Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer

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 $(\theta = \arccos(r^2),\phi)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.