How Do You Derive the Average Absolute Y-Value on a Unit Sphere?

[kmarinas86]

How do you derive the average abs(y) value of points on the surface of a sphere with a radius of 1 centered at (x,y,z)=(0,0,0)?

I'm assuming that the points are sampled evenly, or that you can integrate for all points without any distortions that result by averaging over the xy-plane.

How do you do this?

Motivation: I wanted to know the average cos(theta) for a unit vector of a random direction, where theta is the angle away from the xy-plane. I know that I can't just arrive at this by averaging the height over the xy-plane because doing so under-represents points on the sphere with a larger x^2+y^2. I strongly suspect that this value will be less than 2/3. I am guessing that when it's solved, it will be 1/2, but I don't really know. I need to know how to do this. Thank you!

 

[Bill Simpson]

Imagine your sphere is a thin shell. Then I believe can think of your problem as wanting to find the centroid of the upper half shell.

http://en.wikipedia.org/wiki/Centroid

The "By integral formula" section looks most promising.
Note: I don't believe the answer is 1/2.

 

[HallsofIvy]

It should be obvious, by symmetry, that E(x)= E(y)= E(z) and, with standard spherical coordinates, it is slightly easier to find E(z).

for a point on the unit sphere, z= cos(\phi) so that is |z|= |cos(\phi)|. That is, a given value of z corresponds to a given \phi. And, we can weight each z value by the circumference of the "circle of latitude" at that z which is 2\pi z sin(\phi).

So the integral is
\int_{\phi= 0}^\pi |cos(\phi)|sin(\phi) d\phi
divided by the surface area of the unit sphere, 4\pi[/tex].<br /> <br /> Of course, that is the same as <b>twice</b> the z-coordinate of the centroid of the upper half sphere.<br /> <br /> <br /> And, of course, that is exactly

 

[kmarinas86]

It should be obvious, by symmetry, that E(x)= E(y)= E(z) and, with standard spherical coordinates, it is slightly easier to find E(z).

for a point on the unit sphere, z= cos(\phi) so that is |z|= |cos(\phi)|. That is, a given value of z corresponds to a given \phi. And, we can weight each z value by the circumference of the "circle of latitude" at that z which is 2\pi z sin(\phi).

So the integral is
\int_{\phi= 0}^\pi |cos(\phi)|sin(\phi) d\phi
divided by the surface area of the unit sphere, 4\pi[/tex].<br /> <br /> Of course, that is the same as <b>twice</b> the z-coordinate of the centroid of the upper half sphere. And, of course, that is exactly

<i><b>If I take &quot;phi&quot; as the angle from the xy-plane and z as the vertical axis, for a unit sphere centered at the origin (x,y,z)=(0,0,0):</b></i><br /> <br /> The domain for the top hemisphere should be phi=[0,pi/2].<br /> The domain for the bottom hemisphere should be phi=[-pi/2,0].<br /> <br /> I integrate &quot;shortest distance from the xy-plane*1D set of points arranged in a circle&quot; i.e. &quot;|sin(phi)|*circumference(phi)&quot; with respect to phi.<br /> <br /> |sin(phi)|*circumference(phi) expands to:<br /> <br /> |sin(phi)|*(2*pi*cos(phi))<br /> <br /> 2*pi is constant, so:<br /> <br /> 2*pi * integral(|sin(phi)|*cos(phi))<br /> <br /> If I integrate &quot;|sin(phi)|*cos(phi)&quot; for the top hemisphere (phi=0 to phi=pi/2), I get (1/2). Thus the integral of the above would equal 2*pi*(1/2), or simply pi.<br /> <br /> The area gives us a measure of the points which were integrated for all points on phi=[0,pi/2]. This area can arrived at by integrating circumference(phi) from phi=0 to phi=pi/2. The integral of 2*pi*cos(phi) from phi=0 to phi=pi/2 is 2*pi. Given that r=1, 2*pi also equals the top half area of the unit sphere.<br /> <br /> Here we can see that &quot;circumference(phi)&quot; is our weight <i>per phi</i>, and the integral of this weight with respect to <i>phi</i> gives us the total weight over the <i>interval of phi integrated</i>, which is the total area. So we shall take the integral of &quot;|sin(phi)|*circumference(phi)&quot; from phi=0 to phi=pi/2, which is 2*pi*(1/2), and divide by the area, to figure out the average |sin(phi)| for the top hemisphere:<br /> <br /> integral / weight = 2*pi*(1/2) / (2*pi) = 1/2<br /> <br /> I should get the same thing for the bottom hemisphere as well (i.e. integration from phi=-pi/2 to phi=0).<br /> <br /> Is my logic mathematically correct?<br /> <br /> Edit: The following link seems to assume a solid hemisphere rather than a hollow one.[/color]<br /> <br /> I see that Wolfram has a different answer for the centroid of a hemisphere:<br /> <a href="http://mathworld.wolfram.com/Hemisphere.html" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://mathworld.wolfram.com/Hemisphere.html</a><br /> <br /> He comes up with 3/8 times the radius. This would give us 3/8 instead of my 1/2. I don&#039;t know how to explain this difference. He integrates by z instead, which is what I tried to avoid. What&#039;s the catch?

 

[kmarinas86]

If I take "phi" as the angle from the xy-plane and z as the vertical axis, for a unit sphere centered at the origin (x,y,z)=(0,0,0):

The domain for the top hemisphere should be phi=[0,pi/2].
The domain for the bottom hemisphere should be phi=[-pi/2,0].

I integrate "shortest distance from the xy-plane*1D set of points arranged in a circle" i.e. "|sin(phi)|*circumference(phi)" with respect to phi.

|sin(phi)|*circumference(phi) expands to:

|sin(phi)|*(2*pi*cos(phi))

2*pi is constant, so:

2*pi * integral(|sin(phi)|*cos(phi))

If I integrate "|sin(phi)|*cos(phi)" for the top hemisphere (phi=0 to phi=pi/2), I get (1/2). Thus the integral of the above would equal 2*pi*(1/2), or simply pi.

The area gives us a measure of the points which were integrated for all points on phi=[0,pi/2]. This area can arrived at by integrating circumference(phi) from phi=0 to phi=pi/2. The integral of 2*pi*cos(phi) from phi=0 to phi=pi/2 is 2*pi. Given that r=1, 2*pi also equals the top half area of the unit sphere.

Here we can see that "circumference(phi)" is our weight per phi, and the integral of this weight with respect to phi gives us the total weight over the interval of phi integrated, which is the total area. So we shall take the integral of "|sin(phi)|*circumference(phi)" from phi=0 to phi=pi/2, which is 2*pi*(1/2), and divide by the area, to figure out the average |sin(phi)| for the top hemisphere:

integral / weight = 2*pi*(1/2) / (2*pi) = 1/2

I should get the same thing for the bottom hemisphere as well (i.e. integration from phi=-pi/2 to phi=0).

Is my logic mathematically correct?

I see that Wolfram has a different answer for the centroid of a hemisphere:
http://mathworld.wolfram.com/Hemisphere.html

He comes up with 3/8 times the radius. This would give us 3/8 instead of my 1/2. I don't know how to explain this difference. He integrates by z instead, which is what I tried to avoid. What's the catch?

I found something very interesting:

If I integrate "2*pi * integral(|sin(phi)|*cos(phi))" with respect to phi from phi=0 to phi=pi/2, I do get pi. However, if I substitute phi for asin(z), which gives me "2*pi * integral(|sin(asin(z))|*cos(asin(z)))" and integrate with respect to z from z=0 to z=1, I end up with 2 pi / 3.

Additionally, if I integrate the function "circumference(phi)" i.e. "2*pi*cos(phi)" from phi=0 to phi=pi/2, I get 2 pi. However, if I integrate the function "circumference(z)" i.e. "2*pi*cos(asin(z))" from z=0 to z=1, I get pi^2/2, which does not correspond to the area of the top half of the sphere.

In the first case, our answer was:
integral / weight = pi / 2 pi = 1/2

However, in the second case, our answer was:
integral / weight = (2 pi / 3) / (pi^2 / 2) = (4/3) / pi

In the first case, integrating the circumference over phi gave us the value for area of the top half of the sphere. In the second case, integrating the circumference over z gave us a value that does not correspond to the area of the top half of the sphere.

It is clear that the value derived from the first case via integrating phi gives us a weight corresponding to what we would intuit from integrating points on the sphere. The first case would appear to be appropriate, while integrating with respect to z would seem to give us the wrong answer.

Major concern:[/color] Given the fact that I can substitute phi for asin(z) (or, alternatively, sin(phi) for z) and get different values by simply swapping from integration of phi to integration of z, this must mean we must study very carefully what exactly what we are "weighing" when doing integrations such as this. We should not be oblivious to this. Perhaps this would have been avoided if we maintained that integration of points on a surface must be done along paths tangentially to that surface. Doing so would have established phi as the domain of integration of these points, as opposed to z.

 

[kmarinas86]

I see that Wolfram has a different answer for the centroid of a hemisphere:
http://mathworld.wolfram.com/Hemisphere.html

He comes up with 3/8 times the radius. This would give us 3/8 instead of my 1/2. I don't know how to explain this difference. He integrates by z instead, which is what I tried to avoid. What's the catch?

I just noticed that the page on "hemisphere" from Wolfram MathWorld does not make it clear as to whether the centroid is that of a hollow hemisphere of zero thickness or that of a solid hemisphere. If it is the centroid of a solid hemisphere, then the z value of the centroid of a hollow hemisphere must be greater than 3/8 * r.