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Math Help - converting rectangular coord. to spherical coord. II

  1. #1
    gl7
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    converting rectangular coord. to spherical coord. II

    \iiint \sqrt{x^2+y^2+z^2}

    1st integral is 0 to \sqrt{1 -x^2 -y^2} (innermost)
    2nd integral is 0 to \sqrt{1-x^2}
    3rd integral is 0 to 1 (outermost)

    plse help me out
    having trouble inputing question hope this is clear

    i am having trouble converting this rectangular triple integral to spherical integrals
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by gl7 View Post
    \iiint \sqrt{x^2+y^2+z^2}

    1st integral is 0 to \sqrt{1 -x^2 -y^2} (innermost)
    2nd integral is 0 to \sqrt{1-x^2}
    3rd integral is 0 to 1 (outermost)

    plse help me out
    having trouble inputing question hope this is clear

    i am having trouble converting this rectangular triple integral to spherical integrals
    You need to listen to your own title!!

    You are NOT using spherical coordinates!

    Waring 1: Depending on which text you are using and your prof. the representation of spherical coordinates will be different here I will use
    x=\rho \sin(\theta)\cos(\phi)
    y=\rho \sin(\theta)\sin(\phi)
    z=\rho \cos(\theta)
    x^2+y^2+z^2=\rho^2

    Where \phi is the plane angle and \theta is the angle measured from the positive z axis.

    So in Spherical coordinates the integrand is

    \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho why don't we need an absolute value?

    Now Sketch the region of integration and you will see the it is the top half of a sphere of radius 1.

    Now to trace a sphere is radius 1 \rho \in [0,1]
    \theta \in \left[0,\frac{\pi}{2}\right]
    \phi \in [0,\2pi]

    Don't forget to change the differential of volume to spherical coordinates as well.
    dzdydx=\rho^2\sin(\theta)d\rho d\theta d\phi

    Fixed my typo above Thanks Ackbeet
    Last edited by TheEmptySet; April 28th 2011 at 08:21 AM. Reason: Typo
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    You need to listen to your own title!!

    You are NOT using spherical coordinates!

    Waring 1: Depending on which text you are using and your prof. the representation of spherical coordinates will be different here I will use
    x=\rho \sin(\theta)\cos(\phi)
    y=\rho \sin(\theta)\sin(\phi)
    z=\rho \cos(\phi)
    x^2+y^2+z^2=\rho^2

    Where \phi is the plane angle and \theta is the angle measured from the positive z axis.
    In which case you should have

    z=\rho\cos(\theta).

    The angle appearing in the variable z is always the polar angle (measured down from the z-axis) and the angle not appearing in z is always the azimuthal angle (measured in the xy-plane, counter-clockwise from the x-axis).

    So in Spherical coordinates the integrand is

    \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho why don't we need an absolute value?

    Now Sketch the region of integration and you will see the it is the top half of a sphere of radius 1.

    Now to trace a sphere is radius 1 \rho \in [0,1]
    \theta \in \left[0,\frac{\pi}{2}\right]
    \phi \in [0,\2pi]

    Don't forget to change the differential of volume to spherical coordinates as well.
    dzdydx=\rho^2\sin(\theta)d\rho d\theta d\phi
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