Recent content by Djokara

That worked but solution is messy.

Yeah, this is problem from ED R is radius and h is height of cone.

Any ideas how to solve this \int_0^h {\frac{1}{\sqrt{1+(r({\frac{1}{z}}-{\frac{1}{h}}))^2}}}\,dz Don't have an idea from where to begin

Well you could use Laplace transformation. It's good for this type of equations.

Well what you saying is true. But I think they meant this is the rule if you have some vector with this argument, because in the rest of proof there isn't this mistake.

Regarding to surface integral equal to zero as ##|\vec r-\vec r'|\rightarrow\;\infty##: \frac {1}{4\pi}\nabla\int_{v'}\nabla'\cdot\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)d\tau'=\int_{s'}\left(\frac{\vec F(\vec r')}{|\vec r-\vec r'|}\right)\cdot d\vec s' where this term goes to...

Lagrangian is defined that way. Ende.

I'll try to answer your questions, but have in mind that English is not my native language. 1)Volume integral doesn't have to go to zero, it could be but doesn't have to. It could be just some constant, and derivative of a const is 0. 2) I'm not completely sure about this one, but I would...