Proof sketch. Suppose first $A$ is invertible. For any $w \in \R^3$ we note \begin{align} Au \times Av \cdot w &= \det \begin{pmatrix} \uparrow & \uparrow & \uparrow \\ Au & Av & A (A^{-1} w) \\ \downarrow & \downarrow & \downarrow \end{pmatrix} =\det(A) \det \begin{pmatrix} \uparrow & \uparrow & \uparrow \\ u & v & A^{-1} w \\ \downarrow & \downarrow & \downarrow \end{pmatrix} \\ &= u \times v \cdot \adj(A) w = \cof(A) (u \times v) \cdot w \,. \end{align}

Proof of Proposition 6. Let $\sigma \colon U \to \Sigma_1$ be a parametrization of $(\Sigma_1, \hat n_2)$. Using the previous lemma, we verify $\sigma \circ \varphi$ is a parametrization of $(\Sigma_2, \hat n_2)$ and so (by the previous lemma again) \begin{align} \int_{\Sigma_2} F \cdot \hat n_2 \, dS &= \int_{U} F \circ \varphi \circ \sigma \cdot \partial_1 (\varphi \circ \sigma) \times \partial_2 (\varphi \circ \sigma) \, dS \\ &= \int_{\Sigma_1} F \circ \varphi \cdot \paren[\big]{ \cof(D \varphi_\sigma) (\partial_1 \sigma \times \partial_2 \sigma) } \, dS\,, \end{align} which proves \eqref{e:sint-transform} as desired.

Proof. This can directly be brute force derived using the product rule, chain rule and Clairaut’s theorem. However, the calculation is a little tedious if done directly. Here’s a slightly more conceptual way of doing it. It’s probably not any shorter, but you may find it easier to internalize. For convenience let $\adj(D\varphi) = (a_{i,j})$. By the product rule \begin{equation}\label{e:adj1}\tag{A} \dv \paren[\big]{ \adj(D\varphi) \, v \circ \varphi } = \sum_{i,j} \partial_i a_{i,j} v_j \circ \varphi + \sum_{i,j} a_{i,j} \partial_i (v_j \circ \varphi)\,. \end{equation} We claim the first sum on the right vanishes, and the second sum gives us what we want.

Let’s study the second term first. By the chain rule, \begin{equation} \sum_{i,j} a_{i,j} \partial_i (v_j \circ \varphi) = \sum_{i,j,k} a_{i,j} (\partial_k v_j) \circ \varphi \, \partial_i \varphi_k = \sum_{j, k} b_{k,j} (\partial_k v_j) \circ \varphi\,, \end{equation} where \begin{equation} b_{k,j} \defeq \sum_i a_{i,j} \partial_i \varphi_k = (D\varphi)_{k,i} \adj(D\varphi)_{i,j}\,. \end{equation} This is simply the $k, j$-th entry of the matrix product $D\varphi \adj(D\varphi)$, and hence $b_{k,j} = \det(D\varphi)$ if $k = j$ and is $0$ otherwise. This immediately shows \begin{equation} \sum_{i,j} a_{i,j} \partial_i (v_j \circ \varphi) = \det(D\varphi) (\dv v) \circ \varphi\,, \end{equation} as desired.

Now let’s study the first sum on the right of \eqref{e:adj1}. Note, by the formula for the adjugate, \begin{equation} \adj(D\varphi) = \begin{pmatrix} \uparrow & \uparrow & \uparrow\\ \grad \varphi_2 \times \grad \varphi_3 & \grad \varphi_3 \times \grad \varphi_1 & \grad \varphi_1 \times \grad \varphi_2 \\ \downarrow & \downarrow & \downarrow \end{pmatrix} \end{equation} By the vector calculus identities \begin{equation} \curl \grad f = 0 \qquad\text{and}\qquad \dv( G \times H ) = (\curl G) \cdot H - G \cdot (\curl H)\,, \end{equation} we see that each column of $\adj(D\varphi)$ has divergence $0$. (If you’re not familiar with these identities, they can be quickly checked using Clairaut’s theorem and the product rule.) This means for any $j$, \begin{equation} \sum_i \partial_i a_{i,j} v_j \circ \varphi = \dv\paren[\big]{ \adj(D\varphi) e_j } v_j \circ \varphi = 0\,, \end{equation} as desired. This finishes the proof.

Proof sketch. This is just Fubini’s theorem and the fundamental theorem of calculus.

Proof sketch. By Lemma 8, \begin{align} \int_{\partial U} v \cdot \hat n \, dS &= \int_{\partial S} \adj(D\varphi) v \circ \varphi \cdot \hat n \, dS = \int_{\partial S} \dv(\adj(D\varphi) v \circ \varphi) \, dV \\ &= \int_{\partial S} \det(D \varphi) \dv v \circ \varphi \, dV = \int_{U} \dv v \, dV \end{align}

Proof sketch of Theorem 3. Break up $U$ into smaller domains where the previous lemma applies. (If you have to introduce internal boundaries, the surface integral along them will cancel as it appears twice with opposite orientations.)

Proof sketch. The crux of the matter is the following calculation. Given a fixed $y \in \R^3$, define the vector field $F$ by \begin{equation} F(x) = \frac{x-y}{\abs{x - y}^3}. \end{equation} Then \begin{equation}\label{eqnGauss1}\tag{G} \oint_\Sigma F \cdot \hat n \, dS = \begin{cases} 4\pi & \text{if } y \text{ is in the region enclosed by } \Sigma\,,\\ 0 & \text{otherwise} \,. \end{cases} \end{equation} (This is the three dimensional analog of the winding number of the boundary we did earlier.)

For simplicity, we subsequently assume $y = 0$. To prove \eqref{eqnGauss1}, observe \begin{equation} \dv F = 0, \end{equation} when $x \neq 0$. Let $U$ be the region enclosed by $\Sigma$. If $0 \not\in U$, then the divergence theorem will apply to in the region $U$ and we have \begin{equation} \oint_\Sigma F \cdot \hat n \, dS = \int_U \dv F \, dV = 0. \end{equation}

On the other hand, if $0 \in U$, the divergence theorem will not directly apply, since $F \not \in C^1(U)$. To circumvent this, let $\epsilon > 0$ and $U’ = U - B(0, \epsilon)$, and $\Sigma’$ be the boundary of $U’$. Since $0 \not \in U’$, $F$ is $C^1$ on all of $U’$ and the divergence theorem gives \begin{equation} 0 = \int_{U’} \dv F \, dV = \int_{\partial U’} F \cdot \hat n \, dS, \end{equation} and hence \begin{equation} \oint_{\Sigma} F \cdot \hat n \, dS = -\oint_{\partial B(0, \epsilon)} F \cdot \hat n \, dS = \oint_{\partial B(0, \epsilon)} \frac{1}{\epsilon^2} \, dS = -4\pi, \end{equation} as claimed. (Above the normal vector on $\partial B(0, \epsilon)$ points outward with respect to the domain $U’$, and inward with respect to the ball $B(0, \epsilon)$.)

Now, in the general case, suppose the mass distribution has density $\rho$. Then the gravitational field $g(x)$ will be the super-position of the gravitational fields at $x$ due to a point mass of size $\rho(y) \, dV$ placed at $y$. Namely, this means \begin{equation} g(x) = -G \int_{\R^3} \frac{\rho(y) (x - y)}{\abs{x - y}^3} \, dV(y). \end{equation} Now using Fubini’s theorem, \begin{multline} \int_\Sigma g(x) \cdot \hat n(x) \, dS(x) = -G \int_{y \in \R^3} \rho(y) \int_{x \in \Sigma} \frac{x -y}{\abs{x - y}^3} \cdot \hat n(x) \, dS(x) \, dV(y) \\ = -4 \pi G \int_{y \in U} \rho(y) \, dV(y) = -4 \pi GM, \end{multline} where the second last equality followed from \eqref{eqnGauss1}.

Proof sketch. Using Clairaut’s theorem, we can directly check $\dv (\curl F) = 0$. For the converse, suppose ${\dv v} = 0$. We need to find a $C^1$ vector field $F$ such that $\curl F = v$. Note that $\curl( F + \grad \phi) = \curl F$ for any $C^2$ function $\phi$. Set $G = F + \grad \phi$, and define \begin{equation} \phi(x) = -\int_0^{x_3} F_3(x_1, x_2, t) \, dt\,, \end{equation} and observe now that $G_3 = 0$. Thus if $v = \curl F$ for some $F$, we can always find a vector field $G$ such that $G_3 = 0$ and $v = \curl G$. Since this is simpler, we will directly construct a vector field $G$, with $G_3 = 0$, such that $\curl G = v$.

Note that if $G_3 = 0$, then \begin{equation} \curl G = \colvec{-\partial_3 G_2\\ \partial_3 G_1 \\ \partial_1 G_2 - \partial_2 G_1}\,. \end{equation} Since we want $\curl G = v$, we must have \begin{gather} G_2(x) = -\int_0^{x_3} v_1(x_1, x_2, t) \, dt + C_2(x_1, x_2)\,,\\ G_1(x) = \int_0^{x_3} v_2(x_1, x_2, t) \, dt + C_1(x_1, x_2)\,, \end{gather} where $C_1$, $C_2$ are two functions that only depend on $x_1$ and $x_2$. This shows $(\curl G) \cdot e_i = v_i$ for $i = 1, 2$, and to finish the proof we only need to verify the same identity for $i = 3$.

For this, recall $\dv v = 0$, and hence $\partial_3 v_3 = -\partial_1 v_1 - \partial_2 v_2$. Consequently, \begin{align} \curl G \cdot e_3 &= \partial_1 G_2 - \partial_2 G_1 = \int_0^{x_3} (-\partial_1 v_2 - \partial_2 v_2) \, dt + \partial_1 C_2 - \partial_2 C_1 \\ &= \int_0^{x_3} \partial_3 v_3(x_1, x_2, t) \, dt + \partial_1 C_2 - \partial_2 C_1 \\ &= v_3(x_1, x_2, x_3) - v_3(x_1, x_2, 0) + \partial_1 C_2 - \partial_2 C_1\,. \end{align} To finish the proof we only need to choose $C_1$ and $C_2$ so that \begin{equation} v_3(x_1, x_2, 0) = \partial_1 C_2 - \partial_2 C_1\,. \end{equation} This is easily arranged. Indeed, choose $C_1 = 0$ and \begin{equation} C_2(x_1, x_2) = \int_0^t v_3(t, x_2, 0) \, dt\,. \end{equation} With this choice of $G$, $C_1$ and $C_2$ we have $v = \curl G$ as desired.