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Math Help - triple integration

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    triple integration

    use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls
    Last edited by correct; January 6th 2009 at 02:42 AM. Reason: none
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    Quote Originally Posted by correct View Post
    use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls
    A start would be to note that:

    1. x^2 + y^2 + z^2 = 4z \Rightarrow z^2 - 4z = -x^2 - y^2 \Rightarrow (z - 2)^2 - 4 = -x^2 - y^2 \Rightarrow  (z - 2)^2 = 4 - x^2 - y^2.

    2. The line of intersection is in the plane 4 = 4z \Rightarrow z = 1. Therefore the line of intersection is x^2 + y^2 = 3.
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    thank you. pls what should i do next
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    Quote Originally Posted by correct View Post
    thank you. pls what should i do next
    Using a double integral and as Mr. F the intersection of the two volumes is a circle think polar coordinates.
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    i don't know how to proceed
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    Quote Originally Posted by correct View Post
    i don't know how to proceed
    Do you know how to represent the volume as a triple integral? Before answering, please re-read the thread and look at what you've been told so far ....
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    i know its triple integral dxdydz transforming this to polar coordinates it becomes r2sinthetadthetadpsi. if i use this, i don't know the limits
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    Quote Originally Posted by correct View Post
    i know its triple integral dxdydz transforming this to polar coordinates it becomes r2sinthetadthetadpsi. if i use this, i don't know the limits
    to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4.
    you should calculate the volume enclosed by the two spheres and then subtract this volume from the volume of the sphere x^2+y^2+z^2=4z .

    To get the enclosed volume:

    First draw a simple diagram diagram. Looking at it side-one in the xz-plane (y = 0) it's not hard to see that the top surface is z = \sqrt{4 - x^2 - y^2} and the bottom surface is z = 2 - \sqrt{4 - x^2 - y^2}. This supplies the z-integration terminals.

    The x and y integration terminals define the region in the xy-plane defined by the circle x^2 + y^2 = 3.

    So integrate the triple integral first with respect to z and then switch to polar coordinates.
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  9. #9
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    Quote Originally Posted by correct View Post
    use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls
    You can use the spherical coordinates, it's easy to do for your problem.

    \begin{gathered}\left\{ \begin{gathered}x = r\sin \theta \cos \varphi , \hfill \\y = r\sin \theta \sin \varphi , \hfill \\z = r\cos \theta . \hfill \\ \end{gathered}  \right. \hfill \\dxdydz = {r^2}\sin \left( \theta  \right)drd\theta d\varphi  \hfill \\ \end{gathered}

    Then the volume is V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{{\pi  \mathord{\left/{\vphantom {\pi  3}} \right.\kern-\nulldelimiterspace} 3}} {\sin \left( \theta  \right)} d\theta \int\limits_2^{4\cos \theta } {{r^2}dr}.

    The answer must be \frac{{22\pi }}{3}.
    Last edited by DeMath; January 8th 2009 at 12:44 AM.
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