We have that $$\int_{\Sigma}2\sqrt{1-x^2-y^2}dA=\iint_D2dxdy$$ where $\Sigma (x,y)=(x,y,-\sqrt{1-x^2-y^2})$. What will $D$ be?
Will it be the set of all $(x,y)$ such that the square root is defined? It must be $1-x^2-y^2\geq 0 \Rightarrow y^2\leq 1-x^2 \Rightarrow -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$. So that the square root $\sqrt{1-x^2}$ is defined, it must be $1-x^2\geq 0 \Rightarrow x^2\leq 1\Rightarrow -1\leq x\leq 1$.
So, we get that $D=\{(x,y)\mid -1\leq x\leq 1, -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}\}$.
What tools, ways would you propose for getting the closed form of this integral?
$$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$
EDIT: It took a while since I made this post. I'll give a little bounty for the solver of the problem, 500 points bounty.
Supplementary question:
Ca...
Consider distinct integer polynomials of distinct degree. Also they are all univariate polynomials of the variable $x$.
Let $*$ denote composition and $*$ is the operation for the monoid.
Consider when such polynomials $a,b,c,d$ forms an abelian monoid.
Many questions arise naturally.
For insta...
