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5 votes
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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 ...
lonza leggiera's user avatar
4 votes
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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 ...
LPZ's user avatar
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4 votes
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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 ...
Jean Marie's user avatar
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4 votes

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 ...
Ted Shifrin's user avatar
3 votes
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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 ...
Ted Shifrin's user avatar
3 votes
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Integral for the surface area of a half n-sphere

Yes, fortunately, things work out nicely. The integral in question is exactly $\int_{\Sigma}1\,d\sigma$, where $\Sigma$ is the upper hemisphere of $S^{n-1}$ in $\Bbb{R}^n$. Hence, this evaluates to $\...
peek-a-boo's user avatar
  • 57.4k
3 votes

Computing $\iint_S (y+z)\mathrm{d}y\mathrm{d}z+(z+x)\mathrm{d}z\mathrm{d}x+(x+y)\mathrm{d}x\mathrm{d}y$

The problem is greatly simplified by noting that the vector field $\mathbf{F} =(y+z)\hat{\mathbf{x}} + (x+z)\hat{\mathbf{y}} + (x+y)\hat{\mathbf{z}}$ can also be written as $\mathbf{F} = (x+y+z)(\hat{\...
eyeballfrog's user avatar
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3 votes

Problem with bounds on surface integral.

Just a small remark if permitted, since I was puzzled by LPZ's elegant way of solving this. The correct calculation of $d\omega$ is $$\tag{1} d\omega=3y^2\,dy\wedge dx+x\,dz\wedge dx-2x\,dx\wedge dz=-...
Kurt G.'s user avatar
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2 votes
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Seeking help in understanding the proof of the mean value property for harmonic functions

Note that $\partial\Omega$ is the union of two connected components. The first is the unit sphere $S$, and the second is the sphere $\epsilon S$ of radius $\epsilon$ and centered at the origin. So, ...
peek-a-boo's user avatar
  • 57.4k
2 votes

Calculating the mass of some object in space

Let $(x,y,z)=(u,2v,3w)$ to integrate the mass as $$ M=\iiint_B \mu(x,y,z)dxdydz = 6\iiint_{u^2+v^2+w^2<1}e^{\sqrt{u^2+v^2+w^2}}dudvdw\\ =6\int_0^{2\pi}\int_0^\pi \int_0^1 e^{r}r^2\sin\theta dr d\...
Ace's user avatar
  • 1,173
2 votes
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Surface Integral Over Part of Plane

A parametrization of the surface is $$ [-1,1]\times [0,1]\ni (y,z)\mapsto \boldsymbol{\phi}(y,z)=\begin{pmatrix}-2y-3z\\y\\z\end{pmatrix}\,. $$ I have shown in this answer that $$ \boldsymbol{\phi}_y\...
Kurt G.'s user avatar
  • 14.9k
2 votes
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Area of a surface by means of a double integral

Actually, the previous answer contained a mistake; the integral is not elementary as I'd initially suggested. Computing the area requires elliptic integrals, specifically the incomplete flavors of the ...
user170231's user avatar
  • 19.9k
2 votes

Find the flux of the curl across a surface

I found the error I missed. In the former calculation, I set $x = u \cos v$, $y = (u/2) \sin v$, and $z = u^2$, ($0 \leq u \leq 1$, $0 \leq v \leq 2\pi$). By applying Jacobian metrix, I get $dx\,dy = (...
이심심's user avatar
2 votes
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Gradient and differential surface element

On p. 107 in line 2460 I do not see your formula but instead Stokes' theorem for $\vec{u}=\vec{A}\times\vec{v}$ where $\vec{A}$ must be constant followed by a well-known expression for $\nabla\times(\...
Kurt G.'s user avatar
  • 14.9k
2 votes

Evaluate integral of $\langle xz^2,x^2y-z^3,2xy+y^2z\rangle$ with divergence theorem

The integral in your post is over a closed volume bounded by half a sphere of radius $a$. By divergence theorem, $$\begin{aligned} \oint_{S}\langle xz^2,x^2y-z^3,2xy+y^2z\rangle\cdot\mathrm d\mathbf S&...
Conreu's user avatar
  • 2,294
2 votes
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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 ...
PrincessEev's user avatar
  • 44.5k
1 vote
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Surface integral below $z=9-x^2-y^2$ but above $z=5$

Parameterize the paraboloid with $$\vec s_P = (u \cos v, u \sin v, 9-u^2), \quad u\in[0,2] \land v \in[0,2\pi]$$ and normal vector $$\vec n_P = \frac{\partial \vec s_P}{\partial u} \times \frac{\...
user170231's user avatar
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1 vote
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What is wrong in this derivation of the time derivative of a flux?

I figured out the problem with my derivation! This will benefit from using index notation. We begin with a slight formalization of the problem to be able to apply ideas from continuum mechanics. ...
eweiner's user avatar
  • 21
1 vote
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Find the main part of $\frac{1}{4\pi R^2}\iint_{\Sigma}u(x,y,z)dS-u(u_0,y_0,z_0)$.

Use the linearity of integral, compute the first integrals and bound above the last one. By symmetry $$\iint_{\Sigma} (x-x_0) dS = \iint_{\Sigma} (y-y_0) dS = \iint_{\Sigma} (z-z_0) dS = 0$$ and \...
Christophe Leuridan's user avatar
1 vote

Is there any way I can calculate this double integral?

$$S = \int_{-8}^{8} \int_{-\frac{21}{2}}^{21} \left(1 + \frac{1}{8} x^2 + \frac{64}{3969} y^2\right)^{\frac{1}{2}} dy \, dx= \int_{-8}^{8} \int_{-\frac{21}{2}}^{21} \left(1 + \left(\frac{1}{2\sqrt{2}} ...
Michael Morad's user avatar
1 vote

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 ...
Kurt G.'s user avatar
  • 14.9k
1 vote
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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. ...
Kurt G.'s user avatar
  • 14.9k
1 vote
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Evaluating a surface integral of a cylinder cut by a paraboloid.

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^...
user170231's user avatar
  • 19.9k
1 vote
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Evaluating line integral with Stokes' Theorem

I think the end result is correct but find the way it is derived a bit confusing.
Kurt G.'s user avatar
  • 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 ...
Ninad Munshi's user avatar
  • 35.2k
1 vote
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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, ...
Ninad Munshi's user avatar
  • 35.2k
1 vote
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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' ...
Kurt G.'s user avatar
  • 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}....
Robert Z's user avatar
  • 146k

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