Why isn't the average height of a point on an arc of radius a the same as the average height on a surface of radius a.

Stated another way the first problem is: Find the average height of a point on a unit semicircle with respect to the arc length θ.

The second is: Find the average height above the xy-plane of a point chosen at random on the surface of the hemisphere $x^2 + y^2 + z^2 = a^2$.

I thought both questions should have the same answer but.

For arc $$\bar{y} = \frac{1}{\pi}\int_0^\pi a\sin\theta \ \mathrm{d}\theta = \frac{2a}{\pi}$$

For surface

\begin{align} \bar{z} &= \frac{\int\int_S z\ \mathrm{d}S}{\int\int_S \mathrm{d}S}\\ &= \frac{\int_0^{2\pi}\int_0^{\pi/2}a\cos\phi\ a^2\sin\phi\ \mathrm{d}\phi\ \mathrm{d}\theta}{\int_0^{2\pi}\int_0^{\pi/2}a^2\sin\phi\ \mathrm{d}\phi\ \mathrm{d}\theta} \\ &= \frac{a}{2} \end{align}

I'v posted a link which covers what I mean by average height for an arc as well as problem set(sol) where I originally encountered the half-sphere problem. Check it out as they might explain the problems better.


look for question 6B-12



You shouldn't be too surprised that the results are not the same, apart from the fact that the formulas are clearly different.

The reason is that the distribution of points on an arc is different to the distribution of points on a spherical cap. For every two points at the same height on an arc, there is a circle of points on the spherical cap, the length of which varies according to height.

I know this is a somewhat "hand-wavy" argument, but I sense you are looking for an insight or intuitive explanation.

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  • $\begingroup$ So it makes sense that average height on the sphere is lower, because there are more points on a horizontal circle around the sphere the lower you go. $\endgroup$ – z.f.s Aug 26 '15 at 10:51
  • $\begingroup$ yes I agree.... $\endgroup$ – David Quinn Aug 26 '15 at 11:31

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