Okay i know nothing about math, i didn't read all comments, it was Abit above my head, but i think what OP meant in their question is something along the lines.. i don't even know is what I'm trying to explain here is correct in any way, but myself coming here for wondering pretty much the same t...
If you are interested to find the surface area of the strip $(h)$:
You can use integration:
However,
$V = {\pi}(R^2h-\frac{h^3}{3})$ is your formula
It's for hemisphere: The height$(h)$ is ⊥ to the base of the hemisphere.
Let: h = R
$V(R) = 2{\pi}R^3/3$
Which is of hemisphere.
not following EOQS, answer to low-quality question
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a map that continuous in each variable separately.
Assume that for any compact subset $K\subset\mathbb{R}^2$, $f(K)$ is compact.
Prove that $f$ is continuous.
this relativity theory question (moreover in a situation where it is absurd to apply it) is definitely off-topic on this site. Please help to close and delete it.
@XanderHenderson Both the question and the user account are deleted now (you did the former and perhaps the latter, thanks). – Interestingly, this led to some discrepancy of what my reputation is. Just in case that anybody is interested, I posted that as a question on the main Meta: meta.stackexchange.com/q/370245/196432.
@JoséCarlosSantos Could you explain why you voted to delete my answer here? It was targeted by vandals; stop looking at just the votes when determining deletion. There is nothing wrong with my answer. It doesn't surprise me that @OliverDiaz joined you in voting delete it, as he has made his bias against me clear; neither of you had any justification to vote to delete.
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a map that continuous in each variable separately.
Assume that for any compact subset $K\subset\mathbb{R}^2$, $f(K)$ is compact.
Prove that $f$ is continuous.
