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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 \... 2 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^... 2 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 ... 1 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} $$for small step sizes \varepsilon where \vec{X} is of course postition and \vec{V} is of course velocity. It would probably be a good idea to change the value of \varepsilon at each ... 1 You will do an integral from the point P to the same point. This is true in certain situations. The Fundamental Theorem For Line Integrals says that if your vector function \mathbf F is the gradient of a scalar function f, then you can replace these sorts of line integrals (like \int_C \mathbf F(\mathbf C(t))\cdot \mathbf C'(t)\,\mathrm dt) with a ... 1 Draw it out, pick a point, and move along the boundary of the enclosed region counter clockwise. For example if we start at the point p = (1,-1) then our integral over C_1 is$$ \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)$and move to$...