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2

Yes, $S$ can be a closed surface. In that case $$\iint\limits_S (\nabla \times {\mathbf F}) \cdot {\mathbf n} \, dS = 0$$ because we consider the boundary of $S$ to be empty.


0

Is the $\times$ means the cross product? In usual vector operation, we have two kinds of operations, (inner, outer(or so called cross) products). For basis(kinds of unit vector) $\mathbb{e}_i$, the scalar product just changes its scale of vector. but its vector product(i mean inner, cross product) changes its direction(inner products gives the scalar, and ...


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I refer you (and new readers) to Introduction to Electrodynamics by D.J. Griffiths, third edition, page 39.


2

It might be helpful if you first introduce a new symbol to refer to one of the vector cross-products as a whole. E.g., let's define $(a\times b)=:x$. Using the cyclic property of the scalar triple product, we equate the scalar quadruple product to the dot-product of one of the vectors with the vector triple product of the other three: $$\begin{align} ...


6

I would cite the beautiful method from W.H.Beyer to find the center and radius of the sphere $(x-a)^2+(y-b^2)+(y-c)^2=R^2$


1

For the ones calculating by hand, the error-prone numbers are: the center of the sphere is at $p=(24/19,-16/19,4/19)$ the squared radius is $r^2=4230/361$. Here is my small GNU Maxima script display2d : false; /* * purpose: * given points x_k, * calculate circumsphere center p and squared radius r^2 **/ my_A(d,k,x) := block( [res,i,j], ...


1

Another method is to start with the equation of the sphere: $$(x-u)^2+(y-v)^2+(z-w)^2=r^2$$ $(u,v,w)$ are the coordinates of the center of the sphere and $r$ is the radius. Plug into the given points $p_1, p_2, p_3, p_4$ for $x,y,z$ in the equation and you get four equations with variables $u,v,w,r$. But these equations contain quadratic terms. Subtract ...


1

Hint Using a totally different approach, you can also look at your problem using what JJacquelin (participant to MSE) proposed in his book http://fr.scribd.com/doc/14819165/Regressions-coniques-quadriques-circulaire-spherique Pages 17 and 18 give the full approach for a spherical regression. It is quite simple and reduces to a linear system of four ...


1

The centroid of a triangle is usually not at the same distance from the vertices. In the plane the center of the outer circle (= the intersection of the normals bisecting the sides) has this property. But, because this is in 3D you might as well find the point of intersection of the planes the have $\vec{p_1p_2}$, $\vec{p_2p_3}$ and $\vec{p_3p_4}$ as their ...


2

A lie algebra $\frak g$ is a vector space equipped a bilinear operation $[-,-]$ satisfying $[x,x]=0$ for all $x\in\frak g$ and the Jacobi identity $[x,[y,z]]+[z,[x,y]]+[y,[z,x]]$. The space $(\Bbb R^3,\times)$ is a lie algebra. One should look up the classification of three-dimensional real lie algebras (which is fairly easy and textbook, but I don't have ...


0

For each fixed $x\in\mathbb{R}^3$ you get the skew symmetric matrix $[x]_\times$. This matrix does correspond to a bilinear form. It corresponds to the bilinear form $$H:\mathbb{R}^3\times\mathbb{R}^3\to\mathbb{R} \ , \ (a,b)\mapsto a^T[x]_\times b=a^T\cdot(x\times b)$$I think you may be wondering why the matrix $[x]_\times$ is skew-symmetric? By definition ...



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