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The divergence theorem is in general inapplicable but if $\Omega \subset \mathbb{R}^3$ such that $\operatorname{sing supp}\vec{F} \cap \partial\Omega = \emptyset$ then $\iint \vec{F}\cdot\vec{n}\,dS$ is defined and one can even define $\vec{F} \chi_\Omega$ as a distribution. Take $\rho \in C^\infty_c(\mathbb{R}^n)$ with $\rho\equiv 1$ on a neighborhood of $\...


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For typing convenience, define the vector $$w = A(x\circ x) - b$$ and use a colon to denote the trace/Frobenius product, i.e. $$\eqalign{ A:B &= {\rm Tr}(AB^T) \;\;\;\,=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \\ A:(B\circ C) &= (A\circ B):C \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}C_{ij} \\ A:A &= \big\|A\big\|^2_F \\ A:B &= B:A \;\;=\; B^...


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You are trying to calculate the numbers $u_{i,j}$, where $0 < i,j < n$, where $u_{i,j}$ is your approximation to the value of the function $u$ evaluated at the corresponding point $(\frac{i}{n},\frac{j}{n})$. Note that you do not need to calculate the values $u_{0,j}$, $u_{i,n}$, etc because you already know they are zero. Because these are the numbers ...


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You can do some really easy iterations given a velocity field. The most basic iteration would be something like $$ \vec{X}_{k+1} = \vec{X}_{k} + \varepsilon \vec{V}_{k} $$ for small step sizes $\varepsilon$ where $\vec{X}$ is of course postition and $\vec{V}$ is of course velocity. It would probably be a good idea to change the value of $\varepsilon$ at each ...


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You will do an integral from the point $P$ to the same point. This is true in certain situations. The Fundamental Theorem For Line Integrals says that if your vector function $\mathbf F$ is the gradient of a scalar function $f$, then you can replace these sorts of line integrals (like $\int_C \mathbf F(\mathbf C(t))\cdot \mathbf C'(t)\,\mathrm dt$) with a ...


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Draw it out, pick a point, and move along the boundary of the enclosed region counter clockwise. For example if we start at the point $p = (1,-1)$ then our integral over $C_1$ is $$ \int_{-1}^{1} t^2 \text{d}t =\frac{2}{3}\text{ .}$$ Clearly this runs from $C_1(-1)=(1,-1)$ to $C_1(1)=(1,1).$ Going over the parabola, we start at $C_2(1) = (1,1)$ and move to $...


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