Sections 1 to 7: vector fields, scalar line integrals, work integrals, conservative fields, the Fundamental Theorem for Line Integrals, Green's TheoremCALC III第 1 至 7 节：向量场、标量曲线积分、功积分、保守场、曲线积分基本定理、格林定理CALC III
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PART I · CORE TECHNIQUES第 I 部分 · 核心技术Computational fluency · 28 marks计算熟练度 · 28 分
Parametrising and Integrating参数化与积分
Show all working. Parametrise the curve explicitly. For scalar line integrals, compute $ds=|\mathbf{r}'(t)|\,dt$ clearly before integrating. For vector line integrals, write out $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$ before integrating.写出所有计算步骤。明确给出曲线的参数化。对于标量曲线积分，在积分前清楚地计算 $ds=|\mathbf{r}'(t)|\,dt$。对于向量曲线积分，在积分前写出 $\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)$。
Q1MEDIUMCOREscalar line integral: mass of a wire标量曲线积分：导线的质量[8 marks]
A wire occupies the curve $C$: the quarter-circle of radius $3$ centred at the origin, from $(3,0)$ to $(0,3)$, traversed counterclockwise. Its linear density at a point $(x,y)$ is $\delta(x,y)=x^{2}+y^{2}$.一根导线占据曲线 $C$：以原点为圆心、半径为 $3$ 的四分之一圆，从 $(3,0)$ 到 $(0,3)$，沿逆时针方向遍历。其在点 $(x,y)$ 处的线密度为 $\delta(x,y)=x^{2}+y^{2}$。
(a)Parametrise $C$ and compute $ds$.对 $C$ 进行参数化并计算 $ds$。[2]
(b)Evaluate the scalar line integral $\displaystyle\int_{C}\delta\,ds$ to find the total mass.计算标量曲线积分 $\displaystyle\int_{C}\delta\,ds$ 以求总质量。[4]
(c)Explain in one sentence why the value of $\displaystyle\int_{C}\delta\,ds$ does not change if $C$ is traversed clockwise instead.用一句话解释为什么当 $C$ 改为顺时针方向遍历时，$\displaystyle\int_{C}\delta\,ds$ 的值不变。[2]
Q2MEDIUMCOREvector line integral: work along a piecewise path向量曲线积分：沿分段路径的功[8 marks]
Let $\mathbf{F}(x,y)=y\,\mathbf{i}+x^{2}\,\mathbf{j}$. Let $C$ be the piecewise path from $(0,0)$ to $(1,1)$ consisting of $C_{1}$: the horizontal segment along $y=0$, followed by $C_{2}$: the vertical segment along $x=1$.设 $\mathbf{F}(x,y)=y\,\mathbf{i}+x^{2}\,\mathbf{j}$。设 $C$ 为从 $(0,0)$ 到 $(1,1)$ 的分段路径，由 $C_{1}$：沿 $y=0$ 的水平线段，以及 $C_{2}$：沿 $x=1$ 的竖直线段组成。
(c)State the total work $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{C_{1}}\mathbf{F}\cdot d\mathbf{r}+\int_{C_{2}}\mathbf{F}\cdot d\mathbf{r}$.写出总功 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{C_{1}}\mathbf{F}\cdot d\mathbf{r}+\int_{C_{2}}\mathbf{F}\cdot d\mathbf{r}$。[2]
Q3HARDCOREconservative test and finding a potential function保守场判别与求势函数[12 marks]
Let $\mathbf{F}(x,y)=(2xy+e^{x})\,\mathbf{i}+(x^{2}+3y^{2})\,\mathbf{j}$.设 $\mathbf{F}(x,y)=(2xy+e^{x})\,\mathbf{i}+(x^{2}+3y^{2})\,\mathbf{j}$。
(a)Apply the conservative criterion $\partial P/\partial y = \partial Q/\partial x$ to determine whether $\mathbf{F}$ is conservative on $\mathbb{R}^{2}$. Show all partial derivatives.利用保守场判别准则 $\partial P/\partial y = \partial Q/\partial x$ 判断 $\mathbf{F}$ 在 $\mathbb{R}^{2}$ 上是否为保守场。写出全部偏导数。[3]
(b)Find a potential function $f$ such that $\nabla f = \mathbf{F}$. Start from $f_{x}=P$ and integrate with respect to $x$, then differentiate with respect to $y$ to determine the arbitrary function of $y$.求势函数 $f$ 使得 $\nabla f = \mathbf{F}$。从 $f_{x}=P$ 出发，对 $x$ 积分，再对 $y$ 求导以确定关于 $y$ 的任意函数。[5]
(c)Use the Fundamental Theorem for Line Integrals to evaluate $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ where $C$ is any smooth curve from $(0,0)$ to $(1,2)$.利用曲线积分基本定理计算 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$，其中 $C$ 为从 $(0,0)$ 到 $(1,2)$ 的任意光滑曲线。[4]
PART II · DEFINITIONS AND PROOF第 II 部分 · 定义与证明Rigorous arguments · 26 marks严格论证 · 26 分
Path Independence, Potential Theory, and Green's Theorem路径无关性、势理论与格林定理
These items are graded on the logic of the argument. State every hypothesis before invoking a theorem. When proving path independence, use the equivalences listed in the unit; do not assume the result. When using Green's Theorem, verify that the region is simply connected, the curve is positively oriented (counterclockwise), and all partial derivatives are continuous.本部分按论证逻辑评分。在使用定理之前，先陈述所有假设条件。在证明路径无关性时，使用本单元列出的等价关系，不要假设结论成立。在使用格林定理时，验证区域单连通、曲线正向（逆时针）定向，且所有偏导数连续。
Q4MEDIUMPROOFpath independence equivalent to conservative路径无关性等价于保守场[8 marks]
Let $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ be a vector field with continuous partial derivatives on a simply connected open region $D$.设 $\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ 为在单连通开区域 $D$ 上具有连续偏导数的向量场。
(a)State precisely what it means for $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ to be path independent on $D$.精确陈述 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ 在 $D$ 上路径无关的含义。[2]
(b)Prove that if $\mathbf{F}$ is conservative (i.e. $\mathbf{F}=\nabla f$ for some $f$) then the line integral is path independent. Your argument must use the Fundamental Theorem for Line Integrals.证明：若 $\mathbf{F}$ 为保守场（即存在 $f$ 使得 $\mathbf{F}=\nabla f$），则曲线积分路径无关。论证必须使用曲线积分基本定理。[3]
(c)Explain why, on a simply connected domain, the conservative criterion $\partial P/\partial y=\partial Q/\partial x$ is equivalent to path independence. Identify the role of simple connectivity.解释为什么在单连通区域上，保守场判别准则 $\partial P/\partial y=\partial Q/\partial x$ 等价于路径无关性。说明单连通性的作用。[3]
Q5HARDPROOFGreen's Theorem and area via a line integral格林定理与曲线积分求面积[10 marks]
Let $D$ be a simply connected region in the plane with boundary $C$ traversed counterclockwise, and let $P,Q$ have continuous first partial derivatives on an open set containing $D$.设 $D$ 为平面上的单连通区域，其边界 $C$ 沿逆时针方向遍历，且 $P,Q$ 在包含 $D$ 的某开集上具有连续的一阶偏导数。
(a)State the circulation form of Green's Theorem.陈述格林定理的环流形式。[2]
(b)By choosing $P$ and $Q$ so that $\partial Q/\partial x - \partial P/\partial y = 1$, derive the formula $\displaystyle A = \tfrac{1}{2}\oint_{C} x\,dy - y\,dx$ for the area of $D$. Show the three valid choices of $P$ and $Q$ that each yield this formula.通过选取 $P$ 和 $Q$ 使得 $\partial Q/\partial x - \partial P/\partial y = 1$，推导出 $D$ 的面积公式 $\displaystyle A = \tfrac{1}{2}\oint_{C} x\,dy - y\,dx$。给出满足此公式的三种有效的 $P$ 和 $Q$ 选取方式。[5]
(c)Use the formula to compute the area of the ellipse $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ by parametrising the boundary and evaluating the line integral directly.利用该公式，对椭圆 $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$ 的边界进行参数化，直接计算曲线积分，求椭圆的面积。[3]
Q6HARDPROOFvortex field and failure of path independence on a non-simply-connected domain涡旋场与非单连通区域上路径无关性的失效[8 marks]
Consider the vortex field $\mathbf{F}(x,y)=\dfrac{-y}{x^{2}+y^{2}}\,\mathbf{i}+\dfrac{x}{x^{2}+y^{2}}\,\mathbf{j}$, defined on $D=\mathbb{R}^{2}\setminus\{(0,0)\}$.考虑涡旋场 $\mathbf{F}(x,y)=\dfrac{-y}{x^{2}+y^{2}}\,\mathbf{i}+\dfrac{x}{x^{2}+y^{2}}\,\mathbf{j}$，定义在 $D=\mathbb{R}^{2}\setminus\{(0,0)\}$ 上。
(a)Show by direct computation that $\partial P/\partial y = \partial Q/\partial x$ on $D$.通过直接计算证明在 $D$ 上 $\partial P/\partial y = \partial Q/\partial x$。[3]
(b)Let $C$ be the unit circle traversed counterclockwise. Compute $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}$ directly by parametrising $C$.设 $C$ 为沿逆时针方向遍历的单位圆。通过对 $C$ 参数化，直接计算 $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}$。[3]
(c)Your answers to (a) and (b) appear to contradict the path-independence theorem. Identify the hypothesis that fails, and explain what this implies about the vortex field.(a) 与 (b) 的结果看似与路径无关性定理矛盾。指出失效的假设条件，并解释这对涡旋场意味着什么。[2]
PART III · APPLICATIONS AND SYNTHESIS第 III 部分 · 应用与综合Extended problems · 28 marks综合题 · 28 分
Work, Circulation, and Area功、环流与面积
Set up each integral clearly. Carry exact values throughout. For Green's Theorem problems, convert the line integral to a double integral over the enclosed region and integrate the double integral fully.清晰建立每个积分。全程保持精确值。对于格林定理题目，将曲线积分转化为所围区域上的二重积分，并完整计算该二重积分。
Q7HARDAPPLIEDwork via direct parametrisation vs FTLI shortcut直接参数化求功与曲线积分基本定理的对比[8 marks]
Let $\mathbf{G}(x,y)=(2xy^{3})\,\mathbf{i}+(3x^{2}y^{2})\,\mathbf{j}$, and let $C$ be the curve $y=\sin(\pi x/2)$ from $(0,0)$ to $(1,1)$.设 $\mathbf{G}(x,y)=(2xy^{3})\,\mathbf{i}+(3x^{2}y^{2})\,\mathbf{j}$，设 $C$ 为从 $(0,0)$ 到 $(1,1)$ 的曲线 $y=\sin(\pi x/2)$。
(a)Show that $\mathbf{G}$ is conservative and find a potential function $f$ with $\nabla f = \mathbf{G}$.证明 $\mathbf{G}$ 为保守场，并求势函数 $f$ 使得 $\nabla f = \mathbf{G}$。[4]
(b)Use the Fundamental Theorem for Line Integrals to evaluate $\displaystyle\int_{C}\mathbf{G}\cdot d\mathbf{r}$ without parametrising $C$.利用曲线积分基本定理计算 $\displaystyle\int_{C}\mathbf{G}\cdot d\mathbf{r}$，无需对 $C$ 参数化。[2]
(c)Verify by computing $\displaystyle\int_{C}\mathbf{G}\cdot d\mathbf{r}$ directly, using the parametrisation $x=t$, $y=\sin(\pi t/2)$, $t\in[0,1]$. Show that both methods give the same answer.通过参数化 $x=t$，$y=\sin(\pi t/2)$，$t\in[0,1]$，直接计算 $\displaystyle\int_{C}\mathbf{G}\cdot d\mathbf{r}$ 加以验证，证明两种方法结果一致。[2]
Q8HARDAPPLIEDGreen's Theorem: converting a hard line integral to a double integral格林定理：将复杂曲线积分转化为二重积分[8 marks]
Let $C$ be the boundary of the region $D$ bounded by $y=x^{2}$ and $y=x$ for $0\le x\le 1$, traversed counterclockwise. Let $\mathbf{H}(x,y)=(x^{3}-y^{3})\,\mathbf{i}+(x^{3}+y^{3})\,\mathbf{j}$.设 $C$ 为由 $y=x^{2}$ 和 $y=x$（$0\le x\le 1$）围成的区域 $D$ 的边界，沿逆时针方向遍历。设 $\mathbf{H}(x,y)=(x^{3}-y^{3})\,\mathbf{i}+(x^{3}+y^{3})\,\mathbf{j}$。
(a)Verify that Green's Theorem applies to $\mathbf{H}$ on $D$. Apply the theorem to express $\displaystyle\oint_{C}\mathbf{H}\cdot d\mathbf{r}$ as a double integral over $D$.验证格林定理可用于 $D$ 上的 $\mathbf{H}$。应用该定理，将 $\displaystyle\oint_{C}\mathbf{H}\cdot d\mathbf{r}$ 表示为 $D$ 上的二重积分。[3]
(b)Set up the double integral with explicit limits and evaluate it exactly.列出含明确积分限的二重积分，并精确计算。[3]
(c)State the answer and explain why evaluating the two boundary segments directly (along $y=x^{2}$ and along $y=x$) would have been significantly harder.写出答案，并解释为什么直接计算两段边界（沿 $y=x^{2}$ 和沿 $y=x$）会困难得多。[2]
Q9HARDAPPLIEDGreen's Theorem: flux form and area computation格林定理：通量形式与面积计算[6 marks]
Let $D$ be the region enclosed by the triangle with vertices $(0,0)$, $(2,0)$, and $(0,4)$, with boundary $C$ traversed counterclockwise. Let $\mathbf{K}(x,y)=(x+y^{2})\,\mathbf{i}+(x^{2}-y)\,\mathbf{j}$.设 $D$ 为顶点为 $(0,0)$、$(2,0)$ 和 $(0,4)$ 的三角形所围成的区域，边界 $C$ 沿逆时针方向遍历。设 $\mathbf{K}(x,y)=(x+y^{2})\,\mathbf{i}+(x^{2}-y)\,\mathbf{j}$。
(a)Use the flux form of Green's Theorem to express the outward flux $\displaystyle\oint_{C}\mathbf{K}\cdot\hat{\mathbf{n}}\,ds$ as a double integral, and evaluate it.利用格林定理的通量形式，将向外通量 $\displaystyle\oint_{C}\mathbf{K}\cdot\hat{\mathbf{n}}\,ds$ 表示为二重积分并计算。[3]
(b)Use the area formula $A=\tfrac{1}{2}\oint_{C}x\,dy-y\,dx$ to verify that the area of $D$ is $4$.利用面积公式 $A=\tfrac{1}{2}\oint_{C}x\,dy-y\,dx$ 验证 $D$ 的面积为 $4$。[3]
Q10HARDAPPLIEDvector field classification and synthesis向量场分类与综合[6 marks]
Four vector fields are listed below. For each, state (i) whether it is conservative on $\mathbb{R}^{2}$ (or on the given domain), giving the conservative test value, and (ii) if conservative, the exact value of $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ for $C$ the straight line from $(1,0)$ to $(0,1)$.下面列出四个向量场。对每个向量场，陈述：(i) 它在 $\mathbb{R}^{2}$（或给定区域）上是否为保守场，并给出保守场判别结果；(ii) 若为保守场，给出 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ 的精确值，其中 $C$ 为从 $(1,0)$ 到 $(0,1)$ 的直线段。