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Homework Help: Spherical coordinates angles of phi

  1. Jul 20, 2005 #1
    im having trouble determining the angles of phi in spherical coordinates when asked to convert a triple integral into spherical, and find the limits of the phi integral. can anybody point out any hints/tips/tricks how this may be done??Please...i have an exam tomorrow and im tryn to prepare for it...
  2. jcsd
  3. Jul 20, 2005 #2
    well... phi is the angle measured from the positive z-axis. a solid sphere consists of points whose value of phi vary from 0 to pi.

    cones, on the other hand, have points whose phi values vary from 0 to some specified or implied angle.

    umm... this is what you're asking for? :confused:
  4. Jul 20, 2005 #3
    for example im given a triple integral for the function w=z
    outermost integral is:
    the integral from -2 to 2, dx

    middle integral is:
    the integral from -sqrt(4-x^2) to sqrt(4-x^2), dy

    and inner most integral is:
    the integral from x^+y^2 to 4

    they say to convert this to spherical.
    i have a hard time finding the new limits of integration

    the way i interpret this is to say that z=4, z=x^2+y^2
    y=-sqrt(4-x^2) and y=sqrt(4-x^2)
    and x=-2 and x=2.
    then from here i use the conversion factors to get the limits???

    man im stumped..
  5. Jul 20, 2005 #4


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    Okay, have you drawn the graph?
    First draw a 2 dimensional graph and draw vertical lines at x= -2, x= 2.
    Now y is between [tex]y=-\sqrt{4-x^2}[/tex] and [tex]y= /sqrt{4-x^2}[/tex] which you should recognise immediately as both giving [tex]y^2= 4- x^2[/tex] or [tex]x^2+y^2= 4[/tex], the circle centered at (0,0) with radius 2 (and so fitting nicely between x= -2 and x= 2).
    The innermost integral has z between [tex]z= x^2+ y^2[/tex] and z= 4, a paraboloid and a horizontal plane.

    Since everything is circularly symmetric, obviously [tex]\theta[/tex] runs from 0 to [tex]2\pi[/tex].
    The paraboid [tex]z= x^2+y^2[/tex] is tangent to the xy-plane at (0,0) so [tex]\phi[/tex] starts at [tex]\frac{\pi}{2}[/tex].

    The hard part (are you really required to do this in spherical coordinates? Cylindrical coordinates would be much easier.) is determining what [tex]\phi[/tex] is when we change from the paraboloid to the plane. [tex]z= x^2+ y^2[/tex] to z= 4 when, of course, [tex]x^2+ y^2= 4[/tex]. Looking along the x-axis (which we can do because of the symmetry), when z= 4 and x= 2. The line through (0,0,0) to (2, 0, 4) has slope 4/2= 2 so cot(φ)= 2 (remember that φ is measured from the z-axis).\
    That means we need to break the integral into 2 parts, one with φ ranging from 0 to arccot(2), the other from arccot(2) to [tex]\frac{\pi}{2}[/tex]. Now we need to calculate ρ for each of those. For the first integral, ρ is measured along the line from (0,0,0) to the plane z= 4 along the line with slope cot(φ). Use cos(φ)= cos(arccot(2))= 4/ρ so that [tex]\rho= 2\sqrt{5}[/tex].
    Doing the same for the paraboloid, φ from arccot(2) to [tex]\frac{\pi}{2}[/tex] is going to be harder.
  6. Jul 20, 2005 #5


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    Hey guys. Always nice to draw a plot of what you're trying to integrate. As I see it, you're integrating f(x,y,z)=z over the volume enclosed inside the paraboloid and below the surface z=4 as shown below. :smile:

    Edit: I tell you what, if I were attempting to convert this into spherical coordinates, I might try to flip it over and work from the flat surface UP to the paraboloid. This to me seems more amendable to spherical coordinates or am I making it more difficult?

    Attached Files:

    Last edited: Jul 20, 2005
  7. Jul 21, 2005 #6


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    I modified the definition of a volume element of a spherical partition in 3-D in order to work this problem in terms of spherical coordinates. The best way to envision this is to look at the standard definition of the partition and then just turn it upside down.

    Rho then becomes the line segment from the point (0,0,4) to the point (x,y,z). Phi is the angle between rho and the z axis with phi=0 when rho is vertical and phi=pi/2 when rho is horizontal. We then have the following definitions:

    [tex]x=\rho Sin[\phi]Cos[\theta][/tex]

    [tex]y=\rho Sin[\phi]Sin[\theta][/tex]

    [tex]z=4-\rho Cos[\phi][/tex]

    I then expressed the paraboloid [itex]z=x^2+y^2[/tex] in terms of this new coordinate system:

    [tex]\rho[\phi]=\frac{-Cos[\phi]+\sqrt{Cos^2(\phi)+16 Sin^2(\phi)}}{2Sin^2(\phi)}[/tex]

    The integral in cartesian corrdinates is:

    [tex]\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{x^2+y^2}^4 z dz dy dz[/tex]

    The integral in this new spherical coordinate system is then:

    [tex]\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\rho[\phi]} (4-\rho Cos(\phi)\ \rho^2 Sin(\phi)d\rho d\theta d\phi[/tex]

    The value in both cases is:

    [tex]\frac{64 \pi}{3}[/tex]
    Last edited: Jul 22, 2005
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