Recent content by sszabo

Agree. Agree. Of course. Certainly a similar calculation gives \int\int_S n_2\,dS=\int\int_S \left\langle \mathbf{j},\hat{\mathbf{n}} \right\rangle \,dS=\int_V div\, \mathbf{j}\, dV=0, and \int\int_S n_3\,dS=\int\int_S \left\langle...

Sorry, I absolutely don't understand what you think. Please, give more details, for example point out the wrong step in the calculations. Thanks.

I see. Yes, indeed, the answer -intuitively- is the 0 vector. It means \int\int_S \hat{\mathbf{n}}\,dS=\left( \int\int_S n_1\,dS,\int\int_S n_2\,dS,\int\int_S n_3\,dS\right), where \hat{\mathbf{n}}=(n_1,n_2,n_3). Now n_1=\left\langle \mathbf{i},\hat{\mathbf{n}} \right\rangle...

Right. However in the l.h.s. we have a vector \hat{\mathbf{n}} and not an inner product which would be a scalar. I can integrate only vector-variable scalar-valued functions on a surface and not a vector-valued function. So what does it mean?

In h.m. schey, div grad curl and all that, II-25: Use the divergence theorem to show that \int\int_S \hat{\mathbf{n}}\,dS=0, where S is a closed surface and \hat{\mathbf{n}} the unit vector normal to the surface S. How should I understand the l.h.s. ? Coordinatewise? The r.h.s. is not...

By definition z^{a}=\exp (a\ln z). Denote u:=\ln z. So you have to solve \exp(a u)=1. Hence au=i2k\pi, where k is arbitrary integer. It gives u=\frac{i2k\pi}{a}. Now we solve \ln z=\frac{i2k\pi}{a}=\frac{i2k\pi\bar{a}}{\left|a\right|^2}= \frac{2k\pi...

At this moment there is no Sigma icon. What happened?