2 votes

Derivation of the formula for parametrized surface area element

So, your surface $S$ is presumably parameterized by a function $\sigma(u,v)$ for $(u,v) \in R$. Note that, in particular, the values of $f$ on $S$ are given by $f(\sigma(u,v))$, akin to line integrals ...
PrincessEev's user avatar
  • 44.1k
1 vote

Calculating Electric Flux Through a Closed Surface

Three issues: $S$ should be split up into $3$ components, not $2$ (although ultimately it's as you say, the planar faces of $S$ will contribute nothing) The normal vector to $S_1$ is incorrect: $$\...
user170231's user avatar
  • 19.6k
1 vote

The surface integral $\iint_S (z^2 + y^2 + x^2) \, dS$ over the cube $S$?

By symmetry of the integrand, $$\iint_S(x^2+y^2+z^2)\,\mathrm dS=24\int_0^a\int_0^a(x^2+y^2+a^2)\, \mathrm dx\,\mathrm dy=24\int_0^a\left(\frac{4}{3}a^3+ay^2\right)\mathrm dy=40a^2$$
Conreu's user avatar
  • 1,831
1 vote

Restriction of a compactly supported function on a bounded domain in a surface

I am sure is no. In fact we have trace theorem $[f\mapsto f|_S]:H^1(\mathbb R^n)\to H^{1/2}(S)$ whenever $S$ is a bounded smooth surface. Here $H^1$ is $W^{1,2}$ Sobolev space, $H^{1/2}(\mathbb R^m)$ ...
Liding Yao's user avatar
  • 1,821

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