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Rectangular to sphere coordinates

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Hello,
    the problem ask you to pass this integral
    [tex]\displaystyle\int_{-2}^{2}\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}\int_{x^2+y^2}^{4} x {dz}{dy}{dx}[/tex]
    to sphere coordinates, but I don't really know how

    2. Relevant equations

    Well, I know the basics formulas,

    x=rho*sin(phi)*cos(theta)
    y=rho*sin(phi)*sin(theta)
    z=rho*cos(theta)

    rho^2=x^2+y^2+z^2
    tan theta=x/y

    and all the average problems I know how to do them, but this one I can't see it


    3. The attempt at a solution
    The solution of the book is this, but I don't know how to get there, any clues???

    [tex]\displaystyle\int_{0}^{2{\pi}}\int_{0}^{arctan(1/2)}\int_{0}^{4sec\phi} \rho^3 sin^2\phi cos\vartheta {d\rho}{d\phi}{d\vartheta}[/tex]+[tex]\displaystyle\int_{0}^{2{\pi}}\int_{arctan(1/2)}^{\pi/2}\int_{0}^{cot\phi csc\phi} \rho^3sin^2\phi cos\phi{d\rho}{d\phi}{d\vartheta}[/tex]

    Thank you
     
  2. jcsd
  3. Sep 4, 2009 #2

    Mark44

    Staff: Mentor

    One of the trickier things for iterated integrals is understanding what the region over which integration takes place looks like. Do you have a good sense of what the region in your problem looks like?
     
  4. Sep 4, 2009 #3
    Hello, I think it's an elliptic paraboloid
    like this
    attachment.php?attachmentid=20401&stc=1&d=1252102176.jpg

    but from that to the sum of integrals with csc, cot etc it's where I get really confused

    any hint will be great, thank you
     

    Attached Files:

  5. Sep 5, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]\displaystyle\int_{-2}^{2}\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}\int_{x^2+y^2}^{4} x {dz}{dy}{dx}[/tex]
    Okay, x must vary from -2 to 2 so draw vertical lines on an xy- graph at x= -2, x= 2. Then, for each x, y goes from [itex]y= -\sqrt{4- x^2}[/itex] to [itex]y= \sqrt{4- x^2}[/itex], two semi-circles. That gives the circle [itex]x^2+ y^2= 4[/itex]. Finally, for each (x,y), z goes from [itex]z= x^2+ y^2[/itex] to 4. Yes, that is the region inside the paraboloid you show up to z= 4.4

    Are you really asked to convert to spherical coordinates? Cylindrical coordinates would be most reasonable. In polar coordinates, the circle is just r varying from 0 to 2, [itex]\theta[/itex] from 0 to [itex]2\pi[/itex]. The integral in cylindrical coordinates would be
    [tex]\int_{r= 0}^2\int_{\theta= 0}^{2\pi}\int_{z= 0}^4 r^2 cos(\theta)d\theta dr dz[/tex]

    To change to spherical coordinates, note that the points most distant from the origin are on the circle where [itex]z= x^2+ y^2[/itex] intersects z= 4 and there distance is [itex]\sqrt{2^2+ 4^2}= 2\sqrt{4}[/itex]. [itex]\rho[/itex] will vary from 0 to [itex]2\sqrt{5}[/itex]. Of course, because of the circular symmetry, [itex]\theta[/itex] varies from 0 to [itex]\2\pi[/itex]. [itex]\phi[/itex] is the hard one.

    Imagine a line from (0,0,0) to [itex](\rho, 0, \rho^2)[/itex] (because of the symmetry, it enough to look at y= 0). That has slope [itex]\rho^2/\rho= \rho[/itex]. Since \phi is the complement of the angle from the x-axis, [itex]\phi= cot^{-1}(\rho)[/itex]. [itex]\phi[/itex] varies from 0 to [itex]cot^{-1}(\rho)[/itex].
     
  6. Sep 6, 2009 #5
    Thank you HallsofIvy and Mark44, now understand everything except how to get [tex]cot\phi csc\phi[/tex] in [tex]
    \displaystyle\int_{0}^{2{\pi}}\int_{arctan(1/2)}^{\pi/2}\int_{0}^{cot\phi csc\phi} \rho^3sin^2\phi cos\phi{d\rho}{d\phi}{d\vartheta}
    [/tex]

    this integral, say in the zy plane, is something like

    attachment.php?attachmentid=20432&stc=1&d=1252240003.jpg

    what I dont understand is how z=x^2+y^2 becomes [tex]cot\phi csc\phi[/tex] in sphere coordinates

    please this is the last thing :smile:
     

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