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Thread: converting rectangular coord. to spherical coord. II

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

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

    1st integral is 0 to $\displaystyle \sqrt{1 -x^2 -y^2} $ (innermost)
    2nd integral is 0 to $\displaystyle \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
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    Quote Originally Posted by gl7 View Post
    $\displaystyle \iiint $$\displaystyle \sqrt{x^2+y^2+z^2} $

    1st integral is 0 to $\displaystyle \sqrt{1 -x^2 -y^2} $ (innermost)
    2nd integral is 0 to $\displaystyle \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
    $\displaystyle x=\rho \sin(\theta)\cos(\phi)$
    $\displaystyle y=\rho \sin(\theta)\sin(\phi)$
    $\displaystyle z=\rho \cos(\theta)$
    $\displaystyle x^2+y^2+z^2=\rho^2$

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

    So in Spherical coordinates the integrand is

    $\displaystyle \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ 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 $\displaystyle \rho \in [0,1]$
    $\displaystyle \theta \in \left[0,\frac{\pi}{2}\right]$
    $\displaystyle \phi \in [0,\2pi]$

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

    Fixed my typo above Thanks Ackbeet
    Last edited by TheEmptySet; Apr 28th 2011 at 07:21 AM. Reason: Typo
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  3. #3
    A Plied Mathematician
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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
    $\displaystyle x=\rho \sin(\theta)\cos(\phi)$
    $\displaystyle y=\rho \sin(\theta)\sin(\phi)$
    $\displaystyle z=\rho \cos(\phi)$
    $\displaystyle x^2+y^2+z^2=\rho^2$

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

    $\displaystyle z=\rho\cos(\theta).$

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

    So in Spherical coordinates the integrand is

    $\displaystyle \sqrt{x^2+y^2+z^2}=\sqrt{\rho^2}=\rho$ 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 $\displaystyle \rho \in [0,1]$
    $\displaystyle \theta \in \left[0,\frac{\pi}{2}\right]$
    $\displaystyle \phi \in [0,\2pi]$

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