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I've been wondering about something, and it might be nonsense (if so I apologize!). Consider the unit disk in $\mathbb{R}^2$ and a function $f$ defined on the disk. I can compute its double integral as

$$\int_{D(0,1)} f dA = \int_{0}^{2\pi} \int_0^1 f(r,\theta) r dr d\theta$$

by polar coordinates. Now, separately, consider a function $g$ defined on the upper hemisphere of $S^2$ (the sphere in $\mathbb{R}^3$. In spherical coordinates, I can compute its integral over this region as

$$\int_0^{2\pi} \int_0^{\frac{\pi}{2}} f(\theta,\phi) \sin(\theta) d\theta d\phi$$

What I'm wondering about is the following: given a function $f$ on the unit disc, can I find a function $g$ on the hemisphere such that $$\int_{D(0,1)} f dA = \int_{0}^{2\pi} \int_0^{\frac{\pi}{2}} g(\theta,\phi) \sin(\theta) d\theta d\phi$$

Of course I could just choose some $g$ that satisfies that, but I was wondering if there was a systematic way for each $f$ on the unit disk to associate it with a $g$ on the sphere such that their integrals are the same.

So far, I was thinking about the following. I know I can map the disk onto the upper hemisphere by the map $F(x,y) = (x,y, \sqrt{1-x^2-y^2})$. At first, I naively just thought of assigning each point on the sphere the value of $f(x,y)$ at the corresponding point beneath it, but that fails. This would send a constant function on the disk to a constant function on the hemisphere, but their integrals are different. Integrating a constant on the unit disk would just yield the constant times $\pi$, whereas a constant on the upper hemisphere when integrated would yield a $2\pi$. I would be okay with this if this process always differed just by a factor of two (i.e., I could identify $g$ with $\frac{1}{2} f$), but I don't think that works.

I thought maybe some change of coordinates might work, but I can't seem to get it to pan out. If anyone could give a suggestion or a pointer in the right direction, that would be very helpful. I do not want a full solution, just a pointer or a reference that discusses relevant ideas would be great. Thanks!

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2 Answers

up vote 1 down vote accepted

You've made life a bit difficult for yourself by giving the corresponding angles different names and unrelated angles the same name. If you rename $\theta$ to $\phi$ on the disk and put together your equations, you have

$$ \int_{0}^{2\pi} \int_0^1 f(r,\phi) r\,\mathrm dr\,\mathrm d\phi=\int_0^{2\pi} \int_0^{\frac{\pi}{2}} g(\theta,\phi) \sin(\theta)\,\mathrm d\theta\,\mathrm d\phi\;. $$

Now if you equate $r$ with $\sin\theta$ to associate points vertically, the bounds come out right, and you need $f\,\mathrm dr=g\,\mathrm d\theta$, so

$$g(\theta,\phi)=f(r,\phi)\frac{\mathrm d r}{\mathrm d\theta}=f(\sin\theta,\phi)\cos\theta\;.$$

This is directly related to this answer.

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An "equal-area projection" is what you're looking for. The spherical cap $\theta > \alpha$ has area $\int_0^{2\pi} \int_\alpha^{\pi/2} \sin(\theta) \ d\theta \ d\phi = 2 \pi \cos \alpha$. The disk $r < \beta$ has area $\pi \beta^2$. The cap has twice the area of the disk if $\cos \alpha = \beta^2$. Thus you want to map the point on the disk with polar coordinates $(r,\phi)$ to the point in the upper hemisphere with spherical coordinates $(\theta = \arccos(r^2),\phi)$.

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