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How to prove $\oint_{A}\mathbf{E}\cdot d\mathbf{A}=\frac{q}{\epsilon_{0}}$ mathematically and rigorously?

You can obtain a rigorous proof by first obtaining the result for $\ A=RS^2 ,$ the surface of a sphere of radius $R\$ centred on the origin, and then using Gauss's divergence theorem to extend it to ...
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Problem with bounds on surface integral.

A correct candidate of $M$ would be the surface parametrised by: $$(r,\phi)\in [0,1)\times[0,2\pi)\to\begin{pmatrix} r\cos\phi \\ r\sin\phi \\ r^2\cos\phi\sin\phi \end{pmatrix}$$ (Bruno B only gave ...
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Finding a volume of region $0 \geq \cos(\theta_1 - \theta_2) + \cos(\theta_1 - \theta_3) + \cos(\theta_2 - \theta_3)$

First of all, three 3D representations of surface $(S)$ with implicit equation : $$\cos(x_1 - x_2) + \cos(x_1 - x_3) + \cos(x_2 - x_3)=0 \tag{1}$$ the first one in $(0, 2 \pi)^3$ with a main ...
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What exactly does "outward normal vector" means when talking about an orientation?

I agree with @Kurt that you should be looking at polar/spherical coordinates and not stereographic projection. This is true in general, but particularly because of the form of your integrand. But I'm ...
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Scalar integrals in higher dimensions

Ultimately, your question boils down to two ways of calculating the $k$-dimensional volume of a $k$-dimensional parallelepiped in $\Bbb R^n$. The more standard one is the Gram determinant and it ...
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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 ...
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Calculating surface of a solid bounded by cone and plane

The first transformation I would make is to replace $y-3$ with $y$ which turns the equations into $$x^2=z^2+y^2\,,\quad 2x+y=9\,.$$ Then let's rename the variables to have compatibility with this ...
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confusion about surface integral using inverse of parameterized surface

Presumably you are more interested in the integral $$\int_S\,dS$$ which is the surface area rather than rather than $\iint_S(x^2+y^2)\,dS$ but this does not matter for the resolution of the riddle. ...
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Continuing with your approach, we let $x(y,z) = \sqrt{\dfrac y2-y^2}$ so that $$\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} = \frac1{2\sqrt{2y-4y^... • 19.9k 1 vote Accepted Evaluating line integral with Stokes' Theorem I think the end result is correct but find the way it is derived a bit confusing. • 14.9k 1 vote \int_D \left(\frac{1}{\Bigr((x+1)^2+(y+1)^2+(z+2)^2 \Bigr)^{10}} - \frac{1}{\Bigr(1^2+1^2+2^2 \Bigr)^{10}} \right) dS is positive or negative? Consider translating to a coordinate system centered at (-1,-1,-2) and rotated such that the new z unit vector is parallel to the old (1,0,2) vector and y remains unrotated. This gives us the ... • 35.2k 1 vote Accepted How to Figure out the bounds of a Surface for Surface Integral Just for completeness' sake, from the comments we clarified that the surface of integration is the portion of that plane located in the first octant only. All of your work up to this point is correct, ... • 35.2k 1 vote Accepted To evaluate \int \int_{S} \hat{n}×(\bar{a} × \bar{r}) dS According to (1) in this answer the surface integral equals$$ \int_V\nabla\times(\mathbf{a}\times\mathbf{r})\,dV\,. $$The proof is simple and essentially a componentwise application of Gauss' ... • 14.9k 1 vote How to calculate \int_{x^2+y^2=R^2\\ 0\leq z\leq R}(x^2+z^2)dS? The correct result is yours:$$\int_{0}^{2\pi}\left(\int_0^RR(R^2\cos^2\theta+z^2)dz\right)d\theta = \int_{0}^{2\pi}(R^4\cos^2\theta+\frac{R^4}{3})d\theta=\pi R^4+\frac{2\pi R^4}{3}=\frac{5\pi R^4}{3}....
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