Let f be a continuous real-valued function on R3. Suppose that for every sphere S of radius 1, the integral of f(x, y, z) over the surface of S equals 0. Must f(x, y, z) = 0 for all points (x, y, z)?
Maybe my math is not the best, but wouldn't a simple function like f = cos(pi * z) work?
Say the sphere is centred at the origin. If I make this an integral in z, the area element is dA = 2*pi dz (the curvature of the sphere exactly compensates for the chord becoming smaller as we go from z=0 to z=1, so the area element is dA = 2*pi*cos(theta) / cos(theta), where cos(theta)=sqrt(1-z^2) and the first part is the radius of the circle defined by intersecting the sphere with a plane of constant z, and the second part is to take into account the slantedness of the area element relative to z)
Say the sphere center has coordinate z=alpha (the other coordinates don't matter). Now the integral becomes: integral from -1+alpha to 1+alpha of 2*pi*f(z). or, equivalently: integral from -1 to 1 of 2*pi*f(z - alpha).
And now it's clear that whatever the center of the sphere, this integral is zero for our function f = cos(pi*z) because of its periodicity.
This answer seems suspiciously simple, so I'm ready to be corrected here
It's a question from a math competition, so if you'd answered it then you would get something
Dont trust me I havent taken math for a few years. Any non-zero f with div f = 0 should satisfy the condition, by divergence theorem. So no.
I think this wouldn't work since it looks like f is from R3 -> R. Divergence would only apply if the range of f was also in R3 afaik. I havent been able to think up a counterexample personally.
I'd say it is. Here is my intuition at a proof.
Imagine that we shift the sphere continuously along any direction.
During this shift, any point for which f increases must be matched with another point for which f decreases commensurably.
Now take the circle of points where the associated spoke is perpendicular to the direction of travel.
By adding an infinitesimal scalar product of the vector in the direction of travel to the spoke, in negative and positive terms, two more circles are produced, infinitesimally shiftsd from each the original circle in the direction of travel
After the shift is applied, one circle will switch positions to the position of the next, and the other circle will shift positions still.
Now there are two possibilities for the integral not to change. Either the function is infinitely symmetrical, which is only true of constant functions, in which case the function must be null.
Or, the change of the values of the two circles must be constant, which means that the derivative of the integral of a circle over any direction must be zero. By repeating the same process with spokes, you come to find that either there is infinite symmetry or the derivative of the integral for a point is constant, in which case the function is constant.
In conclusion, the function must be constant. The only constant function for which the surface integral of a sphere is zero is the null function, therefore f is the null function.
I'm sure I'm being unrigorous somewhere or missing something, didn't take multivariate calculus yet. But this is my intuition at an answer :)
A single circle can change as the sphere moves even if the integral over the whole thing doesn't, though.
If it helps, the final answer is no, f doesn't have to be 0.
Hmm, a single circle isn't enough, though. It has to be the case for every single circle. I'll give it a rethink, though.
Not every circle independently, though, the parts as a whole.
this sounds like a conservative vector field problem so im gonna say no ?
