Unit C8: Surface Integrals, Stokes’, and the Divergence Theorem第 C8 单元：曲面积分（surface integral）、斯托克斯定理（Stokes' theorem）与散度定理（divergence theorem）

Let $\mathbf{F} = \langle xz,\; xyz,\; -y^2 \rangle$. Compute the curl and the divergence.

For the curl, use the determinant formula with $P = xz$, $Q = xyz$, $R = -y^2$:

$$ \nabla \times \mathbf{F} = \left\langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \right\rangle = \left\langle -2y - xy,\; x - 0,\; yz - 0 \right\rangle. $$

So $\nabla \times \mathbf{F} = \langle -2y - xy,\; x,\; yz \rangle$. For the divergence,

$$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(xyz) + \frac{\partial}{\partial z}(-y^2) = z + xz + 0 = z(1 + x). $$

设 $\mathbf{F} = \langle xz,\; xyz,\; -y^2 \rangle$。求旋度与散度。

求旋度，用行列式公式，取 $P = xz$，$Q = xyz$，$R = -y^2$：

$$ \nabla \times \mathbf{F} = \left\langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \right\rangle = \left\langle -2y - xy,\; x - 0,\; yz - 0 \right\rangle. $$

故 $\nabla \times \mathbf{F} = \langle -2y - xy,\; x,\; yz \rangle$。求散度：

$$ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(xz) + \frac{\partial}{\partial y}(xyz) + \frac{\partial}{\partial z}(-y^2) = z + xz + 0 = z(1 + x). $$

Decide whether $\mathbf{F} = \langle 2xy + z^2,\; x^2,\; 2xz \rangle$ can be conservative, and if so find a potential.

Because the curl of a gradient is zero, a necessary condition for $\mathbf{F} = \nabla f$ is $\nabla \times \mathbf{F} = \mathbf{0}$. Compute with $P = 2xy + z^2$, $Q = x^2$, $R = 2xz$:

$$ \nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle = \langle 0 - 0,\; 2z - 2z,\; 2x - 2x \rangle = \mathbf{0}. $$

The curl vanishes, and the domain is all of space, which is simply connected, so $\mathbf{F}$ is conservative. To recover $f$, integrate $P$ in $x$: $f = x^2 y + xz^2 + h(y, z)$. Then $f_y = x^2 + h_y$ must equal $Q = x^2$, so $h_y = 0$ and $h = k(z)$. Finally $f_z = 2xz + k'(z)$ must equal $R = 2xz$, so $k'(z) = 0$. A potential is $f(x,y,z) = x^2 y + x z^2$.

判断 $\mathbf{F} = \langle 2xy + z^2,\; x^2,\; 2xz \rangle$ 是否可能为保守场，若是则求出一个势函数（potential function）。

由于梯度的旋度为零，$\mathbf{F} = \nabla f$ 的必要条件是 $\nabla \times \mathbf{F} = \mathbf{0}$。取 $P = 2xy + z^2$，$Q = x^2$，$R = 2xz$ 计算：

$$ \nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle = \langle 0 - 0,\; 2z - 2z,\; 2x - 2x \rangle = \mathbf{0}. $$

旋度为零，且定义域为整个空间，是单连通的，故 $\mathbf{F}$ 是保守场。为求 $f$，将 $P$ 对 $x$ 积分：$f = x^2 y + xz^2 + h(y, z)$。则 $f_y = x^2 + h_y$ 须等于 $Q = x^2$，故 $h_y = 0$，$h = k(z)$。最后 $f_z = 2xz + k'(z)$ 须等于 $R = 2xz$，故 $k'(z) = 0$。一个势函数为 $f(x,y,z) = x^2 y + x z^2$。

For $\mathbf{F} = \langle -y,\; x,\; 0 \rangle$, compute both operators and interpret them.

Divergence: $\nabla \cdot \mathbf{F} = \partial_x(-y) + \partial_y(x) + \partial_z(0) = 0$. The field neither creates nor destroys fluid; the flow is incompressible. Curl: with $P = -y$, $Q = x$, $R = 0$,

$$ \nabla \times \mathbf{F} = \langle 0 - 0,\; 0 - 0,\; 1 - (-1) \rangle = \langle 0, 0, 2 \rangle. $$

The curl is a constant vector pointing along the positive $z$-axis with length $2$. A paddle wheel dropped anywhere in this field spins about the $z$-axis with angular velocity $1$, since the curl is twice the angular velocity. This is the prototype rotational field, and its nonzero curl confirms at a glance that it cannot be a gradient.

对 $\mathbf{F} = \langle -y,\; x,\; 0 \rangle$，计算两个算子并加以解释。

散度：$\nabla \cdot \mathbf{F} = \partial_x(-y) + \partial_y(x) + \partial_z(0) = 0$。该场既不产生也不消灭流体，流动是不可压缩的。旋度：取 $P = -y$，$Q = x$，$R = 0$，

$$ \nabla \times \mathbf{F} = \langle 0 - 0,\; 0 - 0,\; 1 - (-1) \rangle = \langle 0, 0, 2 \rangle. $$

旋度是沿 $z$ 轴正向、长度为 $2$ 的常向量。在此场中任意位置放入桨轮，它都绕 $z$ 轴以角速度 $1$ 旋转，因为旋度是角速度的两倍。这是旋转场的原型，其非零旋度一眼便确认它不可能是梯度。

Let $f$ have continuous second partials. The third component of $\nabla \times (\nabla f)$ is

$$ \frac{\partial}{\partial x}\!\left(\frac{\partial f}{\partial y}\right) - \frac{\partial}{\partial y}\!\left(\frac{\partial f}{\partial x}\right) = f_{yx} - f_{xy}. $$

By Clairaut's Theorem the mixed partials are equal, so this component is zero. The same cancellation occurs in the other two components, giving $\nabla \times (\nabla f) = \mathbf{0}$. This is the differential form of the statement that conservative fields are irrotational.

The companion identity $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ is proved the same way. Write $\nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle$ and take its divergence:

$$ \nabla \cdot (\nabla \times \mathbf{F}) = (R_y - Q_z)_x + (P_z - R_x)_y + (Q_x - P_y)_z. $$

Expanding gives six second-order terms, $R_{yx} - Q_{zx} + P_{zy} - R_{xy} + Q_{xz} - P_{yz}$. Provided $\mathbf{F}$ has continuous second partials, Clairaut pairs them: $R_{yx} = R_{xy}$, $Q_{zx} = Q_{xz}$, and $P_{zy} = P_{yz}$. Every term cancels against its mate, so the sum is identically zero. Geometrically this says a curl field is always source-free, which is why the curl flux of a field through a closed surface is automatically zero, a fact we will see again in the Divergence Theorem.

设 $f$ 有连续二阶偏导。$\nabla \times (\nabla f)$ 的第三个分量为

$$ \frac{\partial}{\partial x}\!\left(\frac{\partial f}{\partial y}\right) - \frac{\partial}{\partial y}\!\left(\frac{\partial f}{\partial x}\right) = f_{yx} - f_{xy}. $$

由克莱罗定理（Clairaut's Theorem），混合偏导相等，故此分量为零。其余两个分量发生同样的抵消，于是 $\nabla \times (\nabla f) = \mathbf{0}$。这是"保守场无旋"这一论断的微分形式。

配套恒等式 $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ 以同样方式证明。写出 $\nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle$ 并取其散度：

$$ \nabla \cdot (\nabla \times \mathbf{F}) = (R_y - Q_z)_x + (P_z - R_x)_y + (Q_x - P_y)_z. $$

展开得六个二阶项 $R_{yx} - Q_{zx} + P_{zy} - R_{xy} + Q_{xz} - P_{yz}$。只要 $\mathbf{F}$ 有连续二阶偏导，克莱罗定理将它们两两配对：$R_{yx} = R_{xy}$，$Q_{zx} = Q_{xz}$，$P_{zy} = P_{yz}$。每一项都与其配对项抵消，故和恒为零。几何上这说明旋度场总是无源的，这正是场穿过闭曲面的旋度通量自动为零的原因，我们在散度定理中会再次见到这一事实。

Find the area of the part of the paraboloid $z = x^2 + y^2$ lying below the plane $z = 4$.

With $g(x,y) = x^2 + y^2$ we have $g_x = 2x$, $g_y = 2y$, so

$$ A = \iint_D \sqrt{1 + 4x^2 + 4y^2}\; dA. $$

The region $D$ is the disk $x^2 + y^2 \le 4$. In polar coordinates the integrand becomes $\sqrt{1 + 4r^2}$ and $dA = r\, dr\, d\theta$:

$$ A = \int_0^{2\pi}\!\!\int_0^2 \sqrt{1 + 4r^2}\; r\, dr\, d\theta = 2\pi \cdot \frac{1}{12}\Big[(1 + 4r^2)^{3/2}\Big]_0^2 = \frac{\pi}{6}\left(17^{3/2} - 1\right). $$

求抛物面 $z = x^2 + y^2$ 位于平面 $z = 4$ 下方部分的面积。

取 $g(x,y) = x^2 + y^2$，则 $g_x = 2x$，$g_y = 2y$，故

$$ A = \iint_D \sqrt{1 + 4x^2 + 4y^2}\; dA. $$

区域 $D$ 是圆盘 $x^2 + y^2 \le 4$。在极坐标下被积函数变为 $\sqrt{1 + 4r^2}$，$dA = r\, dr\, d\theta$：

$$ A = \int_0^{2\pi}\!\!\int_0^2 \sqrt{1 + 4r^2}\; r\, dr\, d\theta = 2\pi \cdot \frac{1}{12}\Big[(1 + 4r^2)^{3/2}\Big]_0^2 = \frac{\pi}{6}\left(17^{3/2} - 1\right). $$

Recover the classical area $4\pi a^2$ of a sphere of radius $a$ from the parametrization $\mathbf{r}(\phi,\theta) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi\rangle$.

The two tangent vectors are

$$ \mathbf{r}_\phi = \langle a\cos\phi\cos\theta,\; a\cos\phi\sin\theta,\; -a\sin\phi\rangle, \qquad \mathbf{r}_\theta = \langle -a\sin\phi\sin\theta,\; a\sin\phi\cos\theta,\; 0\rangle. $$

Their cross product has length $\left|\mathbf{r}_\phi \times \mathbf{r}_\theta\right| = a^2 \sin\phi$ (a standard computation; the $z$-component is $a^2\sin\phi\cos\phi$ and the horizontal part contributes $a^2\sin^2\phi$, summing under the root to $a^2\sin\phi$ since $\sin\phi \ge 0$ on $[0,\pi]$). Therefore

$$ A = \int_0^{2\pi}\!\!\int_0^{\pi} a^2 \sin\phi\, d\phi\, d\theta = a^2 (2\pi)\big[-\cos\phi\big]_0^{\pi} = a^2 (2\pi)(2) = 4\pi a^2. $$

由参数化 $\mathbf{r}(\phi,\theta) = \langle a\sin\phi\cos\theta,\; a\sin\phi\sin\theta,\; a\cos\phi\rangle$ 重新得出半径为 $a$ 的球面的经典面积 $4\pi a^2$。

两个切向量为

$$ \mathbf{r}_\phi = \langle a\cos\phi\cos\theta,\; a\cos\phi\sin\theta,\; -a\sin\phi\rangle, \qquad \mathbf{r}_\theta = \langle -a\sin\phi\sin\theta,\; a\sin\phi\cos\theta,\; 0\rangle. $$

其叉积长度为 $\left|\mathbf{r}_\phi \times \mathbf{r}_\theta\right| = a^2 \sin\phi$（标准计算；$z$ 分量为 $a^2\sin\phi\cos\phi$，水平部分贡献 $a^2\sin^2\phi$，在根号下相加得 $a^2\sin\phi$，因为在 $[0,\pi]$ 上 $\sin\phi \ge 0$）。因此

$$ A = \int_0^{2\pi}\!\!\int_0^{\pi} a^2 \sin\phi\, d\phi\, d\theta = a^2 (2\pi)\big[-\cos\phi\big]_0^{\pi} = a^2 (2\pi)(2) = 4\pi a^2. $$

Find the area of the part of the cylinder $x^2 + y^2 = 1$ between $z = 0$ and the slanted plane $z = 2 + x$.

Parametrize by $\mathbf{r}(\theta, z) = \langle \cos\theta, \sin\theta, z\rangle$, so $dS = d\theta\, dz$. On the cylinder $x = \cos\theta$, so the surface runs from $z = 0$ up to $z = 2 + \cos\theta$, which is positive for every $\theta$. Hence

$$ A = \int_0^{2\pi}\!\!\int_0^{2 + \cos\theta} dz\, d\theta = \int_0^{2\pi} (2 + \cos\theta)\, d\theta = 2(2\pi) + 0 = 4\pi. $$

The $\cos\theta$ term integrates to zero over a full revolution, so the slanted top contributes the same area as a flat cut at the average height $z = 2$.

求柱面 $x^2 + y^2 = 1$ 介于 $z = 0$ 与斜平面 $z = 2 + x$ 之间部分的面积。

用 $\mathbf{r}(\theta, z) = \langle \cos\theta, \sin\theta, z\rangle$ 参数化，故 $dS = d\theta\, dz$。在柱面上 $x = \cos\theta$，于是曲面从 $z = 0$ 延伸到 $z = 2 + \cos\theta$，后者对每个 $\theta$ 都为正。因此

$$ A = \int_0^{2\pi}\!\!\int_0^{2 + \cos\theta} dz\, d\theta = \int_0^{2\pi} (2 + \cos\theta)\, d\theta = 2(2\pi) + 0 = 4\pi. $$

$\cos\theta$ 项在一整圈上积分为零，故斜顶贡献的面积与在平均高度 $z = 2$ 处的平切相同。

For the graph parametrization $\mathbf{r}(x,y) = \langle x, y, g(x,y)\rangle$, the tangent vectors are $\mathbf{r}_x = \langle 1, 0, g_x\rangle$ and $\mathbf{r}_y = \langle 0, 1, g_y\rangle$. Their cross product is

$$ \mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & g_x \\ 0 & 1 & g_y \end{vmatrix} = \langle -g_x,\; -g_y,\; 1\rangle. $$

Its length is $\sqrt{g_x^2 + g_y^2 + 1}$, which is exactly the area-scaling factor. The geometric content is that a unit square in the $xy$-plane is tilted into a parallelogram on the surface whose area exceeds the flat square by precisely the secant of the tilt angle. Since the surface normal makes angle $\gamma$ with the vertical and $\cos\gamma = 1/\sqrt{1 + g_x^2 + g_y^2}$, the area element is $dA / \cos\gamma = \sqrt{1 + g_x^2 + g_y^2}\, dA$, confirming the formula from pure geometry.

对图像参数化 $\mathbf{r}(x,y) = \langle x, y, g(x,y)\rangle$，切向量为 $\mathbf{r}_x = \langle 1, 0, g_x\rangle$ 与 $\mathbf{r}_y = \langle 0, 1, g_y\rangle$。其叉积为

$$ \mathbf{r}_x \times \mathbf{r}_y = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & g_x \\ 0 & 1 & g_y \end{vmatrix} = \langle -g_x,\; -g_y,\; 1\rangle. $$

其长度为 $\sqrt{g_x^2 + g_y^2 + 1}$，恰是面积缩放因子。几何含义是：$xy$ 平面上的单位正方形被倾斜成曲面上的平行四边形，其面积比平正方形大出恰好倾角的正割。由于曲面法向量与竖直方向成角 $\gamma$ 且 $\cos\gamma = 1/\sqrt{1 + g_x^2 + g_y^2}$，故面积元素为 $dA / \cos\gamma = \sqrt{1 + g_x^2 + g_y^2}\, dA$，从纯几何角度证实了该公式。

A thin shell is the cone $z = \sqrt{x^2 + y^2}$ for $0 \le z \le 1$, with surface density $\rho(x,y,z) = z$. Find its mass $m = \iint_S z\, dS$.

With $g = \sqrt{x^2 + y^2}$ we get $g_x = x/\sqrt{x^2+y^2}$ and $g_y = y/\sqrt{x^2+y^2}$, so $g_x^2 + g_y^2 = 1$ and $\sqrt{1 + g_x^2 + g_y^2} = \sqrt{2}$. On the cone $z = \sqrt{x^2 + y^2} = r$ in polar coordinates, and $D$ is the disk $r \le 1$:

$$ m = \iint_D \sqrt{x^2 + y^2}\, \sqrt{2}\; dA = \sqrt{2}\int_0^{2\pi}\!\!\int_0^1 r \cdot r\, dr\, d\theta = \sqrt{2}\,(2\pi)\,\frac{1}{3} = \frac{2\sqrt{2}\,\pi}{3}. $$

一个薄壳是圆锥 $z = \sqrt{x^2 + y^2}$（$0 \le z \le 1$），面密度为 $\rho(x,y,z) = z$。求其质量 $m = \iint_S z\, dS$。

取 $g = \sqrt{x^2 + y^2}$，得 $g_x = x/\sqrt{x^2+y^2}$，$g_y = y/\sqrt{x^2+y^2}$，故 $g_x^2 + g_y^2 = 1$，$\sqrt{1 + g_x^2 + g_y^2} = \sqrt{2}$。在锥面上，极坐标下 $z = \sqrt{x^2 + y^2} = r$，且 $D$ 是圆盘 $r \le 1$：

$$ m = \iint_D \sqrt{x^2 + y^2}\, \sqrt{2}\; dA = \sqrt{2}\int_0^{2\pi}\!\!\int_0^1 r \cdot r\, dr\, d\theta = \sqrt{2}\,(2\pi)\,\frac{1}{3} = \frac{2\sqrt{2}\,\pi}{3}. $$

Evaluate $\iint_S z^2\, dS$ where $S$ is the unit sphere $x^2 + y^2 + z^2 = 1$.

Use the spherical parametrization, on which $z = \cos\phi$ and $dS = \sin\phi\, d\phi\, d\theta$ since $a = 1$. Then

$$ \iint_S z^2\, dS = \int_0^{2\pi}\!\!\int_0^{\pi} \cos^2\phi \, \sin\phi\, d\phi\, d\theta = 2\pi \left[-\frac{\cos^3\phi}{3}\right]_0^{\pi} = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}. $$

As a sanity check, by symmetry $\iint_S x^2\, dS = \iint_S y^2\, dS = \iint_S z^2\, dS$, and their sum is $\iint_S (x^2 + y^2 + z^2)\, dS = \iint_S 1\, dS = 4\pi$. Dividing by three gives $4\pi/3$, matching the direct computation.

求 $\iint_S z^2\, dS$，其中 $S$ 是单位球面 $x^2 + y^2 + z^2 = 1$。

用球面参数化，其上 $z = \cos\phi$，且由于 $a = 1$，$dS = \sin\phi\, d\phi\, d\theta$。则

$$ \iint_S z^2\, dS = \int_0^{2\pi}\!\!\int_0^{\pi} \cos^2\phi \, \sin\phi\, d\phi\, d\theta = 2\pi \left[-\frac{\cos^3\phi}{3}\right]_0^{\pi} = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}. $$

作为合理性验算，由对称性 $\iint_S x^2\, dS = \iint_S y^2\, dS = \iint_S z^2\, dS$，三者之和为 $\iint_S (x^2 + y^2 + z^2)\, dS = \iint_S 1\, dS = 4\pi$。除以三得 $4\pi/3$，与直接计算一致。

Find the average value of the height $z$ over the upper unit hemisphere $S$.

The average is $\bar z = \dfrac{1}{A(S)} \iint_S z\, dS$. The hemisphere has area $A(S) = 2\pi$. With the spherical parametrization restricted to $0 \le \phi \le \pi/2$, $z = \cos\phi$ and $dS = \sin\phi\, d\phi\, d\theta$:

$$ \iint_S z\, dS = \int_0^{2\pi}\!\!\int_0^{\pi/2} \cos\phi \sin\phi\, d\phi\, d\theta = 2\pi \left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2} = 2\pi \cdot \frac{1}{2} = \pi. $$

Therefore $\bar z = \pi / (2\pi) = \tfrac{1}{2}$. The average height of a unit dome is exactly half its radius, a clean result that students can quote as a check on harder problems.

求高度 $z$ 在上半单位半球面 $S$ 上的平均值。

平均值为 $\bar z = \dfrac{1}{A(S)} \iint_S z\, dS$。半球面面积 $A(S) = 2\pi$。把球面参数化限制在 $0 \le \phi \le \pi/2$，有 $z = \cos\phi$，$dS = \sin\phi\, d\phi\, d\theta$：

$$ \iint_S z\, dS = \int_0^{2\pi}\!\!\int_0^{\pi/2} \cos\phi \sin\phi\, d\phi\, d\theta = 2\pi \left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2} = 2\pi \cdot \frac{1}{2} = \pi. $$

因此 $\bar z = \pi / (2\pi) = \tfrac{1}{2}$。单位穹顶的平均高度恰为其半径的一半，是一个干净的结论，学生可在更难的题目中引用作为验算。

Suppose $\tilde{\mathbf{r}}(s,t) = \mathbf{r}(u(s,t), v(s,t))$ is a smooth reparametrization with nonzero Jacobian $J = \partial(u,v)/\partial(s,t)$. By the chain rule the tangent vectors transform so that $\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t = J\,(\mathbf{r}_u \times \mathbf{r}_v)$, hence $\left|\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t\right| = |J|\,\left|\mathbf{r}_u \times \mathbf{r}_v\right|$. Substituting into the integral and applying the change-of-variables theorem for double integrals, the factor $|J|$ is exactly absorbed by $ds\, dt = dA / |J|$ relative to $du\, dv$. The two integrals are equal:

$$ \iint_{\tilde D} f\,\left|\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t\right|\, ds\, dt = \iint_D f\,\left|\mathbf{r}_u \times \mathbf{r}_v\right|\, du\, dv. $$

So $\iint_S f\, dS$ is a genuine property of the surface and the function, not of how we happened to chart it. The absolute value is essential here, and it is precisely why scalar surface integrals are blind to orientation, in contrast to the flux integrals of the next section.

设 $\tilde{\mathbf{r}}(s,t) = \mathbf{r}(u(s,t), v(s,t))$ 是一个雅可比行列式 $J = \partial(u,v)/\partial(s,t)$ 非零的光滑重参数化。由链式法则，切向量按 $\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t = J\,(\mathbf{r}_u \times \mathbf{r}_v)$ 变换，故 $\left|\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t\right| = |J|\,\left|\mathbf{r}_u \times \mathbf{r}_v\right|$。代入积分并对二重积分应用换元定理，因子 $|J|$ 恰好被相对于 $du\, dv$ 的 $ds\, dt = dA / |J|$ 吸收。两个积分相等：

$$ \iint_{\tilde D} f\,\left|\tilde{\mathbf{r}}_s \times \tilde{\mathbf{r}}_t\right|\, ds\, dt = \iint_D f\,\left|\mathbf{r}_u \times \mathbf{r}_v\right|\, du\, dv. $$

所以 $\iint_S f\, dS$ 是曲面与函数本身的真实属性，与我们恰好如何为它建图无关。这里绝对值至关重要，它正是标量曲面积分对定向不敏感的原因，与下一节的通量积分形成对照。

Find the upward flux of $\mathbf{F} = \langle 0, 0, z \rangle$ through the upper unit hemisphere $z = \sqrt{1 - x^2 - y^2}$.

Here $g = \sqrt{1 - x^2 - y^2}$, $g_x = -x/g$, $g_y = -y/g$, so the upward normal element is $d\mathbf{S} = \langle x/g,\; y/g,\; 1 \rangle\, dA$. On the surface $\mathbf{F} = \langle 0, 0, g \rangle$, hence $\mathbf{F} \cdot d\mathbf{S} = g\, dA$ over the unit disk:

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_{r \le 1} \sqrt{1 - r^2}\; r\, dr\, d\theta = 2\pi \int_0^1 r\sqrt{1 - r^2}\, dr = 2\pi \cdot \frac{1}{3} = \frac{2\pi}{3}. $$

求 $\mathbf{F} = \langle 0, 0, z \rangle$ 穿过上半单位半球面 $z = \sqrt{1 - x^2 - y^2}$ 的向上通量。

此处 $g = \sqrt{1 - x^2 - y^2}$，$g_x = -x/g$，$g_y = -y/g$，故向上的法向元素为 $d\mathbf{S} = \langle x/g,\; y/g,\; 1 \rangle\, dA$。在曲面上 $\mathbf{F} = \langle 0, 0, g \rangle$，因此在单位圆盘上 $\mathbf{F} \cdot d\mathbf{S} = g\, dA$：

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_{r \le 1} \sqrt{1 - r^2}\; r\, dr\, d\theta = 2\pi \int_0^1 r\sqrt{1 - r^2}\, dr = 2\pi \cdot \frac{1}{3} = \frac{2\pi}{3}. $$

Find the upward flux of $\mathbf{F} = \langle x, y, z\rangle$ through the part of the plane $z = 1 - x - y$ lying over the triangle $D$ with vertices $(0,0)$, $(1,0)$, $(0,1)$.

With $g = 1 - x - y$ we have $g_x = g_y = -1$, so the upward element is $d\mathbf{S} = \langle 1, 1, 1\rangle\, dA$. On the surface $z = 1 - x - y$, so

$$ \mathbf{F}\cdot d\mathbf{S} = \big(x + y + (1 - x - y)\big)\, dA = 1\, dA. $$

The integrand collapses to the constant $1$, so the flux is just the area of the triangle $D$, which is $\tfrac{1}{2}$. Therefore $\iint_S \mathbf{F}\cdot d\mathbf{S} = \tfrac{1}{2}$. Recognizing such collapses saves a great deal of integration.

求 $\mathbf{F} = \langle x, y, z\rangle$ 穿过平面 $z = 1 - x - y$ 位于以 $(0,0)$、$(1,0)$、$(0,1)$ 为顶点的三角形 $D$ 上方那部分的向上通量。

取 $g = 1 - x - y$，则 $g_x = g_y = -1$，故向上的元素为 $d\mathbf{S} = \langle 1, 1, 1\rangle\, dA$。在曲面上 $z = 1 - x - y$，于是

$$ \mathbf{F}\cdot d\mathbf{S} = \big(x + y + (1 - x - y)\big)\, dA = 1\, dA. $$

被积函数坍缩为常数 $1$，故通量就是三角形 $D$ 的面积 $\tfrac{1}{2}$。因此 $\iint_S \mathbf{F}\cdot d\mathbf{S} = \tfrac{1}{2}$。认出这类坍缩能省下大量积分。

Find the outward flux of $\mathbf{F} = \langle x, y, 0\rangle$ through the closed cylinder $x^2 + y^2 = 1$, $0 \le z \le 2$, including its top and bottom disks.

The closed surface has three pieces. On the top ($z = 2$) and bottom ($z = 0$) the outward normals are $\pm\langle 0, 0, 1\rangle$, but $\mathbf{F}$ has no $z$-component, so $\mathbf{F}\cdot\mathbf{n} = 0$ and both caps contribute zero flux. On the curved side, parametrize by $\mathbf{r}(\theta,z) = \langle\cos\theta,\sin\theta,z\rangle$; the outward normal element is $\mathbf{r}_\theta \times \mathbf{r}_z = \langle\cos\theta,\sin\theta,0\rangle\, d\theta\, dz$, and on the surface $\mathbf{F} = \langle\cos\theta,\sin\theta,0\rangle$, so $\mathbf{F}\cdot d\mathbf{S} = (\cos^2\theta + \sin^2\theta)\, d\theta\, dz = d\theta\, dz$. Hence

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \int_0^2\!\!\int_0^{2\pi} d\theta\, dz = 2\pi \cdot 2 = 4\pi. $$

As a check, $\nabla\cdot\mathbf{F} = 2$ and the solid has volume $2\pi$, so the Divergence Theorem predicts $2 \cdot 2\pi = 4\pi$, in agreement.

求 $\mathbf{F} = \langle x, y, 0\rangle$ 穿过闭柱面 $x^2 + y^2 = 1$（$0 \le z \le 2$，含上下两个圆盘）的外向通量。

闭曲面有三块。在顶面（$z = 2$）和底面（$z = 0$）上，外法向为 $\pm\langle 0, 0, 1\rangle$，但 $\mathbf{F}$ 无 $z$ 分量，故 $\mathbf{F}\cdot\mathbf{n} = 0$，两个端盖贡献的通量均为零。在侧曲面上，用 $\mathbf{r}(\theta,z) = \langle\cos\theta,\sin\theta,z\rangle$ 参数化；外法向元素为 $\mathbf{r}_\theta \times \mathbf{r}_z = \langle\cos\theta,\sin\theta,0\rangle\, d\theta\, dz$，在曲面上 $\mathbf{F} = \langle\cos\theta,\sin\theta,0\rangle$，故 $\mathbf{F}\cdot d\mathbf{S} = (\cos^2\theta + \sin^2\theta)\, d\theta\, dz = d\theta\, dz$。因此

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \int_0^2\!\!\int_0^{2\pi} d\theta\, dz = 2\pi \cdot 2 = 4\pi. $$

作为验算，$\nabla\cdot\mathbf{F} = 2$，立体体积为 $2\pi$，故散度定理预测 $2 \cdot 2\pi = 4\pi$，一致。

If two parametrizations $\mathbf{r}(u,v)$ and $\tilde{\mathbf{r}}(s,t)$ trace the same surface, the change of variables theorem shows that $\iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)\, dA$ has the same magnitude under both, because the Jacobian of the reparametrization rescales the cross product exactly as it rescales $dA$. What can change is the sign: if the new parametrization induces the opposite normal direction, the Jacobian is negative and the flux changes sign. This is why every flux statement must fix an orientation of $S$.

若两个参数化 $\mathbf{r}(u,v)$ 与 $\tilde{\mathbf{r}}(s,t)$ 描出同一曲面，换元定理表明 $\iint_D \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)\, dA$ 在两者下大小相同，因为重参数化的雅可比行列式对叉积的缩放恰与它对 $dA$ 的缩放相同。可能变的是符号：若新参数化诱导出相反的法向方向，雅可比行列式为负，通量便变号。这正是每个通量陈述都必须为 $S$ 固定一个定向的原因。

Let $\mathbf{F} = \langle -y, x, z \rangle$ and let $S$ be the disk $x^2 + y^2 \le 1$ in the plane $z = 0$, oriented upward, with boundary the unit circle $C$ traversed counterclockwise.

Circulation side: parametrize $C$ by $\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle$, so $\mathbf{r}'(t) = \langle -\sin t, \cos t, 0 \rangle$ and $\mathbf{F} = \langle -\sin t, \cos t, 0 \rangle$. Then

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} (\sin^2 t + \cos^2 t)\, dt = 2\pi. $$

Flux side: $\nabla \times \mathbf{F} = \langle 0, 0, 2 \rangle$, and the upward normal element on the flat disk is $d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$, so

$$ \iint_S (\nabla \times \mathbf{F})\cdot d\mathbf{S} = \iint_{r \le 1} 2\, dA = 2 \cdot \pi = 2\pi. $$

The two sides agree, as Stokes' Theorem requires.

设 $\mathbf{F} = \langle -y, x, z \rangle$，$S$ 为平面 $z = 0$ 内的圆盘 $x^2 + y^2 \le 1$，向上定向，其边界为逆时针绕行的单位圆 $C$。

环量一侧：用 $\mathbf{r}(t) = \langle \cos t, \sin t, 0 \rangle$ 参数化 $C$，则 $\mathbf{r}'(t) = \langle -\sin t, \cos t, 0 \rangle$，$\mathbf{F} = \langle -\sin t, \cos t, 0 \rangle$。于是

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} (\sin^2 t + \cos^2 t)\, dt = 2\pi. $$

通量一侧：$\nabla \times \mathbf{F} = \langle 0, 0, 2 \rangle$，平圆盘上向上的法向元素为 $d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$，故

$$ \iint_S (\nabla \times \mathbf{F})\cdot d\mathbf{S} = \iint_{r \le 1} 2\, dA = 2 \cdot \pi = 2\pi. $$

两侧相等，正如斯托克斯定理所要求。

Compute $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$ for $\mathbf{F} = \langle -y, x, xyz\rangle$, where $S$ is the upper hemisphere $x^2 + y^2 + z^2 = 1$, $z \ge 0$, oriented upward.

Computing the curl flux directly over the curved hemisphere is unpleasant. But $S$ has the same boundary as the flat disk $D: x^2 + y^2 \le 1$ in the plane $z = 0$, namely the unit circle $C$ traversed counterclockwise. By Stokes, both curl fluxes equal $\oint_C \mathbf{F}\cdot d\mathbf{r}$, so we may integrate over the disk instead. On $z = 0$ the field reduces to $\langle -y, x, 0\rangle$, and with $\mathbf{r}(t) = \langle\cos t, \sin t, 0\rangle$,

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \big(\sin^2 t + \cos^2 t\big)\, dt = 2\pi. $$

So the hemisphere curl flux is $2\pi$, obtained without ever parametrizing the hemisphere. The $xyz$ term in $\mathbf{F}$ vanishes on the boundary and therefore never enters the answer.

计算 $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$，其中 $\mathbf{F} = \langle -y, x, xyz\rangle$，$S$ 为上半球面 $x^2 + y^2 + z^2 = 1$（$z \ge 0$），向上定向。

在弯曲的半球面上直接计算旋度通量很麻烦。但 $S$ 与平面 $z = 0$ 内的平圆盘 $D: x^2 + y^2 \le 1$ 共享同一边界，即逆时针绕行的单位圆 $C$。由斯托克斯定理，两个旋度通量都等于 $\oint_C \mathbf{F}\cdot d\mathbf{r}$，故可改在圆盘上积分。在 $z = 0$ 上场化为 $\langle -y, x, 0\rangle$，取 $\mathbf{r}(t) = \langle\cos t, \sin t, 0\rangle$，

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \big(\sin^2 t + \cos^2 t\big)\, dt = 2\pi. $$

所以半球面旋度通量为 $2\pi$，全程无需参数化半球面。$\mathbf{F}$ 中的 $xyz$ 项在边界上为零，故根本不进入答案。

Let $\mathbf{F} = \langle z, x, y\rangle$ and let $C$ be the triangle with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, oriented counterclockwise as seen from the first octant. Find $\oint_C \mathbf{F}\cdot d\mathbf{r}$.

Rather than parametrize three line segments, use Stokes with $S$ the flat triangle in the plane $x + y + z = 1$. The curl is $\nabla\times\mathbf{F} = \langle 1 - 0,\; 1 - 0,\; 1 - 0\rangle = \langle 1, 1, 1\rangle$. The plane has upward unit normal $\mathbf{n} = \tfrac{1}{\sqrt 3}\langle 1, 1, 1\rangle$, so $(\nabla\times\mathbf{F})\cdot\mathbf{n} = 3/\sqrt 3 = \sqrt 3$, a constant. Therefore

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \sqrt 3 \cdot A(S). $$

The triangle has side length $\sqrt 2$ and is equilateral, so its area is $A(S) = \tfrac{\sqrt 3}{4}(\sqrt 2)^2 = \tfrac{\sqrt 3}{2}$. Hence the circulation is $\sqrt 3 \cdot \tfrac{\sqrt 3}{2} = \tfrac{3}{2}$.

设 $\mathbf{F} = \langle z, x, y\rangle$，$C$ 为以 $(1,0,0)$、$(0,1,0)$、$(0,0,1)$ 为顶点的三角形，从第一卦限看为逆时针定向。求 $\oint_C \mathbf{F}\cdot d\mathbf{r}$。

与其参数化三条线段，不如用斯托克斯定理，取 $S$ 为平面 $x + y + z = 1$ 内的平三角形。旋度为 $\nabla\times\mathbf{F} = \langle 1 - 0,\; 1 - 0,\; 1 - 0\rangle = \langle 1, 1, 1\rangle$。该平面的向上单位法向为 $\mathbf{n} = \tfrac{1}{\sqrt 3}\langle 1, 1, 1\rangle$，故 $(\nabla\times\mathbf{F})\cdot\mathbf{n} = 3/\sqrt 3 = \sqrt 3$，是常数。因此

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \sqrt 3 \cdot A(S). $$

该三角形边长为 $\sqrt 2$ 且为等边三角形，故面积为 $A(S) = \tfrac{\sqrt 3}{4}(\sqrt 2)^2 = \tfrac{\sqrt 3}{2}$。于是环量为 $\sqrt 3 \cdot \tfrac{\sqrt 3}{2} = \tfrac{3}{2}$。

The proof reduces the spatial statement to the planar Green's Theorem. Take a surface that is a graph $z = g(x,y)$ over a region $D$, oriented upward, with boundary curve $C$ projecting to the boundary $\partial D$. Parametrize the circulation by pushing the planar boundary up onto the surface. Writing $\mathbf{F} = \langle P, Q, R\rangle$ and substituting $z = g(x,y)$, the line integral $\oint_C \mathbf{F}\cdot d\mathbf{r}$ becomes a planar line integral over $\partial D$ with integrand involving $P$, $Q$, and the chain-rule terms $R\, g_x$ and $R\, g_y$.

Applying Green's Theorem to that planar integral converts it to a double integral over $D$ of a combination of partial derivatives. Collecting terms, that combination is exactly $(\nabla\times\mathbf{F})\cdot\langle -g_x, -g_y, 1\rangle$, which is the integrand of the curl flux through the graph. Hence $\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$ for a graph. A general orientable surface is cut into graph-like patches whose internal boundaries cancel in pairs, leaving only the outer boundary, which extends the result to the surfaces met in practice.

证明把空间陈述归约为平面的格林定理。取一个区域 $D$ 上的图像曲面 $z = g(x,y)$，向上定向，其边界曲线 $C$ 投影到边界 $\partial D$。把平面边界提升到曲面上来参数化环量。写 $\mathbf{F} = \langle P, Q, R\rangle$ 并代入 $z = g(x,y)$，线积分（line integral）$\oint_C \mathbf{F}\cdot d\mathbf{r}$ 变为 $\partial D$ 上的平面线积分，被积量含 $P$、$Q$ 以及链式法则项 $R\, g_x$ 与 $R\, g_y$。

对该平面积分应用格林定理，将其化为 $D$ 上若干偏导数组合的二重积分。整理各项，该组合恰为 $(\nabla\times\mathbf{F})\cdot\langle -g_x, -g_y, 1\rangle$，正是穿过图像的旋度通量的被积量。于是对图像有 $\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$。一般可定向曲面被切成若干图像状的小块，其内部边界两两抵消，只剩外边界，这就把结论推广到实际遇到的曲面。

Find the outward flux of $\mathbf{F} = \langle x, y, z \rangle$ through the sphere $x^2 + y^2 + z^2 = a^2$.

Computing flux directly is tedious, but the divergence is simple: $\nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3$. Let $E$ be the solid ball of radius $a$. Then

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 3\, dV = 3 \cdot \operatorname{Vol}(E) = 3 \cdot \frac{4}{3}\pi a^3 = 4\pi a^3. $$

求 $\mathbf{F} = \langle x, y, z \rangle$ 穿过球面 $x^2 + y^2 + z^2 = a^2$ 的外向通量。

直接算通量很繁琐，但散度很简单：$\nabla \cdot \mathbf{F} = 1 + 1 + 1 = 3$。设 $E$ 为半径 $a$ 的实心球。则

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 3\, dV = 3 \cdot \operatorname{Vol}(E) = 3 \cdot \frac{4}{3}\pi a^3 = 4\pi a^3. $$

Find the outward flux of $\mathbf{F} = \langle x^2,\; y^2,\; z^2\rangle$ through the surface of the cube $0 \le x, y, z \le 1$.

Six face integrals would work, but the divergence makes it one triple integral. Here $\nabla\cdot\mathbf{F} = 2x + 2y + 2z$, so

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E (2x + 2y + 2z)\, dV = \int_0^1\!\!\int_0^1\!\!\int_0^1 (2x + 2y + 2z)\, dx\, dy\, dz. $$

By symmetry each of the three terms integrates to the same value. For the first, $\int_0^1 2x\, dx = 1$ and the other two integrations over the unit cube contribute factors of $1$, giving $1$; the three terms sum to $1 + 1 + 1 = 3$. The total outward flux is $3$.

求 $\mathbf{F} = \langle x^2,\; y^2,\; z^2\rangle$ 穿过立方体 $0 \le x, y, z \le 1$ 表面的外向通量。

分六个面积分也行，但用散度只需一个三重积分。此处 $\nabla\cdot\mathbf{F} = 2x + 2y + 2z$，故

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E (2x + 2y + 2z)\, dV = \int_0^1\!\!\int_0^1\!\!\int_0^1 (2x + 2y + 2z)\, dx\, dy\, dz. $$

由对称性，三项各自积分得到相同的值。对第一项，$\int_0^1 2x\, dx = 1$，另两次在单位立方体上的积分各贡献因子 $1$，得 $1$；三项之和为 $1 + 1 + 1 = 3$。总外向通量为 $3$。

Find the upward flux of $\mathbf{F} = \langle 0, 0, z\rangle$ through the open paraboloid $S: z = x^2 + y^2$, $0 \le z \le 1$, oriented with an upward component.

$S$ is open, so the Divergence Theorem does not apply directly. Close the solid bowl $E$ by adding the top disk $T: z = 1$, $x^2 + y^2 \le 1$, oriented upward. On the closed surface $\partial E = S_{\text{down}} \cup T$, where the paraboloid must be oriented outward (downward, away from the solid) to match the outward convention. With $\nabla\cdot\mathbf{F} = 1$,

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 1\, dV = \operatorname{Vol}(E) = \int_0^{2\pi}\!\!\int_0^1\!\!\int_{r^2}^1 r\, dz\, dr\, d\theta = 2\pi\int_0^1 (r - r^3)\, dr = 2\pi\cdot\tfrac{1}{4} = \tfrac{\pi}{2}. $$

The flux through the top disk is $\iint_T \langle 0,0,1\rangle\cdot\langle 0,0,1\rangle\, dA = \pi$. The closed-surface flux equals the outward paraboloid flux plus the top flux: $\tfrac{\pi}{2} = \Phi_{\text{out}} + \pi$, so the outward (downward) paraboloid flux is $-\tfrac{\pi}{2}$. The problem wants the upward orientation, which is the negative, giving $+\tfrac{\pi}{2}$.

求 $\mathbf{F} = \langle 0, 0, z\rangle$ 穿过开抛物面 $S: z = x^2 + y^2$（$0 \le z \le 1$，带向上分量定向）的向上通量。

$S$ 是开的，故散度定理不能直接套用。给碗状立体 $E$ 添上顶部圆盘 $T: z = 1$，$x^2 + y^2 \le 1$，向上定向，将其封闭。在闭曲面 $\partial E = S_{\text{down}} \cup T$ 上，抛物面必须向外定向（向下，背离立体）以符合外向约定。由 $\nabla\cdot\mathbf{F} = 1$，

$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 1\, dV = \operatorname{Vol}(E) = \int_0^{2\pi}\!\!\int_0^1\!\!\int_{r^2}^1 r\, dz\, dr\, d\theta = 2\pi\int_0^1 (r - r^3)\, dr = 2\pi\cdot\tfrac{1}{4} = \tfrac{\pi}{2}. $$

穿过顶部圆盘的通量为 $\iint_T \langle 0,0,1\rangle\cdot\langle 0,0,1\rangle\, dA = \pi$。闭曲面通量等于抛物面外向通量加顶部通量：$\tfrac{\pi}{2} = \Phi_{\text{out}} + \pi$，故抛物面的外向（向下）通量为 $-\tfrac{\pi}{2}$。题目要的是向上定向，即其相反数，给出 $+\tfrac{\pi}{2}$。

Apply the Divergence Theorem to a small ball $B_\varepsilon$ of radius $\varepsilon$ centered at a point $\mathbf{p}$. Since $\nabla \cdot \mathbf{F}$ is continuous, it is nearly constant on $B_\varepsilon$, so

$$ \iint_{\partial B_\varepsilon} \mathbf{F}\cdot d\mathbf{S} = \iiint_{B_\varepsilon} (\nabla\cdot\mathbf{F})\, dV \approx (\nabla\cdot\mathbf{F})(\mathbf{p}) \cdot \operatorname{Vol}(B_\varepsilon). $$

Dividing by the volume and letting $\varepsilon \to 0$ gives

$$ (\nabla\cdot\mathbf{F})(\mathbf{p}) = \lim_{\varepsilon \to 0} \frac{1}{\operatorname{Vol}(B_\varepsilon)} \iint_{\partial B_\varepsilon} \mathbf{F}\cdot d\mathbf{S}. $$

This is the precise sense in which the divergence is the outward flux per unit volume at a point, independent of any coordinate system.

对以点 $\mathbf{p}$ 为心、半径 $\varepsilon$ 的小球 $B_\varepsilon$ 套用散度定理。由于 $\nabla \cdot \mathbf{F}$ 连续，它在 $B_\varepsilon$ 上近似为常数，故

$$ \iint_{\partial B_\varepsilon} \mathbf{F}\cdot d\mathbf{S} = \iiint_{B_\varepsilon} (\nabla\cdot\mathbf{F})\, dV \approx (\nabla\cdot\mathbf{F})(\mathbf{p}) \cdot \operatorname{Vol}(B_\varepsilon). $$

除以体积并令 $\varepsilon \to 0$，得

$$ (\nabla\cdot\mathbf{F})(\mathbf{p}) = \lim_{\varepsilon \to 0} \frac{1}{\operatorname{Vol}(B_\varepsilon)} \iint_{\partial B_\varepsilon} \mathbf{F}\cdot d\mathbf{S}. $$

这正是散度作为某点单位体积外向通量的精确含义，与任何坐标系无关。

You must evaluate the outward flux of $\mathbf{F} = \langle x^3, y^3, z^3 \rangle$ through the closed unit sphere. Which theorem applies, and what is the value?

The surface is closed and bounds a solid, so the Divergence Theorem is the right tool. Compute $\nabla\cdot\mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3\rho^2$ in spherical coordinates. Then

$$ \iiint_E 3\rho^2\, dV = \int_0^{2\pi}\!\!\int_0^\pi\!\!\int_0^1 3\rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta = 3 \cdot \frac{1}{5} \cdot 2 \cdot 2\pi = \frac{12\pi}{5}. $$

If instead the problem asked for circulation around a single closed curve, Stokes' Theorem would be the tool; for an endpoint difference, the Fundamental Theorem.

需要求 $\mathbf{F} = \langle x^3, y^3, z^3 \rangle$ 穿过闭单位球面的外向通量。该用哪个定理，其值是多少？

曲面是闭的且界出一个立体，故散度定理是正确工具。在球坐标下算 $\nabla\cdot\mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3\rho^2$。则

$$ \iiint_E 3\rho^2\, dV = \int_0^{2\pi}\!\!\int_0^\pi\!\!\int_0^1 3\rho^2 \cdot \rho^2 \sin\phi\, d\rho\, d\phi\, d\theta = 3 \cdot \frac{1}{5} \cdot 2 \cdot 2\pi = \frac{12\pi}{5}. $$

若题目问的是绕单条闭曲线的环量，则用斯托克斯定理；若问端点之差，则用基本定理。

Consider $\mathbf{F} = \dfrac{\langle x, y, z\rangle}{(x^2 + y^2 + z^2)^{3/2}}$, the inverse-square field. Show that $\nabla\cdot\mathbf{F} = 0$ away from the origin, yet the outward flux through any sphere centered at the origin is $4\pi$, not $0$.

A direct computation gives $\nabla\cdot\mathbf{F} = 0$ everywhere except the origin, where $\mathbf{F}$ is undefined. On the sphere of radius $a$, the outward unit normal is $\mathbf{n} = \langle x, y, z\rangle / a$, so on the surface $\mathbf{F}\cdot\mathbf{n} = \dfrac{x^2+y^2+z^2}{a \cdot a^3} = \dfrac{a^2}{a^4} = \dfrac{1}{a^2}$, a constant. Then

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \frac{1}{a^2}\cdot A(S) = \frac{1}{a^2}\cdot 4\pi a^2 = 4\pi. $$

The flux is independent of radius and nonzero, even though the divergence vanishes throughout the shell. The resolution is that the Divergence Theorem fails here because the origin, where $\mathbf{F}$ blows up, lies inside. This is the mathematics behind Gauss' law in electrostatics: the flux measures the enclosed charge, which the smooth divergence away from it cannot see.

考虑平方反比场 $\mathbf{F} = \dfrac{\langle x, y, z\rangle}{(x^2 + y^2 + z^2)^{3/2}}$。证明在原点之外 $\nabla\cdot\mathbf{F} = 0$，然而穿过任何以原点为心的球面的外向通量是 $4\pi$，而非 $0$。

直接计算给出除原点外处处 $\nabla\cdot\mathbf{F} = 0$，而在原点处 $\mathbf{F}$ 无定义。在半径 $a$ 的球面上，外向单位法向为 $\mathbf{n} = \langle x, y, z\rangle / a$，故在曲面上 $\mathbf{F}\cdot\mathbf{n} = \dfrac{x^2+y^2+z^2}{a \cdot a^3} = \dfrac{a^2}{a^4} = \dfrac{1}{a^2}$，是常数。则

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \frac{1}{a^2}\cdot A(S) = \frac{1}{a^2}\cdot 4\pi a^2 = 4\pi. $$

尽管散度在整个壳层中为零，通量却与半径无关且非零。原因在于：散度定理在此失效，因为 $\mathbf{F}$ 在其处爆破的原点落在内部。这正是静电学中高斯定律背后的数学：通量度量的是所围电荷，而远离电荷处的光滑散度看不到它。

Let $\mathbf{F} = \langle y, -x, z^3\rangle$ and let $S$ be the closed boundary of the solid cylinder $x^2 + y^2 \le 1$, $0 \le z \le 2$. Find the outward flux through $S$, and separately the circulation of $\mathbf{F}$ around the top rim $C$ (the circle $x^2 + y^2 = 1$, $z = 2$, counterclockwise from above).

For the flux, the surface is closed, so use the Divergence Theorem. $\nabla\cdot\mathbf{F} = 0 + 0 + 3z^2 = 3z^2$, hence

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_E 3z^2\, dV = \int_0^{2\pi}\!\!\int_0^1\!\!\int_0^2 3z^2\, r\, dz\, dr\, d\theta = (2\pi)\left(\tfrac{1}{2}\right)\big[z^3\big]_0^2 = (2\pi)\left(\tfrac{1}{2}\right)(8) = 8\pi. $$

For the circulation, use Stokes on the flat top disk $D$. The curl is $\nabla\times\mathbf{F} = \langle 0, 0, -2\rangle$, and the upward normal on $D$ is $\langle 0, 0, 1\rangle$, so $(\nabla\times\mathbf{F})\cdot\mathbf{n} = -2$. Thus $\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (-2)\, dA = -2\pi$. The two theorems answer two different questions about the same field on the same solid.

设 $\mathbf{F} = \langle y, -x, z^3\rangle$，$S$ 为实心柱 $x^2 + y^2 \le 1$（$0 \le z \le 2$）的闭边界。求穿过 $S$ 的外向通量，并另外求 $\mathbf{F}$ 绕顶边 $C$（圆 $x^2 + y^2 = 1$，$z = 2$，从上方看逆时针）的环量。

对通量，曲面是闭的，故用散度定理。$\nabla\cdot\mathbf{F} = 0 + 0 + 3z^2 = 3z^2$，因此

$$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_E 3z^2\, dV = \int_0^{2\pi}\!\!\int_0^1\!\!\int_0^2 3z^2\, r\, dz\, dr\, d\theta = (2\pi)\left(\tfrac{1}{2}\right)\big[z^3\big]_0^2 = (2\pi)\left(\tfrac{1}{2}\right)(8) = 8\pi. $$

对环量，在平顶圆盘 $D$ 上用斯托克斯定理。旋度为 $\nabla\times\mathbf{F} = \langle 0, 0, -2\rangle$，$D$ 上向上的法向为 $\langle 0, 0, 1\rangle$，故 $(\nabla\times\mathbf{F})\cdot\mathbf{n} = -2$。于是 $\oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (-2)\, dA = -2\pi$。两个定理回答的是关于同一场、同一立体的两个不同问题。

The repeating pattern is not a coincidence. In the language of differential forms, every one of these results is a special case of a single identity, the generalized Stokes' Theorem,

$$ \int_{\Omega} d\omega = \int_{\partial \Omega} \omega, $$

where $\omega$ is a differential form, $d$ is the exterior derivative, $\Omega$ is an oriented region, and $\partial\Omega$ is its boundary. Choosing $\omega$ to be a $0$-form (a function) recovers the Fundamental Theorem for line integrals; a $1$-form on a planar region recovers Green's Theorem; a $1$-form on a surface recovers the classical Stokes' Theorem; and a $2$-form on a solid recovers the Divergence Theorem. The gradient, curl, and divergence are the three faces of the single operator $d$ acting in dimension three. This is why all four theorems share the form "integral of a derivative over a region equals integral over the boundary," and learning to see them as one statement is what separates a memorized list from genuine understanding.

这个反复出现的模式并非巧合。用微分形式（differential form）的语言，这些结果中的每一个都是同一个恒等式（即广义斯托克斯定理）的特例，

$$ \int_{\Omega} d\omega = \int_{\partial \Omega} \omega, $$

其中 $\omega$ 是一个微分形式，$d$ 是外微分，$\Omega$ 是定向区域，$\partial\Omega$ 是其边界。取 $\omega$ 为 $0$ 形式（一个函数）便得线积分基本定理；平面区域上的 $1$ 形式便得格林定理；曲面上的 $1$ 形式便得经典斯托克斯定理；立体上的 $2$ 形式便得散度定理。梯度、旋度与散度，是单一算子 $d$ 在三维中作用的三副面孔。这正是四个定理都呈"导数在区域上的积分等于在边界上的积分"形式的原因，而学会把它们看作一个陈述，正是把一张死记的清单与真正的理解区分开来的关键。