Formula & theorem reference

Draw \mathbf{u} and \mathbf{v} from a common point O, with tips at U and V. The triangle OUV has two sides |\mathbf{u}|, |\mathbf{v}| meeting at angle \theta, and third side |\mathbf{u}-\mathbf{v}|, so the Law of Cosines gives

Expanding the left side component-by-component, \mathbf{u}-\mathbf{v}=\langle u_1-v_1,\,u_2-v_2,\,u_3-v_3\rangle, so

Both expressions equal |\mathbf{u}-\mathbf{v}|^2; setting them equal, canceling |\mathbf{u}|^2+|\mathbf{v}|^2 from both sides, and dividing by -2 leaves \sum_i u_iv_i = |\mathbf{u}||\mathbf{v}|\cos\theta, which is exactly \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3 on the left.

Direct computation confirms perpendicularity: expanding the dot product \mathbf{u}\cdot(\mathbf{u}\times\mathbf{v}) component-by-component gives u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1), and every term cancels against another, leaving 0; the same cancellation shows \mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})=0.

For the magnitude, Lagrange's identity states |\mathbf{u}\times\mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u}\cdot\mathbf{v})^2 (both sides expand to the same sum of squared component products — a direct algebraic check). Substituting \mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta:

Since 0\le\theta\le\pi makes \sin\theta\ge0, taking square roots gives |\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta.

Write \mathbf{u}=\langle a,b\rangle. The directional-derivative formula already gives D_{\mathbf{u}}f = f_x a + f_y b, and \langle f_x,f_y\rangle\cdot\langle a,b\rangle = f_x a + f_y b by the definition of the dot product — the two expressions are the same number, just written two different ways.

Near (u_0,v_0), the transformation's best linear approximation sends the edge vector (du,0) to \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right)du and the edge vector (0,dv) to \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right)dv — the two columns of the Jacobian matrix, scaled. The image of the small du\times dv rectangle is (to first order) the parallelogram spanned by those two vectors, and the area of a parallelogram spanned by (a,b) and (c,d) is |ad-bc|, exactly the absolute value of that 2\times2 determinant.

By the chain rule, \dfrac{d}{dt}f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). So \int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}f(\mathbf{r}(t))\, dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A), the last step being the ordinary single-variable Fundamental Theorem of Calculus applied to the scalar function t \mapsto f(\mathbf{r}(t)).

(\Rightarrow) If \mathbf{F} = \nabla f, then P=f_x, Q=f_y, R=f_z. Each component of \operatorname{curl}\mathbf{F} is a difference of mixed second partials of f — e.g. the \mathbf{k}-component is f_{yx}-f_{xy} — which vanishes by Clairaut's Theorem since f's second partials are continuous. The same argument kills the other two components, so \operatorname{curl}\mathbf{F}=\mathbf{0}.

(\Leftarrow) The converse needs simple connectedness in space: every closed curve in D must be shrinkable to a point without leaving D (a stronger condition than merely "no holes" — a region built around a solid tube, for instance, is not simply connected even though it has no cavity). Given \operatorname{curl}\mathbf{F}=\mathbf{0} on such a D, Stokes' Theorem (later in this chapter) lets any closed curve C in D be spanned by a surface S entirely inside D, giving \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 0 — zero circulation around every closed loop, which is the closed-loop form of path independence.

Sketch for a region R that is both x-simple and y-simple (the general case patches finitely many such regions together, with the shared interior edges canceling since each is traversed once in each direction). Write R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Then \iint_R -\frac{\partial P}{\partial y}\, dA = -\int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\, dx, by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly \oint_C P\,dx, splitting C into its bottom piece y=g_1(x) (traversed left to right) and top piece y=g_2(x) (traversed right to left, contributing the minus sign). The same argument with R described as y-simple gives \iint_R \frac{\partial Q}{\partial x}\, dA = \oint_C Q\,dy. Adding the two established identities gives the theorem.

Sketch for S a graph z=g(x,y) over a region R in the xy-plane, with C the boundary of R lifted onto the graph. Parametrize S by (x,y)\mapsto(x,y,g(x,y)), so d\mathbf{S} = (-g_x,-g_y,1)\,dA. Writing out \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} in terms of P, Q, R and their partials, then substituting z=g(x,y) throughout, the right side reduces — after applying the multivariable chain rule to track how P and Q change along the lifted curve — to exactly \oint_{\partial R} \big[P(x,y,g(x,y))+R(x,y,g(x,y))g_x\big]\,dx + \big[Q(x,y,g(x,y))+R(x,y,g(x,y))g_y\big]\,dy, which the circulation form of Green's Theorem converts into a double integral over R identical to the one obtained by substituting z=g(x,y) into the left-hand side's curl-flux integrand directly. A general piecewise-smooth S is patched together from finitely many such graphs, with shared interior edges canceling since each is traversed once in each direction.

Sketch for E a solid that is simultaneously x-simple, y-simple, and z-simple (a general E patches together finitely many such pieces, with shared interior faces canceling since each is traversed with opposite outward normals from the two adjoining pieces). It suffices to prove the three component identities separately and add them; take the z-component. Write E=\{(x,y,z): (x,y)\in R,\ h_1(x,y)\le z\le h_2(x,y)\}. Then \iiint_E \frac{\partial R}{\partial z}\, dV = \iint_R \big[R(x,y,h_2(x,y)) - R(x,y,h_1(x,y))\big]\, dA by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly the flux of R\,\mathbf{k} through S's top piece z=h_2(x,y) (outward normal has positive z-component) plus its bottom piece z=h_1(x,y) (outward normal has negative z-component, contributing the minus sign) — the side faces of E, if any, are vertical and contribute nothing to a \mathbf{k}-component flux. The same argument for the other two components, summed, gives the theorem.

Hold v=v_0 fixed and let u vary: \mathbf{r}(u,v_0) traces a curve lying entirely on the surface and passing through \mathbf{r}(u_0,v_0), with velocity vector \mathbf{r}_u(u_0,v_0) — so \mathbf{r}_u is tangent to the surface there. The same argument with u=u_0 fixed shows \mathbf{r}_v is tangent to the surface too, along the perpendicular grid direction. Two non-parallel vectors tangent to the surface at the same point span its entire tangent plane, and the cross product of two vectors spanning a plane is normal to that plane — so \mathbf{r}_u\times\mathbf{r}_v is a normal vector, and the tangent plane is the plane through \mathbf{r}(u_0,v_0) with that normal.

\det(A - \lambda I) = (a-\lambda)(d-\lambda) - b^2 = \lambda^2 - (a+d)\lambda + (ad - b^2), a quadratic in \lambda. Apply the quadratic formula with A=1, B=-(a+d), C=ad-b^2: \lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-b^2)}}{2}. Expanding (a+d)^2 - 4(ad-b^2) = a^2 + 2ad + d^2 - 4ad + 4b^2 = (a-d)^2 + 4b^2 gives the stated discriminant. It is a sum of two squares, so it is \geq 0, and the two roots are always real.

Every real number x satisfies -|x| \le x \le |x| directly from the definition of absolute value. Applying this to a and to b separately and adding the results gives -(|a|+|b|) \le a+b \le |a|+|b|. A number sandwiched between -k and k has absolute value at most k, so |a+b| \le |a|+|b|.

(\Rightarrow) If S is compact, it's bounded. Cover S by open balls of radius 1 centered at every point. Compactness extracts a finite subcover, and a finite union of radius-1 balls is bounded. S is also closed. Suppose \mathbf{p} were a limit point of S not in S. The balls B(\mathbf{x}, \tfrac12\|\mathbf{x}-\mathbf{p}\|) for \mathbf{x} \in S cover S, and no finite subcollection of them can stay away from \mathbf{p} -- contradicting compactness.

(\Leftarrow) If S is closed and bounded, boundedness places S inside some large closed box [-M,M]^n. A sequential argument, applying Bolzano–Weierstrass coordinate by coordinate, shows every sequence in [-M,M]^n has a convergent subsequence. Since S is closed, that limit lands back in S. Sequential compactness and compactness (the open-cover kind) coincide in \mathbb{R}^n, which completes the equivalence.

Apply the one-variable Bolzano–Weierstrass theorem coordinate by coordinate. Boundedness of (\mathbf{x}_k) means each coordinate sequence is bounded. So the first coordinate has a convergent subsequence. Pass to that subsequence, then extract a further subsequence along which the second coordinate also converges. Continue this way through all n coordinates. After n passes, every coordinate converges along the same final subsequence, so the vectors themselves converge -- componentwise convergence in \mathbb{R}^n is exactly convergence in norm.

The single-variable Extreme Value Theorem needs a closed, bounded interval; this is the exact same statement one dimension up, and both hypotheses are load-bearing in the same way. "Bounded" keeps D from letting f's values run off to infinity along some escaping direction. "Closed" — D contains its own boundary curve, not just its interior — keeps a candidate maximum from sitting at a boundary point that's excluded from the region, the way f(x,y) = x on the open disk x^2+y^2<1 has no maximum: any point falls short of the open edge, and a point arbitrarily close to the edge is always beaten by one closer still. A full proof again goes through compactness (a closed, bounded region in \mathbb{R}^n is exactly a compact one) and a convergent-subsequence argument identical in spirit to the one-variable case; what matters here is the payoff — on a closed, bounded region, the candidates test (interior critical points plus the entire boundary) is guaranteed to contain the winner.

Suppose not: there is some \varepsilon>0 for which no single \delta works everywhere. Then for every \delta=\tfrac1k, there exist points \mathbf{x}_k,\mathbf{y}_k \in K with \|\mathbf{x}_k-\mathbf{y}_k\|<\tfrac1k but |f(\mathbf{x}_k)-f(\mathbf{y}_k)|\geq\varepsilon. Compactness lets (\mathbf{x}_k) pass to a convergent subsequence \mathbf{x}_{k_j} \to \mathbf{p}\in K; since \|\mathbf{x}_{k_j}-\mathbf{y}_{k_j}\|\to 0, \mathbf{y}_{k_j}\to\mathbf{p} too. Ordinary continuity at \mathbf{p} forces f(\mathbf{x}_{k_j})\to f(\mathbf{p}) and f(\mathbf{y}_{k_j})\to f(\mathbf{p}), so |f(\mathbf{x}_{k_j})-f(\mathbf{y}_{k_j})|\to 0 — contradicting \geq\varepsilon for every j. No such failure can occur, so f is uniformly continuous.

Uniqueness: suppose two linear maps L_1, L_2 both satisfy the limit condition. Then \|(L_1-L_2)\mathbf{h}\|/\|\mathbf{h}\| \to 0 as \mathbf{h}\to\mathbf{0}. But (L_1-L_2) is linear, so (L_1-L_2)(t\mathbf{h}) = t(L_1-L_2)(\mathbf{h}) for any t>0 and fixed \mathbf{h}\neq\mathbf{0}; substituting \mathbf{h}\to t\mathbf{h} with t\to 0^+ shows the ratio \|(L_1-L_2)\mathbf{h}\|/\|\mathbf{h}\| is actually CONSTANT in t along that direction, yet the limit forces it to 0 — so (L_1-L_2)\mathbf{h}=\mathbf{0} for every \mathbf{h}, meaning L_1=L_2.

That the matrix of Df(\mathbf{a}) is the Jacobian: apply the defining limit along \mathbf{h}=t\mathbf{e}_j (the j-th standard basis vector) as t\to0. The limit condition forces \lim_{t\to0}\frac{f(\mathbf{a}+t\mathbf{e}_j)-f(\mathbf{a})}{t} = Df(\mathbf{a})\mathbf{e}_j — the left side is exactly the j-th partial derivative (vector of partials, for f:\mathbb{R}^n\to\mathbb{R}^m), so Df(\mathbf{a})\mathbf{e}_j is the j-th column of the Jacobian matrix, for every j.

For f:\mathbb{R}^2\to\mathbb{R} (the general case repeats this argument coordinate by coordinate): write f(a+h,b+k)-f(a,b) = \big[f(a+h,b+k)-f(a,b+k)\big] + \big[f(a,b+k)-f(a,b)\big]. Apply the single-variable Mean Value Theorem to each bracket — the first in x (holding y=b+k fixed), the second in y (holding x=a fixed) — giving f_x(a+\theta_1 h,\,b+k)\,h + f_y(a,\,b+\theta_2 k)\,k for some \theta_1,\theta_2\in(0,1). Because f_x,f_y are assumed CONTINUOUS at (a,b), f_x(a+\theta_1h,b+k)\to f_x(a,b) and f_y(a,b+\theta_2k)\to f_y(a,b) as (h,k)\to(0,0) — so the difference f(a+h,b+k)-f(a,b) - [f_x(a,b)h+f_y(a,b)k] is a sum of two terms, each h or k times a quantity going to 0, which is exactly o(\|(h,k)\|). That is the total-derivative limit condition, with Df(a,b) the row vector [f_x(a,b)\ \ f_y(a,b)].

Write L=Dg(\mathbf{a}), M=Df(g(\mathbf{a})), and \mathbf{b}=g(\mathbf{a}). By definition, g(\mathbf{a}+\mathbf{h}) = \mathbf{b} + L\mathbf{h} + \varepsilon_1(\mathbf{h}) with \|\varepsilon_1(\mathbf{h})\|/\|\mathbf{h}\|\to0, and f(\mathbf{b}+\mathbf{k}) = f(\mathbf{b}) + M\mathbf{k} + \varepsilon_2(\mathbf{k}) with \|\varepsilon_2(\mathbf{k})\|/\|\mathbf{k}\|\to0. Substituting \mathbf{k}=L\mathbf{h}+\varepsilon_1(\mathbf{h}): f(g(\mathbf{a}+\mathbf{h})) = f(\mathbf{b}) + M\big(L\mathbf{h} + \varepsilon_1(\mathbf{h})\big) + \varepsilon_2\big(L\mathbf{h}+\varepsilon_1(\mathbf{h})\big) = f(\mathbf{b}) + ML\mathbf{h} + \big[M\varepsilon_1(\mathbf{h}) + \varepsilon_2(\cdots)\big]. The bracketed remainder is o(\|\mathbf{h}\|): M\varepsilon_1(\mathbf{h}) is since \varepsilon_1 is and M is a fixed linear map (bounded), and \varepsilon_2(L\mathbf{h}+\varepsilon_1(\mathbf{h})) is o of its own argument's norm, which itself is O(\|\mathbf{h}\|). So ML satisfies the defining limit condition for D(f\circ g)(\mathbf{a}) — and by the total derivative's uniqueness, D(f\circ g)(\mathbf{a}) = ML = Df(g(\mathbf{a}))\,Dg(\mathbf{a}).

For n=2, \mathbf{h}^{\mathsf T}H\mathbf{h} = f_{xx}h_1^2 + 2f_{xy}h_1h_2 + f_{yy}h_2^2 — exactly the quadratic bracket the Second Partials Test's own proof completes the square on, with D=\det H = f_{xx}f_{yy}-f_{xy}^2 the same discriminant. Writing the classification in terms of H's eigenvalues (always real, since H is symmetric): both eigenvalues positive means H is positive definite (a bowl, local min); both negative means negative definite (a dome, local max); opposite signs means indefinite (a saddle) — the n=2 case's D>0,\,f_{xx}>0 / D>0,\,f_{xx}<0 / D<0 trichotomy is exactly this eigenvalue-sign trichotomy, since for a 2\times2 symmetric matrix, \det H is the PRODUCT of the eigenvalues (same sign iff \det H>0) and f_{xx}'s sign matches the eigenvalues' shared sign when they agree.

Fix a direction \mathbf{h}. Define the single-variable function g(t) = f(\mathbf{a}+t\mathbf{h}). The ordinary single-variable Taylor theorem gives g(1) = \sum_{j=0}^k \frac{g^{(j)}(0)}{j!} + o(1). Now expand g^{(j)}(0) using the multivariable chain rule, applied j times: each application differentiates the composition f(\mathbf{a}+t\mathbf{h}) once more. By the multinomial theorem, this expansion equals exactly \sum_{|\alpha|=j} \frac{j!}{\alpha!}\partial^\alpha f(\mathbf{a})\,\mathbf{h}^\alpha. The multinomial coefficient j!/\alpha! counts one specific thing: the number of ways j ordinary derivatives can split among the n partial-derivative directions to produce that particular \alpha. Substitute this back into the sum for g(1). The j! cancels against the multinomial coefficient's own j!, leaving exactly \sum_{|\alpha|\le k}\frac{\partial^\alpha f(\mathbf{a})}{\alpha!}\mathbf{h}^\alpha, since g(1)=f(\mathbf{a}+\mathbf{h}).

Existence (sketch): apply the Banach fixed-point theorem to the map g(\mathbf{x}) = \mathbf{x} + [Df(\mathbf{a})]^{-1}\big(\mathbf{y} - f(\mathbf{x})\big) for a target \mathbf{y} near f(\mathbf{a}) — a fixed point of g is exactly a solution to f(\mathbf{x})=\mathbf{y}. Because Df is continuous and Df(\mathbf{a}) is invertible, Dg is small (a contraction) on a small enough ball around \mathbf{a}, so g has a unique fixed point there for every \mathbf{y} close enough to f(\mathbf{a}) — this is what produces f^{-1}, locally.

The derivative formula is the easy half, GIVEN differentiability of f^{-1}: differentiate f^{-1}(f(\mathbf{x}))=\mathbf{x} with the chain rule (as a matrix product), giving Df^{-1}(f(\mathbf{a}))\, Df(\mathbf{a}) = I, the identity matrix — so Df^{-1}(f(\mathbf{a})) must be Df(\mathbf{a})'s matrix inverse.

Define G(x,y) = (x,\ F(x,y)), a map \mathbb{R}^2\to\mathbb{R}^2. Its Jacobian is DG(a,b) = \begin{pmatrix}1 & 0\\ F_x & F_y \end{pmatrix}, with determinant F_y(a,b)\neq0 -- so it's invertible. The Inverse Function Theorem applies: G has a local C^1 inverse G^{-1}(x,u) = (x,\ \psi(x,u)) near (a,0). Set u=0: G^{-1}(x,0) = (x,\psi(x,0)) undoes G(x,y)=(x,F(x,y)) exactly when F(x,y)=0. So \phi(x):=\psi(x,0) is the implicit function we want, and it inherits C^1-ness directly from G^{-1}'s.

The derivative formula now follows by implicit differentiation. This step is fully justified, not just formally correct, since \phi is already known to exist and be differentiable. Differentiate F(x,\phi(x))=0 with the chain rule to get F_x + F_y\,\phi'(x) = 0, then solve for \phi'(x) (valid since F_y\neq0).

Sketch, for the case where Df(\mathbf{a}) already has its top-left r\times r block invertible (always arrangeable by relabeling coordinates). Write f=(f_1,\ldots,f_r,f_{r+1},\ldots,f_m) and define \Phi(\mathbf{x}) = (f_1(\mathbf{x}),\ldots,f_r(\mathbf{x}), x_{r+1},\ldots,x_n). Its Jacobian at \mathbf{a} is invertible: the assumed r\times r block, bordered by an identity block. So by the Inverse Function Theorem, \Phi is a local coordinate change. In the new coordinates \mathbf{u}=\Phi(\mathbf{x}), the first r output components of f\circ\Phi^{-1} are just u_1,\ldots,u_r, by construction. Constant rank r then forces something specific: every remaining output component f_{r+1},\ldots,f_m, expressed in the new coordinates, must have all its partials with respect to u_{r+1},\ldots,u_n vanish identically near \mathbf{a}. If it didn't, the Jacobian would pick up rank beyond r somewhere nearby, contradicting the constant-rank hypothesis. So those output components depend only on u_1,\ldots,u_r. A second coordinate change in the CODOMAIN, subtracting off that dependence, reduces them to the constant 0 -- giving the normal form.

Parametrize the constraint curve near (a,b) by arc length, \mathbf{r}(t) = (x(t), y(t)), with \mathbf{r}(0) = (a,b) — this is possible precisely because \nabla g(a,b) \neq \mathbf{0}, which keeps the level curve g=k smooth with a well-defined tangent there. Restrict f to the curve: h(t) = f(x(t), y(t)). If f has a local extremum at (a,b) subject to the constraint, then h has an ordinary single-variable local extremum at t=0, so h'(0) = 0. The chain rule expands that as h'(0) = f_x(a,b)\,x'(0) + f_y(a,b)\,y'(0) = \nabla f(a,b)\cdot \mathbf{r}'(0) = 0, meaning \nabla f(a,b) is perpendicular to the curve's tangent vector \mathbf{r}'(0) at that point. But \nabla g(a,b) is also perpendicular to the same tangent vector — that's what it means for g=k to be a level curve of g. Two vectors both perpendicular to the same tangent line in the plane must be parallel to each other, so \nabla f(a,b) = \lambda \nabla g(a,b) for some scalar \lambda.

Sketch: by the Implicit Function Theorem, since \nabla g\neq\mathbf 0 at the point, the constraint g(x,y)=k locally parametrizes y as a function of x (or vice versa) — exactly the justification chapter 03's own Lagrange-multiplier section supplies for the "restrict to the curve" argument. Substituting that parametrization into f turns the CONSTRAINED problem into an ordinary one-variable optimization; applying the single-variable second-derivative test to the resulting function h(x)=f(x,\phi(x)), and expanding h'' using implicit differentiation (the Implicit Function Theorem's own derivative formula, applied twice), produces exactly the bordered-Hessian determinant above after algebraic simplification — the border row/column are bookkeeping for the constraint eliminating one degree of freedom.

d\mathbf{r} = \sum_i \dfrac{\partial\mathbf{r}}{\partial u_i}\,du_i by the ordinary multivariable chain rule (chapter 00). Then ds^2 = d\mathbf{r}\cdot d\mathbf{r} = \sum_{i,j} \dfrac{\partial\mathbf{r}}{\partial u_i}\cdot\dfrac{\partial\mathbf{r}} {\partial u_j}\,du_i\,du_j. Orthogonality means every cross term (i\neq j) has \dfrac{\partial\mathbf{r}}{\partial u_i}\cdot\dfrac{\partial\mathbf{r}}{\partial u_j}=0, leaving only the diagonal terms \left\|\dfrac{\partial\mathbf{r}}{\partial u_i}\right\|^2 du_i^2 = h_i^2\,du_i^2.

\nabla f is defined by df = \nabla f\cdot d\mathbf{r} for every displacement d\mathbf{r} (the total-derivative characterization, chapter 02). Writing d\mathbf{r} = \sum_i h_i\,du_i\,\hat{\mathbf e}_i (each coordinate step scaled to physical length by its own h_i, the previous theorem) and df = \sum_i \dfrac{\partial f}{\partial u_i}\,du_i (ordinary chain rule), matching the \hat{\mathbf e}_i component on both sides of df=\nabla f\cdot d\mathbf{r} forces the \hat{\mathbf e}_i-component of \nabla f to be \dfrac{1}{h_i} \dfrac{\partial f}{\partial u_i} — the h_i in the denominator exactly compensates for d\mathbf{r}'s own h_i factor.

Sketch, via the divergence theorem's own defining property (chapter 00): \nabla\cdot\mathbf{F} at a point is the limiting flux per unit volume out of a shrinking box. Build the box from small coordinate steps du_1,du_2,du_3; its physical edge lengths are h_1du_1, h_2du_2, h_3du_3 (the scale-factor theorem), and its volume is h_1h_2h_3\,du_1du_2du_3. The flux through the pair of faces perpendicular to \hat{\mathbf e}_1 is (h_2h_3F_1)'s value at u_1+du_1 minus its value at u_1 — because h_2h_3 is exactly the AREA of that face (edges h_2du_2 and h_3du_3) — which is \dfrac{\partial}{\partial u_1}(h_2h_3F_1)\,du_1 to first order. Summing all three face-pairs and dividing by the box's volume h_1h_2h_3\,du_1du_2du_3 gives exactly the stated formula in the limit.

Sketch, via Stokes' theorem's own defining property (chapter 00): (\nabla\times\mathbf{F})\cdot\hat{\mathbf e}_3 at a point is the limiting circulation per unit area around a small loop in the u_1u_2-plane there. Build that loop from coordinate steps du_1, du_2; its physical edge lengths are h_1du_1 and h_2du_2 (scale-factor theorem), so its enclosed area is h_1h_2\,du_1du_2. Circulation around the four edges works out (the same face-pair bookkeeping the divergence formula's proof uses, one dimension down) to \left[\dfrac{\partial}{\partial u_1}(h_2F_2) - \dfrac{\partial}{\partial u_2}(h_1F_1)\right]du_1du_2, and dividing by the enclosed area h_1h_2\,du_1du_2 gives the \hat{\mathbf e}_3-component of the stated determinant formula. The other two components follow by the same argument in the other two coordinate planes; the determinant is simply a compact way to write all three component formulas' cyclic pattern at once.

\nabla^2 f = \nabla\cdot(\nabla f) — apply the divergence formula directly to the vector field \nabla f, whose curvilinear components are \dfrac{1}{h_i}\dfrac{\partial f}{\partial u_i} (the gradient formula). Substituting F_i = \dfrac{1}{h_i}\dfrac{\partial f}{\partial u_i} into \dfrac{1}{h_1h_2h_3}\dfrac{\partial}{\partial u_1}(h_2h_3F_1)+\cdots turns h_2h_3F_1 into h_2h_3\cdot \dfrac{1}{h_1}\dfrac{\partial f}{\partial u_1} = \dfrac{h_2h_3}{h_1}\dfrac{\partial f}{\partial u_1}, and likewise for the other two terms — exactly the stated formula.

Both sides vanish unless \{j,k\} and \{l,m\} are the same pair of distinct indices (from \{1,2,3\}\setminus\{i\}, whichever value i happens to be summed over) — if j=k or l=m, the corresponding \varepsilon is already 0. Given \{j,k\}=\{l,m\} as sets, there are only two cases: j=l,k=m (matching order) or j=m,k=l (swapped order). Summing the single surviving term (i is forced to be the one remaining index) in each case: matching order contributes \varepsilon_{ijk}^2=1 to the left side while the right side gives \delta_{jl}\delta_{km} = 1\cdot1=1 and \delta_{jm}\delta_{kl}=0; swapped order contributes \varepsilon_{ijk}\varepsilon_{ikj}=-1 to the left side (one is the negative of the other, from a single transposition) while the right side gives \delta_{jl}\delta_{km}=0 and \delta_{jm}\delta_{kl}=1\cdot1=1, net 0-1=-1. Both cases agree, term by term.

The defining requirement is that a tensor represent the SAME physical object regardless of which coordinate frame describes it — concretely, that any equation built from tensors (like F_i = T_{ij}a_j, a stress tensor acting on a direction) hold in every rotated frame if it holds in one. Substituting the transformation law for F, T, and a separately: F'_i = R_{ik}F_k = R_{ik}T_{kl}a_l. Substituting the transformation law's INVERSE for T and a in terms of their primed values (T_{kl}=R_{mk}R_{nl}T'_{mn}, a_l=R_{pl}a'_p, using R^{-1}=R^{\mathsf T}) and simplifying with R_{ik}R_{mk}=\delta_{im} (orthogonality) confirms F'_i = T'_{ij}a'_j — the same equation, unchanged in form, in the new frame. This consistency is exactly what an arbitrary array of numbers indexed the same way would NOT automatically have; it's a real constraint, not a bookkeeping formality.

Geometrically, \alpha\wedge\beta measures signed area swept by the pair (\alpha,\beta), oriented by which one comes first — exactly like \mathbf u\times\mathbf v = -\mathbf v\times\mathbf u (chapter 00). Swapping the order reverses the orientation, which reverses the sign. Setting \beta=\alpha: \alpha\wedge\alpha = -\alpha\wedge\alpha \implies 2(\alpha\wedge\alpha)=0 \implies \alpha\wedge\alpha=0 — a 1-form wedged with itself sweeps out a degenerate parallelogram (zero area), consistent with two identical vectors spanning no area at all.

This is a definition, not a derived fact — but it's forced to look this way by two requirements any reasonable "derivative for forms" must satisfy: agreeing with the ordinary differential on functions (k=0), and obeying a graded product rule d(\omega\wedge\eta) = d\omega\wedge \eta + (-1)^k\omega\wedge d\eta for a k-form \omega. Applying d to f\,dx_{i_1}\wedge\cdots\wedge dx_{i_k} (a product of the function f and the constant-coefficient form dx_{i_1}\wedge\cdots\wedge dx_{i_k}, whose own d vanishes — d(dx_j)=0 trivially since dx_j already has constant coefficient 1) via that product rule gives exactly df\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_k}, matching the stated formula.

For a 0-form f: df = \sum_i \partial_if\,dx_i, so d(df) = \sum_{i,j}\partial_j\partial_if\,dx_j\wedge dx_i. Split the sum into i<j and i>j pairs; relabeling the second group swaps i,j, giving \sum_{i<j}\big(\partial_j\partial_if - \partial_i\partial_jf\big)\, dx_j\wedge dx_i after using dx_j\wedge dx_i = -dx_i\wedge dx_j to match orientations. Clairaut's theorem (chapter 00) makes \partial_j\partial_if=\partial_i\partial_jf exactly, so every coefficient vanishes: d(df)=0. For a general k-form, apply this same coefficient-by-coefficient argument to each df_I term in the exterior-derivative formula — the same symmetric-mixed-partial cancellation kills every term.

df = f_x\,dx+f_y\,dy+f_z\,dz is exactly \omega^1_{\nabla f} by definition of \nabla f. For d\omega^1_{\mathbf F}, expand d(F_1dx)=dF_1\wedge dx = (F_{1y}dy+F_{1z}dz)\wedge dx = F_{1y}\,dy \wedge dx + F_{1z}\,dz\wedge dx = -F_{1y}\,dx\wedge dy + F_{1z}\,dz\wedge dx; doing the same for the F_2dy and F_3dz terms and collecting coefficients of dy\wedge dz, dz\wedge dx, dx\wedge dy gives exactly (F_{3y}-F_{2z}), (F_{1z}-F_{3x}), (F_{2x}-F_{1y}) — the three components of \nabla\times\mathbf F. For d\omega^2_{\mathbf F}: d(F_1\,dy\wedge dz) = F_{1x}\,dx\wedge dy\wedge dz (the dy, dz partials of F_1 contribute terms with a repeated dy or dz, which vanish by \alpha\wedge\alpha=0); summing the three analogous terms gives (F_{1x}+F_{2y}+F_{3z})\,dx\wedge dy\wedge dz = (\nabla\cdot\mathbf F)\,dx\wedge dy\wedge dz.

Only in \mathbb{R}^3 does k and 3-k pair up 1-forms with 2-forms (both 3-dimensional spaces of forms, by the \binom{3}{k} dimension count) — in \mathbb{R}^2, \star pairs 1-forms with 1-forms instead (\binom{2}{1}=2=\binom{2}{1}), and in \mathbb{R}^4 it would pair 1-forms with 3-forms, never with other 1-forms. This 1-form/2-form coincidence specific to three dimensions is exactly why the cross product and curl — both 3D-specific constructions with no n-dimensional analogue built the same way — exist at all: \star is the precise bridge the coordinate formula for \mathbf u\times\mathbf v is secretly built from.

Substitute y_i=\varphi_i(\mathbf x) and expand each dy_i = \sum_j\frac{\partial\varphi_i}{\partial x_j}\,dx_j (the ordinary differential, chapter 00) directly into dy_1\wedge\cdots\wedge dy_n. Expanding this wedge product, every term with a repeated dx_j vanishes (dx_j\wedge dx_j=0); the surviving terms are exactly a signed sum over permutations \sigma of \prod_i \frac{\partial\varphi_i}{\partial x_{\sigma(i)}}, with sign \pm1 matching the permutation's parity — which is precisely the Leibniz cofactor-expansion formula for \det(D\varphi), multiplying the single top-degree basis form dx_1\wedge\cdots\wedge dx_n.

\partial S consists of the outer edges of S's cells — each such edge belongs to exactly one cell of S (that's what makes it a BOUNDARY edge rather than a canceled interior one, chapter 07's own chain-boundary construction). Taking the boundary of \partial S in turn means looking at the ENDPOINTS of those boundary edges. But every endpoint where two boundary edges meet is a single point shared by exactly two edges of \partial S, traversed in a consistent orientation around the loop — each such point is entered by one edge and exited by the next, so it contributes +1 and -1 (entering and leaving) that cancel exactly, the same edge-cancellation mechanism one dimension down. Since \partial S is itself a closed loop (or a union of closed loops) with no loose ends, every point cancels this way, leaving nothing: \partial (\partial S) = \varnothing.

Sketch, via the chain-boundary picture (chapter 07's own construction): subdivide S into small oriented cells, each nearly flat. On each tiny cell, \int_{\partial(\text{cell})}\omega \approx (\text{cell's own size}) \cdot d\omega evaluated at the cell — this is essentially d\omega's own definition, the local rate at which \omega's integral around a shrinking boundary compares to the cell's volume. Summing this approximate identity over every cell in the subdivision gives \sum_{\text{cells}}\int_{\partial(\text{cell})}\omega \approx \int_Sd\omega. On the left, every INTERIOR edge shared by two adjacent cells is traversed twice, once from each cell, in OPPOSITE orientations — exactly the cancellation chapter 07's chain-boundary-lab widget makes visible — leaving only the edges on S's own outer boundary, i.e. exactly \int_{\partial S}\omega. Refining the subdivision (shrinking every cell) makes the approximation exact in the limit.

Sketch, for a closed 1-form \omega=P\,dx+Q\,dy with P_y=Q_x (closedness) on a simply connected region: fix a base point \mathbf a, and define f(\mathbf x) = \int_{\mathbf a}^{\mathbf x}\omega along ANY path from \mathbf a to \mathbf x staying in the region. Simple connectivity guarantees this integral doesn't depend on which path is chosen — two different paths bound a region entirely inside the simply connected domain (since any loop there can shrink to a point), and Green's theorem (a special case of the generalized Stokes theorem, recovered a few sections back) applied to that bounded region gives \oint\omega = \iint d\omega = 0 exactly because \omega is closed. With the integral path-independent, f is a well-defined function, and direct differentiation confirms df=\omega — exactly the exactness claimed. On a domain that ISN'T simply connected, two paths around a hole bound a region that ISN'T entirely inside the domain (it contains the missing hole), so Green's theorem can't be applied, and the path-independence argument breaks down — precisely the gap the punctured-plane counterexample (this chapter's closing sections) exploits.

Fix x and let A(x) = \int_c^d f(x,y)\,dy, the (signed) area of the cross-section cut by the vertical plane at that x. Slicing the solid region under z=f(x,y) into thin slabs of thickness dx, each slab has volume approximately A(x)\,dx, and summing all the slabs gives \int_a^b A(x)\,dx as the total volume — the definition of the double integral itself. This is exactly the cross-sectional-area argument behind volume by slicing from single-variable calculus, run in the x-direction instead of a fixed axis. The same slicing argument in the y-direction produces the other iterated order, and continuity of f on a closed rectangle guarantees both limits exist and agree.

No elementary antiderivative of e^{-x^2} exists, so a direct one-variable approach fails — the trick squares the integral and switches to polar coordinates. Let I=\int_{-\infty}^\infty e^{-x^2}dx. Then I^2 = \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty} ^\infty e^{-y^2}dy\right) = \iint_{\mathbb R^2} e^{-(x^2+y^2)}\,dA. Switching to polar coordinates (chapter 00's own Jacobian, dA=r\,dr\, d\theta): I^2 = \int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta. The inner integral, with u=r^2, du=2r\,dr, becomes \int_0^\infty e^{-u}\,\tfrac12du = \tfrac12. So I^2 = 2\pi\cdot\tfrac12=\pi, giving I=\sqrt\pi (positive, since the integrand is everywhere positive).

By definition, F'(t) = \lim_{h\to0}\dfrac{F(t+h)-F(t)}{h} = \lim_{h\to0}\int_a^b\dfrac{f(x,t+h)-f(x,t)}{h}\,dx. The Mean Value Theorem writes \dfrac{f(x,t+h)-f(x,t)}{h} = f_t(x,t+\theta h) for some \theta\in(0,1) depending on x. Because f_t is continuous on the compact rectangle [a,b]\times[c,d], it's uniformly continuous there (chapter 01's Heine–Cantor theorem) — so f_t(x,t+\theta h)\to f_t(x,t) UNIFORMLY in x as h\to0, which is exactly what's needed to swap the limit and the integral: \lim_{h\to0}\int_a^bf_t(x,t+\theta h)\,dx = \int_a^b\lim_{h\to0}f_t(x,t+\theta h)\,dx = \int_a^bf_t(x,t)\,dx.

Write F(t)=G(a(t),b(t),t) for G(u,v,t)=\int_u^vf(x,t)\,dx, and apply the multivariable chain rule (chapter 02) treating u=a(t), v=b(t), and t as the three inputs to G: F'(t) = G_u\,a'(t) + G_v\,b'(t) + G_t\cdot1. By the Fundamental Theorem of Calculus, G_v = f(v,t) and G_u = -f(u,t) (differentiating the lower bound flips sign, since \int_u^v = -\int_v^u). By the fixed-bounds Leibniz rule (previous section), G_t = \int_u^vf_t(x,t)\,dx. Substituting u=a(t), v=b(t) gives exactly the stated formula.

\Gamma(n)=(n-1)!: integration by parts on \Gamma(s)=\int_0^\infty t^{s-1}e^{-t}dt gives \Gamma(s) = (s-1)\Gamma(s-1) (the boundary term vanishes at both ends); starting from \Gamma(1)=\int_0^\infty e^{-t}dt=1 and applying this recursion n-1 times gives \Gamma(n)=(n-1)(n-2)\cdots1\cdot\Gamma(1) = (n-1)!.

\Gamma(1/2)=\sqrt\pi: substitute t=u^2, dt=2u\,du into \Gamma(1/2)=\int_0^\infty t^{-1/2}e^{-t}dt, giving 2\int_0^\infty e^{-u^2}du — exactly HALF the Gaussian integral (the u>0 half), which is \sqrt\pi by symmetry of the full integral around 0.

B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q): write \Gamma(p)\Gamma(q) as a double integral \iint_{[0,\infty)^2}x^{p-1}y^{q-1}e^{-x-y}\,dx\,dy (product of two 1D integrals), substitute x=rt, y=r(1-t) (r>0, t\in(0,1) — polar-like coordinates adapted to the first quadrant), compute the Jacobian (=r), and the r-integral separates out as \Gamma(p+q), leaving the t-integral as exactly B(p,q).

Compute \int_{\mathbb R^n}e^{-\|\mathbf x\|^2}\,dV two ways. In Cartesian coordinates it separates into a product of n ordinary Gaussian integrals: \left(\int_{-\infty}^\infty e^{-x_1^2}dx_1\right)\cdots = (\sqrt\pi)^n = \pi^{n/2}. In "radial" coordinates, writing the integral as \int_0^\infty e^{-r^2}\,dV_n(r) where dV_n(r) = nV_n(1)r^{n-1}\,dr is the volume of a thin shell of radius r (the surface area of the sphere of radius r, which scales as r^{n-1}, times dr): substituting u=r^2 turns this into nV_n(1)\int_0^\infty e^{-u}u^{n/2-1}\,\tfrac12du = \tfrac{n}{2}V_n(1)\,\Gamma(n/2). Setting the two computations equal, \pi^{n/2} = \tfrac n2 V_n(1)\Gamma(n/2), and using \tfrac{n}{2}\Gamma(n/2)=\Gamma(n/2+1) (the Gamma function's own recursion) gives V_n(1) = \pi^{n/2}/\Gamma(n/2+1).

Let g(r) = \frac{1}{2\pi r}\oint_{\|\mathbf x-\mathbf a\|=r}u\,ds = \frac{1}{2\pi}\int_0^{2\pi}u(\mathbf a+r(\cos\theta,\sin\theta))\, d\theta. Differentiate under the integral sign (chapter 08) with respect to r: g'(r) = \frac{1}{2\pi}\int_0^{2\pi}\nabla u\cdot(\cos\theta,\sin\theta)\,d\theta = \frac{1}{2\pi r}\oint \nabla u\cdot\hat{\mathbf n}\,ds (recognizing (\cos\theta,\sin\theta) as the outward unit normal). By the Divergence Theorem (chapter 00), this boundary flux equals \frac{1}{2\pi r}\iint_{\text{disk}} \nabla^2u\,dA = 0, since u is harmonic. So g'(r)=0 for every r — g is CONSTANT — and \lim_{r\to0}g(r)=u(\mathbf a) (the average over a shrinking circle approaches the center value by continuity), giving g(r)=u(\mathbf a) for every r.

Suppose u attains an interior maximum at \mathbf a\in D, with u(\mathbf a) = M the largest value u takes anywhere on the closure. By the Mean Value Property, u(\mathbf a) equals the average of u over any small circle centered at \mathbf a. But every value on that circle is \le M (since M is the overall maximum) — an average of values \le M equaling exactly M forces EVERY value on the circle to equal M too (otherwise the average would be strictly less). This holds for circles of every radius small enough to stay in D, so u \equiv M on an entire neighborhood of \mathbf a. The set where u=M is therefore open; it's also closed (as the preimage of a point under a continuous function); on a connected region, a nonempty set that's both open and closed must be the WHOLE region — so u\equiv M everywhere, the constant case. If u isn't constant, no interior maximum can occur, forcing the maximum onto the boundary.

First identity: apply the Divergence Theorem (chapter 00) to \mathbf F = u\nabla v. Its divergence is \nabla\cdot(u\nabla v) = \nabla u\cdot\nabla v + u\,\nabla^2v (the ordinary product rule for divergence, chapter 05's index-notation form), and its flux through S is \oiint_S u\nabla v\cdot\hat{\mathbf n}\,dS = \oiint_S u\,\partial v/\partial n\,dS. Setting the Divergence Theorem's two sides equal gives exactly the first identity.

Second identity: apply the first identity twice — once as written, once with u and v swapped — and subtract. The \nabla u\cdot \nabla v terms are identical in both (dot products commute) and cancel, leaving exactly the stated antisymmetric combination.

Let w=u_1-u_2, harmonic (the difference of two harmonic functions is harmonic, by linearity of \nabla^2) and zero on the entire boundary. Apply Green's first identity with u=v=w: \iiint_V\|\nabla w\|^2\,dV = \oiint_S w\,\partial w/\partial n\,dS = 0 (since w=0 on S). The left side is an integral of a NON-NEGATIVE quantity (\|\nabla w\|^2 \geq0) equaling 0, which forces \nabla w\equiv\mathbf 0 throughout V — w is constant. Since w=0 on the boundary and w is constant, w\equiv0 everywhere, so u_1\equiv u_2.

Neumann case: the same argument with \partial w/\partial n=0 on S (rather than w=0) still forces \iiint_V\|\nabla w\|^2\,dV=0 via Green's first identity, hence \nabla w\equiv\mathbf 0 — w is constant, but this time nothing pins the constant to 0.

\nabla^2\Phi=0 away from the origin is a direct computation (the curvilinear Laplacian formula, chapter 04, applied in spherical coordinates to \Phi=-1/(4\pi\rho), a function of \rho alone). Concentrated entirely AT the origin: integrate \nabla^2\Phi over a small ball of radius \varepsilon and apply the Divergence Theorem — \iiint_{B_\varepsilon}\nabla^2\Phi\,dV = \oiint_{\partial B_\varepsilon}\nabla\Phi\cdot\hat{\mathbf n}\,dS. Direct computation gives \nabla\Phi = \dfrac{\mathbf x}{4\pi\|\mathbf x\|^3}, so on the sphere of radius \varepsilon, \nabla\Phi\cdot\hat{\mathbf n} = \dfrac{1}{4\pi\varepsilon^2} (constant), and the flux integral is \dfrac{1}{4\pi\varepsilon^2}\times(\text{area }4\pi\varepsilon^2)=1, independent of \varepsilon — exactly the defining property of a unit point source, \delta(\mathbf x).

The integral formula follows from Green's second identity applied to u and G(\cdot,\mathbf x_0) on V minus a small ball around \mathbf x_0 (to avoid G's own singularity), then shrinking that ball to zero — the singular contribution converges to exactly u(\mathbf x_0) by the same point-source argument above, while G=0 on \partial V kills the other boundary term.

The solid-angle claim is exactly the fundamental solution's flux computation (previous theorem) rescaled: \nabla(1/\|\mathbf x\|) = -\mathbf x/\|\mathbf x\|^3, so d\Omega is precisely -4\pi times the flux density of \nabla\Phi through dS. Integrating over any closed surface enclosing the origin gives 4\pi (the same point-source computation as before); integrating over a closed surface NOT enclosing the origin gives 0, since 1/\|\mathbf x\| is harmonic throughout the enclosed region (no singularity inside), so its flux integral vanishes by the Divergence Theorem applied to a harmonic function's gradient (itself divergence-free, \nabla^2\Phi=0). Gauss's Law for \mathbf E = kq\,\mathbf x/\|\mathbf x\|^3 (the electric field of a point charge, chapter 10) is exactly this computation with the constant kq multiplying d\Omega.

Sketch: define \phi(\mathbf x) = -\dfrac{1}{4\pi}\iiint \dfrac{\nabla\cdot\mathbf F(\mathbf y)}{\|\mathbf x-\mathbf y\|}\,dV_y and \mathbf A(\mathbf x) = \dfrac{1}{4\pi}\iiint\dfrac{ \nabla\times\mathbf F(\mathbf y)}{\|\mathbf x-\mathbf y\|}\,dV_y — both built directly from the fundamental solution \Phi (this chapter's own theorem), convolved against \mathbf F's divergence and curl respectively. A direct computation (differentiating under the integral sign, chapter 08, and using \nabla^2\Phi=\delta) confirms -\nabla\phi+\nabla\times\mathbf A reproduces \nabla(\nabla\cdot \mathbf F) - \nabla\times(\nabla\times\mathbf F) convolved against \Phi's own gradient structure, which the vector identity \nabla\times(\nabla\times\mathbf F) = \nabla(\nabla\cdot\mathbf F) - \nabla^2\mathbf F (provable by chapter 05's epsilon-delta identity) simplifies to exactly \mathbf F itself, using \nabla^2\Phi=\delta one more time to collapse the convolution back to \mathbf F at each point.

Let \mathbf x(t) trace a fluid parcel's actual path, so \mathbf x'(t)=\mathbf v(\mathbf x(t),t). The quantity an observer riding with that parcel experiences is g(t)=f(\mathbf x(t),t); by the multivariable chain rule (chapter 02), $g'(t) = \nabla f\cdot \mathbf x'(t) + \dfrac{\partial f}{\partial t} = \mathbf v\cdot\nabla f

• \dfrac{\partial f}{\partial t} — exactly the stated formula, with Df/Dt defined as g'(t)$.

This is exactly the moving-boundary Leibniz rule (chapter 08), one dimension up: V(t)'s motion generalizes a moving interval's endpoints to a moving BOUNDARY SURFACE, and \mathbf v\cdot\hat{\mathbf n} (how fast the boundary sweeps outward in the normal direction) generalizes b'(t). The first term tracks f changing in place; the second tracks new volume being swept up by the moving boundary, weighted by f's value there — exactly the same two-term structure the 1D moving-boundary formula has, applied to a surface instead of two endpoints.

Apply the Reynolds Transport Theorem to f=\rho over a FIXED region V in space (not moving with the flow — take \mathbf v\to\mathbf 0 in the transport theorem's own derivation, or equivalently, apply the ordinary Leibniz rule to \iiint_V\rho\,dV directly): \dfrac{d}{dt} \iiint_V\rho\,dV = \iiint_V\dfrac{\partial\rho}{\partial t}\,dV. Mass conservation says the ONLY way mass inside V changes is by flowing across \partial V: \dfrac{d}{dt}\iiint_V\rho\,dV = -\oiint_{ \partial V}\rho\mathbf v\cdot\hat{\mathbf n}\,dS (flux OUT decreases the enclosed mass). Applying the Divergence Theorem to the right side and equating integrands (since V was arbitrary) gives exactly \partial\rho/\partial t = -\nabla\cdot(\rho\mathbf v).

Each is a direct physical statement, not derived from the others — they are Maxwell's postulated laws of electromagnetism, and this course's role is showing how they translate between languages (integral, differential, and forms), not deriving them from more basic physics. Gauss's Law for \mathbf E is exactly chapter 09's Gauss's Law theorem, with q\to Q_{\text{enc}}; Gauss's Law for \mathbf B is the same solid-angle argument applied to a field with NO sources (hence always 0 net flux, consistent with no magnetic point charges ever being observed).

Each differential form follows from its integral counterpart by the same "arbitrary region, equate integrands" argument the continuity equation's proof used, via the Divergence Theorem (for the two Gauss's Law equations) or Stokes' Theorem (for Faraday's and Ampère–Maxwell's, converting the boundary circulation integral into a surface integral of curl, then equating integrands over an arbitrary surface). This is exactly why chapter 07's generalized Stokes theorem matters here: it's the single tool converting ALL FOUR integral laws into their differential counterparts, not four separate conversions.

dF=0 is a direct consequence of d^2=0 (chapter 07) once F is written as F=dA for a 4-potential A (chapter 09's scalar/vector potentials, unified into a single 4-component object in spacetime) — d(dA)=0 automatically. dG=J is the differential-forms translation of the two source-driven equations, exactly the way \nabla\cdot\mathbf F (chapter 06) is the forms-translation of ordinary divergence; the full derivation requires spacetime index conventions beyond this course's scope, but the PATTERN — four vector equations collapsing into two form equations — is exactly chapter 06 and 07's grad/curl/div unification, one level more abstract.

Take the curl of Faraday's Law: \nabla\times(\nabla\times\mathbf E) = -\dfrac{\partial}{\partial t}(\nabla\times\mathbf B). The vector identity \nabla\times(\nabla\times\mathbf F) = \nabla(\nabla\cdot \mathbf F) - \nabla^2\mathbf F (provable via chapter 05's epsilon-delta identity) simplifies the left side, using \nabla\cdot\mathbf E=0 in vacuum, to -\nabla^2\mathbf E. On the right, substitute the Ampère–Maxwell Law (also with \mathbf J=\mathbf 0): \nabla\times\mathbf B = \mu_0\varepsilon_0\,\partial\mathbf E/\partial t, giving -\nabla^2\mathbf E = -\mu_0\varepsilon_0\,\partial^2\mathbf E/\partial t^2. Canceling the negative signs gives exactly the wave equation.

Sketch: differentiate \Gamma(t) using the material derivative (section 01) applied along the moving loop, which produces a term \oint D\mathbf v/Dt\cdot d\mathbf l (the fluid's acceleration, circulated) plus a term that vanishes identically by the same \oint\nabla f\cdot d\mathbf l=0 argument chapter 00's conservative-field theorem uses. For an ideal fluid, Euler's equation gives D\mathbf v/Dt = -\nabla p/\rho + \mathbf g (pressure gradient plus a conservative body force like gravity) — both terms on the right are gradients (using \rho constant to write \nabla p/\rho = \nabla(p/\rho)), so their circulation integral vanishes by the same argument. With both pieces of d\Gamma/dt vanishing, \Gamma is constant.

Let y(x) be the extremizer and consider a nearby competitor y(x)+\varepsilon\eta(x), where \eta is any function with \eta(a)=\eta(b)=0 (fixed endpoints). Define \Phi(\varepsilon) = J[y+\varepsilon\eta]; since y extremizes J, \Phi'(0)=0 for EVERY choice of \eta. Differentiating under the integral sign (chapter 08): \Phi'(0) = \int_a^b\left(\frac{\partial L}{\partial y}\eta + \frac{\partial L}{\partial y'}\eta'\right)dx. Integrate the second term by parts: \int_a^b\frac{\partial L}{\partial y'}\eta'\,dx = \left[\frac{\partial L}{\partial y'}\eta\right]_a^b - \int_a^b\frac{d}{dx}\!\left(\frac{\partial L}{\partial y'}\right)\eta\,dx. The boundary term vanishes (\eta(a)=\eta(b)=0), leaving \Phi'(0) = \int_a^b\left(\frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'}\right)\eta\,dx = 0 for EVERY \eta vanishing at the endpoints. The fundamental lemma of the calculus of variations (if a continuous function integrated against every such \eta gives 0, that function must itself be identically 0) forces the bracketed quantity to vanish everywhere — exactly the Euler–Lagrange equation.

Both generalize the single-variable proof identically. For several dependent variables, perturb EACH y_i independently (y_i+\varepsilon\eta_i, all others fixed) and repeat the single-variable argument for each — since the perturbations are independent, each gives its own separate Euler–Lagrange equation. For several independent variables, perturb u\to u+\varepsilon\eta (with \eta=0 on \partial D) and apply the SAME integration-by-parts argument, now using the Divergence Theorem (chapter 00) instead of ordinary 1D integration by parts to move the derivatives off \eta in each of u_x,u_y's terms — the boundary terms vanish since \eta=0 on \partial D, and the fundamental lemma (this time in 2D) forces the bracketed PDE.

This is chapter 03's Lagrange multiplier method, in infinite dimensions. Perturb y\to y+\varepsilon\eta, but only among perturbations preserving the constraint to first order: K[y+\varepsilon\eta] = K[y] + \varepsilon\!\int M_y\eta + M_{y'}\eta'\,dx + O(\varepsilon^2) = c forces \int_a^b\!\left(M_y - \frac{d}{dx}M_{y'}\right)\eta\,dx = 0 to first order — the SAME "orthogonal to the constraint's own Euler–Lagrange expression" condition finite-dimensional Lagrange multipliers use (chapter 03), just with the ordinary gradient \nabla g replaced by the constraint's own Euler–Lagrange expression. The same argument that produced \nabla f=\lambda\nabla g there produces L_y - \frac{d}{dx}L_{y'} = \lambda\!\left(M_y - \frac{d}{dx}M_{y'}\right) here — exactly the Euler–Lagrange equation for L-\lambda M.

Sketch, for L independent of x explicitly (a "time-translation" symmetry, L=L(y,y') only): compute \frac{d}{dx}\left(y'\frac{\partial L}{\partial y'} - L\right) = y''\frac{\partial L}{\partial y'} + y'\frac{d}{dx}\!\left(\frac{ \partial L}{\partial y'}\right) - \frac{\partial L}{\partial y}y' - \frac{\partial L}{\partial y'}y''. The y''\partial L/\partial y' terms cancel; the remaining terms combine to y'\!\left(\frac{d}{dx}\frac{\partial L}{\partial y'} - \frac{\partial L}{\partial y}\right), which is exactly -y' times the Euler–Lagrange expression — ZERO along any extremal. So y'\partial L/\partial y' - L (the "Hamiltonian," in mechanics) is CONSTANT along every solution — a conserved quantity, produced entirely from the symmetry (L not depending on x), with no separate physical input.

For a single particle, L = \tfrac12m\dot x^2 - V(x). The Euler–Lagrange equation, \partial L/\partial x - \dfrac{d}{dt} \left(\partial L/\partial\dot x\right)=0, computes \partial L/\partial x = -V'(x) and \partial L/\partial\dot x = m\dot x, so \dfrac{d}{dt}(m\dot x) = -V'(x). Since -V'(x) is the force (the negative gradient of potential energy) and \dfrac{d}{dt}(m\dot x) = m\ddot x for constant mass, this is exactly m\ddot x = F — Newton's second law, derived from a variational principle rather than assumed as a separate axiom.