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Small Section of a Sphere

Date: 01/10/2002 at 21:26:00
From: Anonymous
Subject: Volume and surface area of a small section of a sphere

I would like to know how to find the volume and the areas of each of 
the surfaces/faces of a small section of a sphere with "dimensions" 
delta r, delta theta, delta phi, in spherical coordinates.

If that isn't clear, maybe I can explain better by analogy in 
rectangular and cylindrical coordinates. In rectangular coordinates, a 
small section of a rectangle would have dimensions delta x, delta y, 
delta z. Its volume would be delta x * delta y * delta z, and its 
surfaces' areas would be delta x * delta y, delta x * delta z, and 
delta y * delta z.

In cylindrical coordinates, a small section of cylinder would have 
"dimensions" delta r, delta theta, delta z. I think its volume 
would be [(delta r)^2/2] * delta theta * delta z. I think its 
surfaces' areas would be r/2 * delta theta * delta z, (r + delta 
r)/2 * delta theta * delta z, delta r * delta z, and (delta r)^2/2 * 
delta theta.

Spherical coordinates are more complex, and I can't figure them out.  
Any help would be appreciated. Thanks.

Date: 01/11/2002 at 00:45:42
From: Doctor Tim
Subject: Re: volume and surface area of a small section of a sphere

Nice work on the cylinder, but there are problems - the solutions to 
which will help you with the sphere. So we'll do the cylinder first, 
or better yet, an area on the plane in polar coordinates.


I'll use little d for delta; it will end up looking like calculus, 
which is what this quickly becomes.

We want the area of a little area defined by dr and d(theta). Okay?

The main problem is that, unlike when you do dx and dy, these things 
are not the same size in different places in the plane. In particular, 
the same size swath in theta is a different physical size if you're 
close to the origin (small r) than if you're far from it (large r). 
You can think of the area of this little arc as being the length of 
the arc times its thickness. That length is r*d(theta). The thickness 
is dr. NOTE that "r" without the "d" is correct! The arc length is 
proportional to the radius, not to any tiny chunk of it. It depends on 
the coordinate. So we say,

dA = r dr d(theta).


Now we can see that the volume of a cylindrical chunk defined by dr 
d(theta) and dz depends on how far the chunk is from the axis of the 
cylinder. It turns out it's

dV = r dr d(theta) dz

The surface areas will be approximately (each)
(inner-outer) dA = r d(theta) dz
(top-bottom)  dA = r d(theta) dr
(left-right)  dA = dr dz


Okay, let's define theta as the angle from the pole, which goes from 
0 to pi, and phi as the azimuthal angle (longitude) which goes from 
0 to 2pi.

Now (as you rightly guessed) it's trickier, but it's the same 
principle. We have to figure out how physically big each dimension of 
your tiny volume is. As with the cylinder, it will be propoartional 
to r. But i also depends on theta, just as the area of something on 
Earth - when you get a square degree of land - depends on latitude: 
the closer to the pole you get, the smaller it is.

For a moment think of the intersection of a plane parallel to the 
equator with the sphere. Where it intersects is a circle of constant 
theta. How long is an arc on that circle with azimuthal extent d(phi)?

Well, if the radius of the sphere is R, the radius of the circle (call 
it S) is given by 

   S = R sin(theta)  -- (draw this until it's clear!)

So the extent in the "phi" direction, dP is

   dP = S d(phi)

(same argument as in polar coordinates above)

or dP = R sin(theta) d(phi)

now, the extent in the "theta" direction, is

   dT = R d(theta)

and in the R direction is simply dR.

So the bit of volume,

   dV = dR dT dP
      = R^2 sin(theta) dR d(theta) d(phi).

the areas are dRdT, dRdP, and dTdP.

Does this make sense? 

One way to check on the R^2: imagine you have one of these little 
boxes, and suddenly the radius increases by a factor of 2. Does it get 
4 times as big? You bet. It's the same proportion of the area of the 
sphere, and area goes as R^2. Is it smaller near the poles? Yes, 
because of the sin(theta) term, and we defined theta to be the polar 

I hope this helps!

- Doctor Tim, The Math Forum   
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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