So now we can set up the triple integral to be \int \int \int f(x,y,z)dV = \int_0^{2\pi} \int_{-z}^z \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r dzdrd\theta. So let's write this as:

    \int_0^{2\pi} \int_{-z}^z \left( \int_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} r dz \right) drd\theta

    =\int_0^{2\pi} \int_{-z}^z \left[rz\right]_{-\sqrt{1-r^2}}^{\sqrt{1-r^2}} drd\theta

    =\int_0^{2\pi} \int_{-z}^z \left(r(\sqrt{1-r^2}) - r(-\sqrt{1-r^2})\right) drd\theta

    =\int_0^{2\pi} \int_{-z}^z \left(2r(\sqrt{1-r^2})\right) drd\theta

    =\int_0^{2\pi}  \left[\frac{-2}{3}(1-r^2)^\frac{3}{2}\right]_{-z}^z d\theta....

    So I would continue here, but I have obviously made a mistake since I do not want this to be in terms of z... Help?
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    Re: Volume of a Solid

    Still confused on this, any thoughts?
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    Re: Volume of a Solid

    The surfaces are given by x^2+ y^2+ z^2= r^2+ z^2= 1 and x^2+ y^2- z^2= r^2- z^2= 0 or z^2= r^2. Where they intersect, r^2+ r^2= 2r^2= 1 so that r= 1/\sqrt{2}= \sqrt{2}/2. r goes from 0 to \sqrt{2}/2. Then for each r z goes from r up to \sqrt{1- r^2}.
    Thanks from vidomagru
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    Re: Volume of a Solid

    Quote Originally Posted by HallsofIvy View Post
    The surfaces are given by x^2+ y^2+ z^2= r^2+ z^2= 1 and x^2+ y^2- z^2= r^2- z^2= 0 or z^2= r^2. Where they intersect, r^2+ r^2= 2r^2= 1 so that r= 1/\sqrt{2}= \sqrt{2}/2. r goes from 0 to \sqrt{2}/2. Then for each r z goes from r up to \sqrt{1- r^2}.

    I might just be missing something obvious here, but why does r not go from -\sqrt{2}/2 and instead begin at zero? And why does z start from r?
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    Re: Volume of a Solid

    In cylindrical coordinates, 0 is the smallest value r can take. Then you have z^2 = r^2 and z^2 = 1-r^2. One of them is the lower bound and one is the upper bound. Since you are integrating over \theta from 0 to 2\pi, you are picking up all of the negatives.
    Last edited by SlipEternal; November 4th 2013 at 11:41 AM.
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