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Homework Help: I don't understand the ranges of the angles in spherical coordinates

  1. Nov 18, 2013 #1
    I'm not sure whether this falls in the homework category, or the standard calculus section, so apologies in advance if this doesn't fall in the right category.

    1. The problem statement, all variables and given/known data

    Evaluate ∫∫∫e^[(x^2 + y^2 + z^2)^3/2]dV, where the region is the unit ball x^2 + y^2 + z^2 ≤ 1.

    (or any variant of this question, where the region is always a ball with a radius of any size)

    2. Relevant equations

    Relevant equations would be the conversion of rectangular coordinates to spherical coordinates, such as ρ^2 = x^2 + y^2 + z^2, as well as intspherertp.gif

    3. The attempt at a solution

    Here, since the region is a whole sphere with a radius of one, I set the ranges for ρ to be from 0 to 1, and initially set the ranges for both angles from 0 to 2∏, and then set up a triple integral while substituting to get e^(r^3)*ρ^2sin∅dρdθd∅. However, I found out that the range for ∅ should be from 0 to ∏, instead of 2∏. Would it be possible to request an explanation as to why ∅ should only be what is essentially half a circle, while the other angle is 2∏? Thank you in advance.
  2. jcsd
  3. Nov 18, 2013 #2


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    You must cover your volume only once. Just draw the usual 3D Cartesian coordinate system, an arbitrary position vector and the spherical coordinates ([itex]r>0[/itex]: length of the vector; [itex]\vartheta \in (0,\pi)[/itex]: angle of the vector with the [itex]z[/itex] axis, which is the standard choice for the polar axis of spherical coordinates; [itex]\varphi \in [0,2 \pi)[/itex]: aximuthal angle, i.e., the angle between the projection of the vector to the [itex]xy[/itex] plane and the [itex]x[/itex] axis). Then you'll see that you cover all space once, except the [itex]z[/itex] axis, where spherical coordinates become singular, because the Jacobi determinant for the transformation from Cartesian to spherical coordinates vanishes along the [itex]z[/itex] axis.
  4. Nov 18, 2013 #3
    I'm sorry, I understood you until right after drawing an arbitrary position vector. Could you possibly "dumb it down" a bit more?
  5. Nov 18, 2013 #4


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    Imagine you are standing on a top of a pole. You want to be able to describe each direction you are looking at with the angles (we will ignore distance for a moment).

    One angle describes your horizontal rotation - you can look to S, N, W, E and any direction between. That means 2π.

    The other direction is vertical. You can look right up, you can look straight ahead, you can look down - that means π.

    Sure, you can look behind (in which case the vertical angle would be negative), but you don't have to. And if you will include all negative angles in your integration you will in fact look into each direction twice - so your integral would be twice the real value.
  6. Nov 18, 2013 #5


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    You can also think about latitude and longitude on a globe, which are similar to the polar and azimuthal angles in spherical coordinates.
  7. Nov 18, 2013 #6


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    Do an experiment with a plate or any nearly circular object. When the plate is flat, you can describe it using polar coordinates: r for the distance out towards the edge of the plate and θ is the angle round the plate (0 <= θ < 2∏).

    Now, rotate the plate vertically. When you have rotated it ∏/2 it will be vertical. Then another ∏/2 it will be horizontal again (now upside down).

    Look what has happened. One half of the plate has described an upper hemisphere and the other half a lower hemisphere. So, if you imagine the volume you have described it is a complete sphere.

    Alternatively if you have a semi-circle, you need to rotate that 2∏ to describe a full sphere.

    This shows that only one of the spherical angles must go from 0 to 2∏ and the other only from 0 to ∏.
  8. Nov 18, 2013 #7


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    It should be noted here, that it matters which one. If you use the standard ##dV=\rho^2\sin\phi
    d\rho d\phi d\theta##, ##\phi## should be restricted to ##[0,\pi]##. Otherwise you need absolute values on the ##\sin\phi##, complicating the integral or, worse, forgotten.
  9. Nov 18, 2013 #8


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    Simple, intuitive, mathematically precise. Pick any two!
  10. Nov 18, 2013 #9
    I had this same mental block a while ago. I'd used the limits dozens of times and one day I just couldn't figure out why.

    They way I figured it out was as follows:

    Picture yourself in a sphere sitting in a swivel chair holding a paint brush so that you're touching the inside of the sphere directly below you (chair is made of non-interacting matter...). If you were to begin spinning your chair and raising your arm you would paint the entire surface after rotating once in the chair and raising your arm to the very top.

    When I did this I was sitting in my home office and my wife came in, looked at me spinning with my arm out and a very intent look on my face, and walked out slowly. I just let it go, I'll never try to explain that to her. Just one more thing for her to be concerned about when it comes to my sanity.
  11. Nov 18, 2013 #10
    I find this one easiest to comprehend. Then based on that logic, you could possibly reverse the limits - so using your swivel chair analogy, I can be sitting on a swivel chair that's attached to a wall, and spinning would look like I'm brushing vertically 2pi, while horizontally it'd appear 1pi.
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