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Math Help - spherical coordinates problem :D

  1. #1
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    Unhappy spherical coordinates problem :D

    hello,

    i'm having little problem with this .... i need to find volume of figure

     (x-1)^2 +y^2 +z^2 = 1

     y\ge 0

    using spherical coordinates ...

    now i realize that this is just a sphere, actually half of the sphere with center in (1,0,0) and radius of 1 ... that is not the problem... and it's volume should be as half of volume of same sphere in center (0,0,0) ...

    but i'm having problem how to figure out limits for variables ...


    i know that

     x = r \cdot \cos {\theta} \cdot \sin {\varphi}

     y = r \cdot \sin {\theta} \cdot \sin {\varphi}

     z = r \cdot \cos {\theta}

     J = r^2 \sin {\varphi}

    but how does them look like that if figure is shifted from (0,0,0) ? or is it something like :
    if shifted

     (x-1)^2 +(y+1)^2 +(z-1 )^2 = 1

    i need to use something like

     x = r \cdot \cos {\theta} \cdot \sin {\varphi} +p

     y = r \cdot \sin {\theta} \cdot \sin {\varphi} -q

     z = r \cdot \cos {\varphi} + t

    and J should be the same ? ? ? ?

    any help will be very much appreciated
    Last edited by sedam7; October 8th 2010 at 02:13 PM.
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  2. #2
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    Maybe just use

    V_{\text{half sphere}} = \frac{1}{2}\times \frac{4}{3}\pi \times r^3

    where r=1

    or do you have to use spherical co-ordinates?
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  3. #3
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    Quote Originally Posted by pickslides View Post
    Maybe just use

    V_{\text{half sphere}} = \frac{1}{2}\times \frac{4}{3}\pi \times r^3

    where r=1

    or do you have to use spherical co-ordinates?
    yes, it's explicitly said that must use spherical... It's not point to find that volume, but the point is to apply spherical coordinates...

    i know that for let's say ellipse, if i need to use polar and it's shifted i'll use

     x = a\cdot (r \cdot \cos {\theta} + p )

     y = b\cdot (r \cdot \sin {\theta} + q )

     J = a\cdot b \cdot r

    but can i apply same thing here ?

    the issue here for that sphere is how to find "r", limit's for "r"


    i try to solve that like this but get nowhere

     (x-1)^2 +y^2 +z^2 = 1
     x^2 -2x+1+y^2+z^2 = 1
     x^2+y^2+z^2 = 2x

    so if apply spherical coordinates

     r^2 = 2 \cdot r \cdot \cos {\theta} \sin {\varphi}
     r= 2 \cdot \cos {\theta} \sin {\varphi}

    so r will go from  0 to  2 \cdot \cos {\theta} \sin {\varphi} ???
    and  \varphi from  -\frac {\pi}{2} to  \frac {\pi}{2}
    and  \theta from  0 to  \frac {\pi}{2}

    where do i go wrong ?
    Last edited by sedam7; October 8th 2010 at 02:03 PM.
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  4. #4
    A Plied Mathematician
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    How about using a slightly modified form of spherical coordinates? Try

    x-1=r\sin(\varphi)\cos(\theta)
    y=r\sin(\varphi)\sin(\theta)
    z=r\cos(\varphi).

    There's nothing says you can't define your coordinates in a slightly different way.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    How about using a slightly modified form of spherical coordinates? Try

    x-1=r\sin(\varphi)\cos(\theta)
    y=r\sin(\varphi)\sin(\theta)
    z=r\cos(\varphi).

    There's nothing says you can't define your coordinates in a slightly different way.
    hmmm... thank you

    that would just mean that i will look at it as the sphere have center in (0,0,0) so than i'll have that my "r" will go from 0 to 1 ? and as you wrote them  \varphi \mid _{-\frac {\pi}{2}} ^{\frac {\pi}{2}} and  \theta \mid _{-\frac {\pi}{2}} ^{\frac {\pi}{2}} ?
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  6. #6
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    Right. The trick is to work your angle limits correctly so that y\ge 0. How can you do that?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Right. The trick is to work your angle limits correctly so that y\ge 0. How can you do that?
    okay if  y\ge 0

    that will mean that

     r \cdot \sin (\varphi) \sin (\theta) \ge 0

    now because i know that r is always  r \ge 0 it will mean that  \sin (\varphi) \sin (\theta) \ge 0 ?

    so from that i find limits ? when the sin is positive ? sin is positive in first and second quadrant ? correct ? and i know natural limits of the  \varphi \mid _0 ^{2\pi} and that  \theta \mid _0 ^{\pi} so it will mean that i have limits

     0<= r <= 1

    0 <= \varphi <=  \pi

     0< = \theta <= \pi ?
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  8. #8
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    I agree with your result, but I think that in the process of getting there, you confused the polar angle with the azimuthal angle. (This is something the mathematicians and physicists wrangle about, and concerning which they use different notation.) You can always tell which one is which by the angle that appears in the z component. That's always the polar angle, as far as I know. Its natural limits are [0,\pi]. The azimuthal angle corresponds to the angle used in cylindrical coordinates and polar coordinates, and is measured positive in the counter-clockwise direction from the positive x axis, as per usual. The azimuthal angle always inhabits [0,2\pi].

    Make sense?
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  9. #9
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    yes thank you very very much

    lol i did make some mistakes actually here we use it like ....

    polar / cylindrical

     x = \rho \cos (\varphi)
     y = \rho \sin (\varphi)
     z = z
     J = \rho

    spherical

     x = \rho \cos (\varphi) \sin (\theta)
     y = \rho \sin (\varphi) \sin (\theta)
     z = \rho \cos (\theta)
     J = \rho ^2  \sin (\theta)

    with natural limits as  \rho \mid _0 ^{\infty} ,  \varphi \mid _0 ^{2\pi} and  \theta \mid _0 ^{\pi}

    or like this

     x = \rho \cos (\varphi) \cos (\theta)
     y = \rho \sin (\varphi) \cos (\theta)
     z = \rho \sin (\theta)
     J = \rho ^2  \cos (\theta)

    with natural limits as  \rho \mid _0 ^{\infty} ,  \varphi \mid _0 ^{2\pi} and  \theta \mid _{-\frac {\pi}{2}} ^{\frac {\pi}{2}}
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  10. #10
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    So, did you get your result? And does it match up with what you know has to be the case?
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    So, did you get your result? And does it match up with what you know has to be the case?
    yes, thank you very very much
    actually i don't have solution, but with your suggestion (x-1 ) = r ... that is very helpful

    thank you
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  12. #12
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    You're welcome. Have a good one!
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  13. #13
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by sedam7 View Post

    spherical

     x = \rho \cos (\varphi) \sin (\theta)
     y = \rho \sin (\varphi) \sin (\theta)
     z = \rho \cos (\theta)
     J = \rho ^2  \sin (\theta)

    with natural limits as  \rho \mid _0 ^{\infty} ,  \varphi \mid _0 ^{2\pi} and  \theta \mid _0 ^{\pi}
    that's if you are to measure angle  \theta from the "north pole"


    Quote Originally Posted by sedam7 View Post
    ..
    or like this

     x = \rho \cos (\varphi) \cos (\theta)
     y = \rho \sin (\varphi) \cos (\theta)
     z = \rho \sin (\theta)
     J = \rho ^2  \cos (\theta)

    with natural limits as  \rho \mid _0 ^{\infty} ,  \varphi \mid _0 ^{2\pi} and  \theta \mid _{-\frac {\pi}{2}} ^{\frac {\pi}{2}}
    that's if you are to measure angle  \theta from the "equator line "


    you probably know that, but just to put it here so, one that reads those equations will know why is it two "different" spherical coordinates that you wrote there (it's the same but... )
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