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Spherical coordinates

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Find [itex] VR_{z}^{2} = \int \!\!\! \int \!\!\! \int_{E} (x^{2} + y^{2})dV[/itex] given a constant density lying above upper half of [itex]x^{2}+y^{2} = 3z^{2}[/itex] and below [itex]x^{2}+y^{2}+z^{2} = 4z[/itex].


    2. Relevant equations



    3. The attempt at a solution
    Why does it say upper half of [itex]x^{2}+y^{2} = 3z^{2}[/itex]? It's not like there are two possible halves.

    Here's what I've done so far but I'm a little unsure about some theory.
    For [itex]x^{2}+y^{2} = 3z^{2}[/itex]
    [itex] \tan^{2}\phi = 3[/itex] which means [itex]\phi=\pm\frac{\pi}{3}[/itex]

    I've seen most questions don't actually have a - value for [itex]\phi[/itex] because it only works in a range of [itex] 0 [/itex] and [itex]\pi[/itex] right? The movement itself is taken care of by [itex]\theta[/itex]? So I can just get rid off that negative [itex] \frac{\pi}{3}[/itex]?


    For [itex]x^{2}+y^{2}+z^{2} = 4z[/itex]:
    [itex]\rho=4cos\phi[/itex] and then usually questions will say [itex]0\leq\rho\leq4\cos\phi[/itex] but I don't really know how we get it ranges from 0... ?

    Assuming I did everything right I'll then get:
    [itex]E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\}[/itex]

    Which I then plug into an integral like this:

    [itex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta[/itex]
     
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  3. May 20, 2013 #2

    HallsofIvy

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    On the contrary, there are "two possible halves", [itex]z= \sqrt{(x^2+ y^2)/3}[/itex] and [itex]z= -\sqrt{(x^2+ y^2)/3}[/itex]. This is a cone with two nappes.

    What negative are you talking about? Yes, in spherical coordinates, [itex]\phi[/itex] is always gbetween 0 and [itex]\pi[/itex]. I don't see a problem with that.

    Where do the two surfaces, [itex]x^2+ y^2= 3z^2[/itex] (a cone) and [itex]x^2+ y^2+ z^2= 4z[/itex] (a sphere with center at (0, 0, 2)) intersect? Subtracting the first from the second we have [itex]z^2= 4z- 3z^2[/itex] which reduces to [itex]z^2- z= z(z- 1)= 0[/tex].
    z= 1 gives. as you say, [itex]tan(\phi)= 3[/itex].

    [/quote]Assuming I did everything right I'll then get:
    [itex]E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\}[/itex]

    Which I then plug into an integral like this:

    [itex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta[/itex][/QUOTE]
    Why do you still have [itex]x^2+ y^2[/itex] after you have converted to spherical coordinates?
     
  4. May 20, 2013 #3
    Cool, I see it now.

    This.

    [itex]x^{2}+y^{2}=3z^{2}[/itex]
    [itex]\rho^{2}\sin^{2}(\phi)\cos^{2}(\theta)+\rho^{2}\sin^{2}(\phi)\sin^{2}(\theta)=3\rho\cos^{2}(\phi)[/itex]
    [itex]\sin^{2}(\phi)=3\cos^{2}(\phi)[/itex]
    [itex]\tan^{2}\phi=3[/itex]
    [itex]\tan\phi=\pm\sqrt{3}[/itex]
    [itex]\phi=\pm\frac{\pi}{3}[/itex]




    Ok. I see you've found the intersection using the rectangular coordinates.
    Can't I do so in spherical co-ordinates?
    Why am I only getting one of the intersections when I do it this way (spherical coordinates)? i.e. [itex]\phi=\pm\frac{\pi}{3}[/itex] and not the 0.
    Also why is z=1 the equivalent of [itex]\tan\phi = \pm\sqrt{3}[/itex] ?
    If I convert z=1 to spherical co-ordinates I get [itex]\rho\sin\phi=1[/itex] ... not sure how to convert the 1 to the spherical equivalent.
    How did you get the centre of the sphere as (0,0,2)? The z on the other side of the equality has thrown me off a little.


    Why do you still have [itex]x^2+ y^2[/itex] after you have converted to spherical coordinates?[/QUOTE]



    Should be...

    [itex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (\rho^{2}\sin^{2}(\phi)) \,\rho^{2}\sin\phi d\rho d\phi d\theta[/itex]
     
    Last edited: May 20, 2013
  5. May 26, 2013 #4
  6. May 26, 2013 #5

    LCKurtz

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    Why the bump? What's left to do?
     
  7. May 29, 2013 #6
    Yeah, I don't really get why z=1 gives [itex]\tan\phi=\pm\sqrt{3}[/itex] I can get this if I convert to spherical co-ordinates straight away but not by solving the intersection of the rectangular coordinates.

    Also while there are no negative in spherical co-ordinate radians I get [itex]\phi=\pm\frac{pi}{3}[/itex]
     
  8. May 29, 2013 #7

    LCKurtz

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    In post #1 you have shown that ##\tan^2\phi = 3##, so ##\tan \phi = \pm\sqrt 3##. If you plug that into a calculator you get ##\phi = \pm \frac \pi 3##. Those are the principle values of the arctangent. But in spherical coordiantes you always want solutions between ##0## and ##\pi##. So if you want the bottom portion of the cone, you need the second quadrant value of ##\phi## whose tangent is ##-\sqrt 3##. That is ##\frac{2\pi} 3##. Of course, you don't need that for your problem because it is only concerned with the upper half of the cone.
     
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