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!