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7 votes
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Can I use divergence theorem right after using Stokes' theorem?

Short Answer Arguably "yes, you can", but certainly not in a useful way. Any application of the combination of the Divergence Theorem and (Kelvin–)Stokes' Theorem results in $0$ for a ...
Mark S.'s user avatar
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6 votes
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Flux calculation - what did I do wrong?

The issue lies in the use of Gauss' theorem (the divergence theorem): it assumes a closed surface. The surface given in cylindrical coordinates by $$ \left\{ \begin{align*} r &= 5 \\ z &\in [0,...
PrincessEev's user avatar
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4 votes

Divergence Theorem when Surface isn't closed

At the bottom of the cylinder, $z=0$. Since $F_z(x,y,0)=0$ we find $$\begin{align} \int_{S_1} \vec F \cdot \hat n\, dS&=\int_{S_1}\vec F(x,y,0)\cdot \hat z\,dS\\\\ &=0 \end{align}$$ and ...
Mark Viola's user avatar
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4 votes
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Find a closed surface for which $\iint_S \textbf{F} \cdot d\vec{S}$ is negative.

As you noted, your approach to the solution of the problem is entirely correct as long as the divergence theorem holds for the domain $D$ you consider, so your question is basically equivalent to ...
Daniele Tampieri's user avatar
4 votes
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Does null Dirichlet condition implies null Neumann condition?

Short answer: No. Counterexample in 1D: Take $\Omega=(0,1)$ and $u(x)=x(1-x)$. This function is in $H^1_0(\Omega)$ but $u’(0)=1$. As indicated in the comments, in general, $\nabla u\cdot n|_{\partial\...
timur's user avatar
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4 votes
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Using the divergence theorem to prove that $\frac{1}{|B_R(0)|} \int_{B_R(0)} M \textbf{y} . \textbf{y} dy = \frac{R^2}{ n + 2} \text{trace}(M)$

Some few notations/properties I use here: $\partial B_1(0)$ denote the unit $n-$sphere of $\mathbb R^n$. For $R>0$, I denote $B_R(0):=\{\sigma _r:=r\sigma \mid r\in (0,R), \sigma \in \partial B_1(...
Surb's user avatar
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3 votes
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How to evaluate the double integral over a non-closed surface?

Notice that the boundary of the surface is the curve $\{(x,y,z):x^2+y^2=1\cap z=0\}$ By Stokes' theorem if two surfaces share the same boundary then the integral of the curl on both surfaces will be ...
Ninad Munshi's user avatar
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3 votes

Measure of the unit ball in $\mathbb{R}^n$

Let $F(x_1, \dots, x_n)= (x_1, \dots, x_n)$. From the Divergence Theorem we have $$\int_{B_1(0)} \nabla \cdot \mathbf{F} \space dV = \int_{\partial B_1(0)} \mathbf{F} \cdot \mathbf{\nu} \space dS, $$ ...
Nicola's user avatar
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3 votes

An integral involving determinant.

We write $$ A=\left( \begin{array} [c]{c} A^{1}\\ \vdots\\ A^{n} \end{array} \right) \text{,} $$ and $M=\left( M^{1},\ldots,M^{n}\right) $, where $M^{i}$ is the $\left( n,i\right) $-cofactor of $\...
Shibo Liu's user avatar
3 votes

Application of the fundamental theorem of calculus in $n$ dimensions: Stokes', divergence or gradient theorem?

What you want is known as the (Generalized) Stokes Theorem. It is stated in the language of differential forms. A good introductory treatment can be found in Spivak's little book Calculus on Manifolds....
Lee Mosher's user avatar
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3 votes
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Calculate the integral with divergence theorem

As already pointed out by Math Lover, the left part is $$\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}(3z+1)\, dV=\iiint_{y\geq 0,x^2+y^2+z^2\leq 4}1\, dV=\frac{1}{2}\left(\frac{4 \pi}{3}\cdot 2^3\right)=\frac{...
Robert Z's user avatar
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3 votes
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Divergence of curl is zero (coordinate free approach)

The problem is, as @Mark S. commented a while ago, very badly written. I am going to reiterate the suggestions of that comment as an answer, because this is an important technique. Yes, surely the ...
Ted Shifrin's user avatar
3 votes
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Prove by the divergence theorem that $\int_{\partial \Omega} u\frac{\partial u}{\partial n}dS \ge 0$. For what harmonic function does equality hold?

Since $\Delta u = 0$ on $\Omega$, therefore by Green's identity, we have $$ \int_{\partial\Omega}u\dfrac{\partial u}{\partial\mathbf{n}}dS = \int_{\Omega} u\Delta udx + \int_{\Omega} |\nabla u|^2dx =...
Bakkune's user avatar
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3 votes
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Proving Hadamard's Variational Formula $\dot{\lambda} = -\int_{\partial U(\tau)}\left|\frac{\partial w}{\partial v}\right|^2v\cdot \nu dS$

For 1): Multiplying the equation $-\Delta w = \lambda w$ by $w$ then integrating over $U(\tau)$ gives $$\lambda(\tau) \int_{U(\tau)} w^2 \, dx = - \int_{U(\tau)}w\Delta w \, dx. $$ Then, by assumption,...
JackT's user avatar
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3 votes
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Verifying the divergence theorem over B

The divergence theorem in spherical coordinates gives $$\begin{align} \iiint_D \text{div}(\mathbf{F})dxdydz&=\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\pi/4}\int_{\rho=0}^1 3\rho^2\sin(\phi) d\rho d\...
Robert Z's user avatar
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3 votes
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Is the Gauss Divergence theorem applicable when the Divergence of a vector field is a dirac delta function?

For any $\phi\in C_C^\infty$, the distribution $\nabla \cdot \vec E$, for $\vec E=\frac{\vec r}{r^3}$ satisfies $$\begin{align} \langle \nabla \cdot \vec E,\phi\rangle &=-\sum_{i=1}^3 \langle E_i, ...
Mark Viola's user avatar
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3 votes
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Divergence theorem in 3D

The limits of integration for $D$ are wrong. The LHS should be $$\begin{align}\iiint_{D} 2dxdydz &=\int_{y=-1}^{1}(\int_{x=-2\sqrt{1-y^2}}^0(\int_{z=0}^{-x}2dz)dx)dy \\&=-\int_{y=-1}^{1}(\int_{...
Robert Z's user avatar
  • 146k
3 votes

Divergence theorem when $\nabla \cdot \vec{F} = 0$

More generally we can evaluate the flux of $\mathbf{F}$ through any closed surface $S$ oriented outwards which does not pass through the origin: $$\iint_S \mathbf{F}\cdot d\mathbf{S}=\begin{cases} 0 &...
Robert Z's user avatar
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2 votes
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Evaluating $\iint_V x^4+y^4+z^4$ with the divergence theorem

The normal vector $\mathrm{n}$ at at point $(x,y,z)$ on a sphere is given by $\frac{(x,y,z)}{\sqrt{x^2+y^2+z^2}}$, hence it you want to write $x^4+y^4+z^4$ as the dot product $\mathrm{F}\cdot\mathrm{n}...
Jack D'Aurizio's user avatar
2 votes
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Calculating flux from an inwards orientation

Flux integrals are defined by calculating how much of the vector $\vec{F}(x,y,z)$ is in the direction of $n$. The divergence theorem, on the other hand, is proven using flux integrals, which are ...
Franklin V's user avatar
2 votes
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Flux of a vector field across an ellipsoid.

No, the vector field is not defined at the origin, so it's divergence is not defined at the origin (some might say its infinite); i.e you only have $\text{div}(F)=0$ on $\Bbb{R}^3\setminus \{0\} $. So,...
peek-a-boo's user avatar
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2 votes
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Expression for potential vector of a central field

Setting $\alpha = 1$, here is a vector potential for $\mathbf F$. It is defined and smooth off the $z$-axis. Set $$\mathbf A = \frac1{(x^2+y^2)(x^2+y^2+z^2)^{1/2}}\big(yz,-xz,0\big).$$ The answer is ...
Ted Shifrin's user avatar
2 votes

Flux of Curl with given function

Since it is always true that $\nabla \cdot (\nabla \times F)=0$, Gauss' Law tells us that the total flux of a curl through any closed surface always vanishes. A closed surface is formed from ... the ...
WW1's user avatar
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2 votes
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Error In Divergence Theorem

The divergence theorem is about a three-dimensional body $B$, its boundary surface $\partial B$, and a vector field ${\bf v}$ with domain $\Omega\supset B$. The theorem says that $$\int_{\partial B}{\...
Christian Blatter's user avatar
2 votes

Finding $\int_S z\,dS$ where $S=\left\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\right\}$

With the same trick, you can use Stokes' theorem in one of two ways. First is directly: $$\nabla \times H = (0,0,a) \to H = (0,ax,0)$$ which gives us the line integral $$\int\limits_{x^2+y^2=a^2\:\cap\...
Ninad Munshi's user avatar
  • 35.2k
2 votes

Finding $\int_S z\,dS$ where $S=\left\{ \big(x,y,z\big):x^2+y^2+z^2=a^2,z\ge 0, a>0\right\}$

But you can apply the Gauss Divergence Theorem if you add to the surface the bottom disk $z=0$, $x^2+y^2\le a^2$, oriented downward. You now have a closed surface and so the flux across that closed ...
Ted Shifrin's user avatar
2 votes
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Find the flux of the vector field $F(x,y,z)=(x+y^2, -x, z-xy)$

Your integral order is wrong. $\displaystyle \int _{0} ^{1} \int _{0} ^{2\pi} \int _{0} ^{\pi} \frac{2}{\sqrt3} \rho^2 \sin \theta \, d \phi \, d\theta \, d\rho = 0$ Based on notation you have used, $\...
Math Lover's user avatar
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2 votes
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Verify Divergence Theorem for $F(x,y,z)=(xz,0,0)$ on a solid bounded by the paraboloid $z=x^2+y^2$ and the plane $z=1$

Please note your second integral is set up incorrectly. Otherwise you working is fine. So it should be $\, \displaystyle \int_{0}^{2 \pi}\int_{0}^{1} \int_{r^2}^{1} \, h \, r \, dh \, dr \, d\theta$ ...
Math Lover's user avatar
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2 votes
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Divergence theorem is not working for this example?

The theorem works: as @Ian wrote in his comment, the flux across the lower surface must be negative. The right computation is $$\int \int_{S_3}(x+y,y,z^2) \cdot n dS = \int_0^{2 \pi} \int_0^1(*,*,1) \...
VoB's user avatar
  • 1,613
2 votes

Triple volume integral of divergence

Nothing forbids you from having a negative flux because it’s just an integral. Through that surface you have portions where the field is outgoing, hence it has the same direction of the external local ...
Gyro's user avatar
  • 105

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