@MartinR The duplicate target is a duplicate itself and I'm not sure if this can be closed. However, I have flagged the post and it's even marked helpful, not yet closed now.
This integral diverges, Yes you're right in saying that the indefinite integration is equal to $\tan x$ but when you apply your limits, it has discontinuity which means it probably does not converge.
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.
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.
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...
Reading this paper by Victor H Moll, I don't understand how does one prove Lemma 6.1. Or is it a standard thing which I should know? It states that $A(z)=\int_0^\infty e^{-zt}B(xt) \mathrm dt$. Then $[z^n]A=n!\times [z^n]B$ where $[z^n]R$ is the coefficient of $z$ in the power series expansion of function $R(z)$.
You should not answer low quality PSQ questions. However, I strongly disagree with that high rep. user suggestion. You are welcomed if you follow the rules. As per my experience with answers flagging for moderator for EOQS, I have not any good experience. You should rather post the links in the Cured Chat room. I have far better experience in the Chatroom.