Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: curl and divergence, parametric surfaces and surface area, scalar surface integrals, flux integrals, Stokes’ Theorem, the Divergence Theorem, and the unified FTC patternCALC III第 1 至 7 节：旋度与散度、参数曲面与曲面面积、标量曲面积分、通量积分、斯托克斯定理、散度定理以及统一的微积分基本定理模式CALC III
Let $\mathbf{F} = \langle xy^2,\; yz^2,\; x^2z \rangle$. (a) Compute $\nabla \times \mathbf{F}$. (b) Compute $\nabla \cdot \mathbf{F}$. (c) Evaluate $\nabla \cdot (\nabla \times \mathbf{F})$ without further computation.
Set $P = xy^2$, $Q = yz^2$, $R = x^2 z$. Expanding the formal determinant: (M1)令 $P = xy^2$，$Q = yz^2$，$R = x^2 z$。展开形式行列式：(M1)
$$ \nabla \times \mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z},\; \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x},\; \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle. $$Component by component (M1): $R_y - Q_z = 0 - 2yz = -2yz$; $P_z - R_x = 0 - 2xz = -2xz$; $Q_x - P_y = 0 - 2xy = -2xy$. (A1 each for correct components)逐分量计算 (M1)：$R_y - Q_z = 0 - 2yz = -2yz$；$P_z - R_x = 0 - 2xz = -2xz$；$Q_x - P_y = 0 - 2xy = -2xy$。（每个正确分量各得 A1）
$$ \nabla \times \mathbf{F} = \langle -2yz,\; -2xz,\; -2xy \rangle. $$$\nabla \cdot \mathbf{F} = \dfrac{\partial}{\partial x}(xy^2) + \dfrac{\partial}{\partial y}(yz^2) + \dfrac{\partial}{\partial z}(x^2 z) = y^2 + z^2 + x^2.$ (M1·A1)
By the identity that the divergence of any curl is identically zero, $\nabla \cdot (\nabla \times \mathbf{F}) = 0$. (R1) This holds because the six second-order mixed partial terms in the expansion cancel in Clairaut pairs: $R_{yx} = R_{xy}$, $Q_{zx} = Q_{xz}$, $P_{zy} = P_{yz}$, provided $\mathbf{F}$ has continuous second partial derivatives. (A1)根据"任意旋度的散度恒为零"这一恒等式，$\nabla \cdot (\nabla \times \mathbf{F}) = 0$。(R1) 这是因为展开式中六个二阶混合偏导数项在克莱罗对中两两抵消：$R_{yx} = R_{xy}$，$Q_{zx} = Q_{xz}$，$P_{zy} = P_{yz}$，前提是 $\mathbf{F}$ 具有连续的二阶偏导数。(A1)
(a) Surface area of the plane $z = 2x + 3y$ over $D = [0,1]\times[0,1]$. (b) Area of the paraboloid $z = x^2 + y^2$ over $x^2 + y^2 \le 1$.
With $g(x,y) = 2x + 3y$ we have $g_x = 2$ and $g_y = 3$. (M1) The area element is对 $g(x,y) = 2x + 3y$，有 $g_x = 2$，$g_y = 3$。(M1) 面积元素为
$$ dS = \sqrt{1 + g_x^2 + g_y^2}\, dA = \sqrt{1 + 4 + 9}\, dA = \sqrt{14}\, dA. $$(A1) Since $\sqrt{14}$ is constant, the surface area equals $\sqrt{14}$ times the area of $D$:(A1) 由于 $\sqrt{14}$ 为常数，曲面面积等于 $\sqrt{14}$ 乘以 $D$ 的面积：
$$ A = \sqrt{14} \iint_D dA = \sqrt{14} \cdot 1 = \sqrt{14}. $$(A1)
With $g = x^2 + y^2$ we have $g_x = 2x$ and $g_y = 2y$, so $g_x^2 + g_y^2 = 4(x^2 + y^2) = 4r^2$. (M1) The area element in polar is对 $g = x^2 + y^2$，有 $g_x = 2x$，$g_y = 2y$，故 $g_x^2 + g_y^2 = 4(x^2 + y^2) = 4r^2$。(M1) 极坐标下的面积元素为
$$ dS = \sqrt{1 + 4r^2}\, dA = \sqrt{1 + 4r^2}\, r\, dr\, d\theta. $$(A1 for the element) The area integral is:（面积元素得 A1）面积积分为：
$$ A = \int_0^{2\pi}\!\!\int_0^1 r\sqrt{1 + 4r^2}\, dr\, d\theta = 2\pi \int_0^1 r\sqrt{1 + 4r^2}\, dr. $$(M1 for pulling out $2\pi$) Set $u = 1 + 4r^2$, $du = 8r\, dr$, so $r\, dr = du/8$. When $r=0$, $u=1$; when $r=1$, $u=5$. (M1)（提出 $2\pi$ 得 M1）令 $u = 1 + 4r^2$，$du = 8r\, dr$，故 $r\, dr = du/8$。当 $r=0$ 时 $u=1$，当 $r=1$ 时 $u=5$。(M1)
$$ 2\pi \cdot \frac{1}{8} \int_1^5 \sqrt{u}\, du = \frac{\pi}{4} \cdot \left[\frac{2}{3} u^{3/2}\right]_1^5 = \frac{\pi}{6}\left(5^{3/2} - 1\right) = \frac{\pi}{6}(5\sqrt{5} - 1). $$(A1)
Cone $z = \sqrt{x^2+y^2}$, $0 \le z \le 2$, density $\rho = z^2$. Find the total mass $m = \iint_S z^2\, dS$.
With $g = \sqrt{x^2 + y^2}$ we compute $g_x = \dfrac{x}{\sqrt{x^2+y^2}}$ and $g_y = \dfrac{y}{\sqrt{x^2+y^2}}$, so $g_x^2 + g_y^2 = \dfrac{x^2 + y^2}{x^2 + y^2} = 1$. (M1) Therefore对 $g = \sqrt{x^2 + y^2}$，计算得 $g_x = \dfrac{x}{\sqrt{x^2+y^2}}$，$g_y = \dfrac{y}{\sqrt{x^2+y^2}}$，故 $g_x^2 + g_y^2 = \dfrac{x^2 + y^2}{x^2 + y^2} = 1$。(M1) 因此
$$ dS = \sqrt{1 + 1}\, dA = \sqrt{2}\, dA. $$(A1)
On the cone $z = \sqrt{x^2 + y^2} = r$, so the density is $z^2 = r^2$. The shadow region is the disk $r \le 2$. (M1)在锥面上 $z = \sqrt{x^2 + y^2} = r$，故密度为 $z^2 = r^2$。投影区域为圆盘 $r \le 2$。(M1)
$$ m = \iint_D r^2 \cdot \sqrt{2}\, dA = \sqrt{2}\int_0^{2\pi}\!\!\int_0^2 r^2 \cdot r\, dr\, d\theta = \sqrt{2}\int_0^{2\pi}\!\!\int_0^2 r^3\, dr\, d\theta. $$(A1)
(M1·A1)
$\mathbf{F} = \langle y,\; z,\; x \rangle$. Surface: $z = 4 - x - y$ over the triangle $D$ with vertices $(0,0)$, $(2,0)$, $(0,2)$, upward orientation. Find $\iint_S \mathbf{F} \cdot d\mathbf{S}$.
For a graph $z = g(x,y) = 4 - x - y$ the upward vector element is $d\mathbf{S} = \langle -g_x, -g_y, 1\rangle\, dA$. With $g_x = -1$ and $g_y = -1$: (M1)对于图形曲面 $z = g(x,y) = 4 - x - y$，向上向量元素为 $d\mathbf{S} = \langle -g_x, -g_y, 1\rangle\, dA$。由 $g_x = -1$，$g_y = -1$：(M1)
$$ d\mathbf{S} = \langle 1,\; 1,\; 1\rangle\, dA. $$(A1) Check: the $z$-component is $+1 > 0$, confirming upward orientation.(A1) 验证：$z$ 分量为 $+1 > 0$，确认为向上方向。
On the surface, substitute $z = 4 - x - y$ into $\mathbf{F} = \langle y,\; z,\; x\rangle = \langle y,\; 4-x-y,\; x\rangle$. (M1)在曲面上，将 $z = 4 - x - y$ 代入 $\mathbf{F} = \langle y,\; z,\; x\rangle = \langle y,\; 4-x-y,\; x\rangle$。(M1)
$$ \mathbf{F} \cdot d\mathbf{S} = (y \cdot 1 + (4-x-y) \cdot 1 + x \cdot 1)\, dA = (y + 4 - x - y + x)\, dA = 4\, dA. $$The integrand is the constant $4$. (A1) This collapse occurs because the components of $\mathbf{F}$ are a cyclic permutation of $\langle x, y, z\rangle$, so $P + Q + R = x + y + z = 4$ on the surface $x + y + z = 4$.被积表达式为常数 $4$。(A1) 这一化简成立，是因为 $\mathbf{F}$ 的各分量是 $\langle x, y, z\rangle$ 的循环置换，故在曲面 $x + y + z = 4$ 上 $P + Q + R = x + y + z = 4$。
The triangle $D$ has vertices $(0,0)$, $(2,0)$, $(0,2)$, so it has base and height both $2$. (M1)三角形 $D$ 的顶点为 $(0,0)$、$(2,0)$、$(0,2)$，底和高均为 $2$。(M1)
$$ A(D) = \tfrac{1}{2} \cdot 2 \cdot 2 = 2, \qquad \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D 4\, dA = 4 \cdot 2 = 8. \quad\text{(A1)} $$$\mathbf{F} = \langle -y^2,\; x,\; z \rangle$. $S$: disk $x^2 + y^2 \le 4$, $z = 1$, upward normal. $C$: circle $x^2+y^2=4$, $z=1$, counterclockwise from above. (a) $\nabla \times \mathbf{F}$. (b) Curl-flux side. (c) Circulation side.
With $P = -y^2$, $Q = x$, $R = z$: (M1)令 $P = -y^2$，$Q = x$，$R = z$：(M1)
$$ \nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y \rangle = \langle 0 - 0,\; 0 - 0,\; 1 - (-2y) \rangle = \langle 0,\; 0,\; 1 + 2y \rangle. $$(A1)
On the flat disk $z = 1$ with upward orientation, $d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$. (M1)在取向上方向的平面圆盘 $z = 1$ 上，$d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$。(M1)
$$ \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{x^2+y^2 \le 4} (1 + 2y)\, dA. $$(M1) Split: $\iint_{D} 1\, dA = \pi(2)^2 = 4\pi$ and $\iint_D 2y\, dA = 0$ by symmetry ($y$ is odd over a disk centred at the origin). (A1)(M1) 拆分：$\iint_{D} 1\, dA = \pi(2)^2 = 4\pi$，由对称性 $\iint_D 2y\, dA = 0$（$y$ 在以原点为中心的圆盘上为奇函数）。(A1)
$$ \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = 4\pi + 0 = 4\pi. $$Parametrize $C$ by $\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 1\rangle$, $0 \le t \le 2\pi$, so $\mathbf{r}'(t) = \langle -2\sin t,\; 2\cos t,\; 0\rangle$. On $C$: $\mathbf{F} = \langle -(2\sin t)^2,\; 2\cos t,\; 1\rangle = \langle -4\sin^2 t,\; 2\cos t,\; 1\rangle$. (M1)用 $\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 1\rangle$，$0 \le t \le 2\pi$ 参数化 $C$，故 $\mathbf{r}'(t) = \langle -2\sin t,\; 2\cos t,\; 0\rangle$。在 $C$ 上：$\mathbf{F} = \langle -(2\sin t)^2,\; 2\cos t,\; 1\rangle = \langle -4\sin^2 t,\; 2\cos t,\; 1\rangle$。(M1)
$$ \mathbf{F} \cdot \mathbf{r}'(t) = (-4\sin^2 t)(-2\sin t) + (2\cos t)(2\cos t) + 1 \cdot 0 = 8\sin^3 t + 4\cos^2 t. $$(M1) Now integrate from $0$ to $2\pi$:(M1) 从 $0$ 到 $2\pi$ 积分：
$$ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} (8\sin^3 t + 4\cos^2 t)\, dt. $$The integral of $\sin^3 t$ over a full period is $0$ (odd power of sine). The integral of $\cos^2 t$ over $[0,2\pi]$ is $\pi$. So the circulation is $0 + 4\pi = 4\pi$. (A1) Both sides equal $4\pi$, confirming Stokes’ Theorem.$\sin^3 t$ 在整个周期上的积分为 $0$（正弦的奇次幂）。$\cos^2 t$ 在 $[0,2\pi]$ 上的积分为 $\pi$。因此环量为 $0 + 4\pi = 4\pi$。(A1) 两侧均等于 $4\pi$，验证了斯托克斯定理。
(a) State the Divergence Theorem with all hypotheses. (b) $\mathbf{F} = \langle x,\; y^2,\; z^3 \rangle$, $E$: ball of radius $a$; find outward flux through $\partial E$. (c) Explain why the Divergence Theorem fails for $\mathbf{G} = \langle x,y,z\rangle/(x^2+y^2+z^2)^{3/2}$ when $E$ is the unit ball.
Let $E$ be a bounded solid region in $\mathbb{R}^3$ whose boundary $\partial E$ is a closed, piecewise smooth, outward-oriented surface. Let $\mathbf{F}$ be a vector field with continuous first partial derivatives on an open set containing $E$ and $\partial E$. Then (R1 for the hypotheses)设 $E$ 为 $\mathbb{R}^3$ 中的有界实心区域，其边界 $\partial E$ 为封闭的分段光滑向外定向曲面。设 $\mathbf{F}$ 为在包含 $E$ 和 $\partial E$ 的开集上具有连续一阶偏导数的向量场。则（前提条件得 R1）
$$ \iint_{\partial E} \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\, dV. $$(A1 for the correct formula; R1 for noting the outward orientation convention)（正确公式得 A1；注明向外方向约定得 R1）
Compute the divergence: $\nabla \cdot \mathbf{F} = \dfrac{\partial x}{\partial x} + \dfrac{\partial y^2}{\partial y} + \dfrac{\partial z^3}{\partial z} = 1 + 2y + 3z^2$. (M1)计算散度：$\nabla \cdot \mathbf{F} = \dfrac{\partial x}{\partial x} + \dfrac{\partial y^2}{\partial y} + \dfrac{\partial z^3}{\partial z} = 1 + 2y + 3z^2$。(M1)
By the Divergence Theorem the outward flux equals $\iiint_E (1 + 2y + 3z^2)\, dV$. (M1) Split into three integrals. Using spherical coordinates $x = \rho\sin\phi\cos\theta$, $y = \rho\sin\phi\sin\theta$, $z = \rho\cos\phi$, $dV = \rho^2\sin\phi\, d\rho\, d\phi\, d\theta$ over $0 \le \rho \le a$, $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$:由散度定理，向外通量等于 $\iiint_E (1 + 2y + 3z^2)\, dV$。(M1) 拆分为三个积分。用球坐标 $x = \rho\sin\phi\cos\theta$，$y = \rho\sin\phi\sin\theta$，$z = \rho\cos\phi$，$dV = \rho^2\sin\phi\, d\rho\, d\phi\, d\theta$，范围 $0 \le \rho \le a$，$0 \le \phi \le \pi$，$0 \le \theta \le 2\pi$：
Total outward flux $= \dfrac{4\pi a^3}{3} + 0 + \dfrac{4\pi a^5}{5}$.总向外通量 $= \dfrac{4\pi a^3}{3} + 0 + \dfrac{4\pi a^5}{5}$。
Away from the origin a direct calculation shows $\nabla \cdot \mathbf{G} = 0$. However, the Divergence Theorem requires $\mathbf{G}$ to have continuous first partial derivatives throughout $E$ and on $\partial E$. The field $\mathbf{G}$ is undefined (and blows up) at the origin, which lies inside the unit ball. Because one hypothesis is violated, the theorem cannot be applied; the conclusion $\iint_{\partial E}\mathbf{G}\cdot d\mathbf{S} = 0$ does not follow. In fact the flux is $4\pi \ne 0$. (R1·A1)在原点以外，直接计算表明 $\nabla \cdot \mathbf{G} = 0$。然而，散度定理要求 $\mathbf{G}$ 在整个 $E$ 及 $\partial E$ 上具有连续的一阶偏导数。向量场 $\mathbf{G}$ 在原点处无定义（且趋于无穷），而原点位于单位球内部。由于一个前提条件被违反，定理无法应用；结论 $\iint_{\partial E}\mathbf{G}\cdot d\mathbf{S} = 0$ 不成立。实际上通量为 $4\pi \ne 0$。(R1·A1)
(a) Write all four boundary theorems in a display, labelling the derivative and dimension for each. (b) Prove $\nabla \cdot (\nabla \times \mathbf{F}) = 0$ via Clairaut’s Theorem.
All four theorems share the form: integral of a derivative of $\mathbf{F}$ over a region equals an integral of $\mathbf{F}$ over its boundary. (A1 each for correct theorem)四个定理具有相同的形式：$\mathbf{F}$ 的某种导数在区域上的积分等于 $\mathbf{F}$ 在其边界上的积分。（每个正确定理各得 A1）
$$\underbrace{\int_C \nabla f \cdot d\mathbf{r} = f(B) - f(A)}_{\text{curve (1D) } \to \text{ endpoints (0D), derivative } = \nabla f}$$ $$\underbrace{\oint_{\partial D} \mathbf{F} \cdot d\mathbf{r} = \iint_D (Q_x - P_y)\, dA}_{\text{region (2D) } \to \text{ curve (1D), derivative } = Q_x - P_y}$$ $$\underbrace{\oint_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}}_{\text{surface (2D) } \to \text{ curve (1D), derivative } = \nabla \times \mathbf{F}}$$ $$\underbrace{\iint_{\partial E} \mathbf{F} \cdot d\mathbf{S} = \iiint_E (\nabla \cdot \mathbf{F})\, dV}_{\text{solid (3D) } \to \text{ surface (2D), derivative } = \nabla \cdot \mathbf{F}}$$Let $\mathbf{F} = \langle P, Q, R\rangle$ have continuous second partial derivatives. Then $\nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y\rangle$. (M1) Take the divergence of this vector:设 $\mathbf{F} = \langle P, Q, R\rangle$ 具有连续的二阶偏导数。则 $\nabla \times \mathbf{F} = \langle R_y - Q_z,\; P_z - R_x,\; Q_x - P_y\rangle$。(M1) 对该向量取散度：
$$ \nabla \cdot (\nabla \times \mathbf{F}) = \frac{\partial}{\partial x}(R_y - Q_z) + \frac{\partial}{\partial y}(P_z - R_x) + \frac{\partial}{\partial z}(Q_x - P_y). $$(M1) Expand:(M1) 展开：
$$ = R_{yx} - Q_{zx} + P_{zy} - R_{xy} + Q_{xz} - P_{yz}. $$By Clairaut’s Theorem (equality of mixed partials for functions with continuous second derivatives): $R_{yx} = R_{xy}$, $Q_{zx} = Q_{xz}$, $P_{zy} = P_{yz}$. (A1) Every term cancels against its Clairaut partner, leaving由克莱罗定理（具有连续二阶偏导数的函数的混合偏导数相等）：$R_{yx} = R_{xy}$，$Q_{zx} = Q_{xz}$，$P_{zy} = P_{yz}$。(A1) 每一项与其克莱罗对应项抵消，结果为
$$ \nabla \cdot (\nabla \times \mathbf{F}) = 0. \qquad \text{(R1 for citing Clairaut)} $$$\mathbf{F} = \langle z^2 - y,\; z^2 + x,\; 0 \rangle$. $S$: upper hemisphere $x^2+y^2+z^2=4$, $z \ge 0$, outward normal. $C$: circle $x^2+y^2=4$, $z=0$, counterclockwise from above. Find $\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}$.
With $P = z^2 - y$, $Q = z^2 + x$, $R = 0$: (M1)令 $P = z^2 - y$，$Q = z^2 + x$，$R = 0$：(M1)
$$ R_y - Q_z = 0 - 2z = -2z, \qquad P_z - R_x = 2z - 0 = 2z, \qquad Q_x - P_y = 1 - (-1) = 2. $$(A1·A1)
$$ \nabla \times \mathbf{F} = \langle -2z,\; 2z,\; 2 \rangle. $$To compute $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$ directly over the hemisphere, we would need the outward spherical normal $\mathbf{n} = \langle x,y,z\rangle/2$ on the surface $\rho=2$, leading to an integral of $(-2z)\frac{x}{2} + (2z)\frac{y}{2} + 2\frac{z}{2}$ times $4\sin\phi$ in spherical coordinates. The first two terms produce cross terms in $\phi$ and $\theta$ that do not simplify easily. (M1)若直接在半球面上计算 $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S}$，需要在曲面 $\rho=2$ 上使用向外球面法向量 $\mathbf{n} = \langle x,y,z\rangle/2$，在球坐标中产生 $(-2z)\frac{x}{2} + (2z)\frac{y}{2} + 2\frac{z}{2}$ 乘以 $4\sin\phi$ 的积分。前两项产生 $\phi$ 和 $\theta$ 的交叉项，不易化简。(M1)
By Stokes’ Theorem, $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \oint_C \mathbf{F}\cdot d\mathbf{r}$, and this circulation depends only on the boundary curve $C$, not on which surface fills it. We may therefore replace $S$ by the flat disk $D: x^2+y^2 \le 4$, $z=0$, which has the same oriented boundary $C$. The right-hand rule check: upward normal on $D$ (positive $z$-direction) pairs with counterclockwise $C$ when viewed from above, consistent with the outward-upward normal on $S$. (A1)由斯托克斯定理，$\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \oint_C \mathbf{F}\cdot d\mathbf{r}$，此环量仅取决于边界曲线 $C$，与填充哪个曲面无关。因此可以用平面圆盘 $D: x^2+y^2 \le 4$，$z=0$ 代替 $S$，它与 $C$ 具有相同的有向边界。右手定则验证：$D$ 上的向上法向量（$z$ 轴正方向）与从上方俯视的逆时针 $C$ 配对，与 $S$ 上的向外向上法向量一致。(A1)
On $C$, $z = 0$, so $\mathbf{F} = \langle -y,\; x,\; 0\rangle$. (M1) Parametrize $C$ by $\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 0\rangle$, $t \in [0,2\pi]$. Then $\mathbf{r}'(t) = \langle -2\sin t,\; 2\cos t,\; 0\rangle$ and (M1)在 $C$ 上，$z = 0$，故 $\mathbf{F} = \langle -y,\; x,\; 0\rangle$。(M1) 用 $\mathbf{r}(t) = \langle 2\cos t,\; 2\sin t,\; 0\rangle$，$t \in [0,2\pi]$ 参数化 $C$，则 $\mathbf{r}'(t) = \langle -2\sin t,\; 2\cos t,\; 0\rangle$，(M1)
$$ \mathbf{F}\big(\mathbf{r}(t)\big) = \langle -2\sin t,\; 2\cos t,\; 0\rangle. $$ $$ \mathbf{F} \cdot \mathbf{r}'(t) = (-2\sin t)(-2\sin t) + (2\cos t)(2\cos t) + 0 = 4\sin^2 t + 4\cos^2 t = 4. $$(M1) Therefore(M1) 因此
$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} 4\, dt = 8\pi. $$(A1) By Stokes’ Theorem, $\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = 8\pi$. (A1)(A1) 由斯托克斯定理，$\iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = 8\pi$。(A1)
$\mathbf{F} = \langle xz,\; yz,\; z^2 \rangle$. $E$: region between paraboloid $z = x^2+y^2$ and plane $z=1$. $S_{\text{par}}$: paraboloid cap. $T$: top disk $z=1$, $x^2+y^2 \le 1$. Find the outward flux through $S_{\text{par}}$, and the upward flux through it.
Compute: $\nabla\cdot\mathbf{F} = \dfrac{\partial(xz)}{\partial x} + \dfrac{\partial(yz)}{\partial y} + \dfrac{\partial(z^2)}{\partial z} = z + z + 2z = 4z$. (M1)计算：$\nabla\cdot\mathbf{F} = \dfrac{\partial(xz)}{\partial x} + \dfrac{\partial(yz)}{\partial y} + \dfrac{\partial(z^2)}{\partial z} = z + z + 2z = 4z$。(M1)
The region $E$ in cylindrical coordinates: $r^2 \le z \le 1$, $0 \le r \le 1$, $0 \le \theta \le 2\pi$. (M1) Apply the Divergence Theorem:区域 $E$ 在柱坐标下：$r^2 \le z \le 1$，$0 \le r \le 1$，$0 \le \theta \le 2\pi$。(M1) 应用散度定理：
$$ \iint_{\partial E} \mathbf{F}\cdot d\mathbf{S} = \iiint_E 4z\, dV = \int_0^{2\pi}\!\!\int_0^1\!\!\int_{r^2}^1 4z\, r\, dz\, dr\, d\theta. $$Integrate in $z$ first: $\int_{r^2}^1 4z\, dz = 2z^2\big|_{r^2}^1 = 2(1 - r^4)$. (M1)先对 $z$ 积分：$\int_{r^2}^1 4z\, dz = 2z^2\big|_{r^2}^1 = 2(1 - r^4)$。(M1)
$$ = 2\pi \int_0^1 2(1 - r^4)\, r\, dr = 4\pi \int_0^1 (r - r^5)\, dr = 4\pi \left[\frac{r^2}{2} - \frac{r^6}{6}\right]_0^1 = 4\pi \left(\frac{1}{2} - \frac{1}{6}\right) = 4\pi \cdot \frac{1}{3} = \frac{4\pi}{3}. $$(A1·A1) This is the total outward flux through the closed surface $\partial E = T \cup S_{\text{par,out}}$.（A1·A1）这是通过封闭曲面 $\partial E = T \cup S_{\text{par,out}}$ 的总向外通量。
On $T$: $z = 1$, outward from $E$ means upward, so $\mathbf{n} = \langle 0, 0, 1\rangle$ and $d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$. (M1)在 $T$ 上：$z = 1$，从 $E$ 向外即向上，故 $\mathbf{n} = \langle 0, 0, 1\rangle$，$d\mathbf{S} = \langle 0, 0, 1\rangle\, dA$。(M1)
$$ \iint_T \mathbf{F}\cdot d\mathbf{S} = \iint_{x^2+y^2 \le 1} z^2\big|_{z=1}\, dA = \iint_{x^2+y^2 \le 1} 1\, dA = \pi. $$(A1)
The total outward flux through $\partial E$ is the sum of the outward flux through $T$ and the outward flux through $S_{\text{par}}$ (with outward from $E$ meaning pointing downward, away from the solid). (M1)通过 $\partial E$ 的总向外通量等于通过 $T$ 的向外通量与通过 $S_{\text{par}}$ 的向外通量之和（从 $E$ 向外即指向固体外侧，即向下）。(M1)
$$ \frac{4\pi}{3} = \pi + \Phi_{\text{par,out}} \implies \Phi_{\text{par,out}} = \frac{4\pi}{3} - \pi = \frac{\pi}{3}. $$(A1) The outward-from-$E$ (downward) flux through the paraboloid is $+\tfrac{\pi}{3}$. The upward flux (opposite orientation) is $-\tfrac{\pi}{3}$. (A1)(A1) 抛物面从 $E$ 向外（向下）的通量为 $+\tfrac{\pi}{3}$。取反方向（向上）的通量为 $-\tfrac{\pi}{3}$。(A1)
Direct verification: on $z = x^2+y^2$, upward $d\mathbf{S} = \langle -2x,-2y,1\rangle dA$. $\mathbf{F}\cdot d\mathbf{S} = xz(-2x)+yz(-2y)+z^2 = z(-2x^2-2y^2+z) = r^2(-2r^2+r^2) = -r^4$. $\iint_{r\le 1}(-r^4)r\,dr\,d\theta = 2\pi\int_0^1(-r^5)dr = -2\pi/6 = -\pi/3$. The upward flux through the paraboloid is $-\tfrac{\pi}{3}$, consistent with the subtraction result, and the outward-from-$E$ (downward) flux is $+\tfrac{\pi}{3}$.直接验证：在 $z = x^2+y^2$ 上，向上 $d\mathbf{S} = \langle -2x,-2y,1\rangle dA$。$\mathbf{F}\cdot d\mathbf{S} = xz(-2x)+yz(-2y)+z^2 = z(-2x^2-2y^2+z) = r^2(-2r^2+r^2) = -r^4$。$\iint_{r\le 1}(-r^4)r\,dr\,d\theta = 2\pi\int_0^1(-r^5)dr = -2\pi/6 = -\pi/3$。抛物面的向上通量为 $-\tfrac{\pi}{3}$，与减法结果一致，从 $E$ 向外（向下）的通量为 $+\tfrac{\pi}{3}$。
$\mathbf{F} = \langle x^3,\; y^3,\; z^3 \rangle$. Closed cylinder $x^2+y^2 \le 1$, $0 \le z \le 2$, outward orientation. (a) Divergence Theorem. (b) Direct surface-by-surface verification.
$\nabla\cdot\mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(r^2 + z^2)$ in cylindrical. (M1) Set up the triple integral over $E$: $0 \le r \le 1$, $0 \le \theta \le 2\pi$, $0 \le z \le 2$. (M1)柱坐标下 $\nabla\cdot\mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(r^2 + z^2)$。(M1) 在 $E$ 上建立三重积分：$0 \le r \le 1$，$0 \le \theta \le 2\pi$，$0 \le z \le 2$。(M1)
$$ \iiint_E 3(r^2 + z^2)\, r\, dr\, d\theta\, dz = 2\pi \int_0^1\!\!\int_0^2 3r(r^2 + z^2)\, dz\, dr. $$ $$ = 2\pi \int_0^1 3r \left[r^2 z + \frac{z^3}{3}\right]_0^2 dr = 2\pi \int_0^1 3r\left(2r^2 + \frac{8}{3}\right) dr = 6\pi \int_0^1 \left(2r^3 + \frac{8r}{3}\right) dr. $$ $$ = 6\pi \left[\frac{r^4}{2} + \frac{4r^2}{3}\right]_0^1 = 6\pi\left(\frac{1}{2} + \frac{4}{3}\right) = 6\pi \cdot \frac{11}{6} = 11\pi. $$(A1·A1)
Lateral surface侧面 $r = 1$, $0 \le z \le 2$: outward normal $\mathbf{n} = \langle\cos\theta,\sin\theta,0\rangle$, $dS = d\theta\, dz$. On the surface $x = \cos\theta$, $y = \sin\theta$, so $\mathbf{F} = \langle\cos^3\theta,\sin^3\theta,z^3\rangle$ and向外法向量 $\mathbf{n} = \langle\cos\theta,\sin\theta,0\rangle$，$dS = d\theta\, dz$。在曲面上 $x = \cos\theta$，$y = \sin\theta$，故 $\mathbf{F} = \langle\cos^3\theta,\sin^3\theta,z^3\rangle$，
$$ \mathbf{F}\cdot\mathbf{n} = \cos^4\theta + \sin^4\theta. $$(M1) Using $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \tfrac{1}{2}\sin^2 2\theta$:(M1) 利用 $\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \tfrac{1}{2}\sin^2 2\theta$：
$$ \int_0^{2\pi}(\cos^4\theta + \sin^4\theta)\, d\theta = 2\pi - \frac{1}{2}\cdot\pi = \frac{3\pi}{2}. $$Lateral flux $= \dfrac{3\pi}{2} \cdot \int_0^2 dz = \dfrac{3\pi}{2} \cdot 2 = 3\pi$. (A1)侧面通量 $= \dfrac{3\pi}{2} \cdot \int_0^2 dz = \dfrac{3\pi}{2} \cdot 2 = 3\pi$。(A1)
Top disk顶部圆盘 $z = 2$, $r \le 1$: outward normal $\mathbf{n} = \langle 0,0,1\rangle$, $\mathbf{F}\cdot\mathbf{n} = z^3 = 8$. Flux $= 8 \cdot \pi(1)^2 = 8\pi$. (M1)向外法向量 $\mathbf{n} = \langle 0,0,1\rangle$，$\mathbf{F}\cdot\mathbf{n} = z^3 = 8$。通量 $= 8 \cdot \pi(1)^2 = 8\pi$。(M1)
Bottom disk底部圆盘 $z = 0$, $r \le 1$: outward normal $\mathbf{n} = \langle 0,0,-1\rangle$, $\mathbf{F}\cdot\mathbf{n} = -z^3\big|_{z=0} = 0$. Flux $= 0$. (A1)向外法向量 $\mathbf{n} = \langle 0,0,-1\rangle$，$\mathbf{F}\cdot\mathbf{n} = -z^3\big|_{z=0} = 0$。通量 $= 0$。(A1)
Total $= 3\pi + 8\pi + 0 = 11\pi$. Both methods agree.总计 $= 3\pi + 8\pi + 0 = 11\pi$。两种方法结果一致。