Companion to the University-Style Practice Set大学风格练习题配套解答
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
Wire on the quarter-circle of radius $3$, from $(3,0)$ to $(0,3)$ counterclockwise, with density $\delta(x,y)=x^{2}+y^{2}$. Find the total mass.导线位于半径为 $3$ 的四分之一圆上，从 $(3,0)$ 到 $(0,3)$ 逆时针方向，线密度为 $\delta(x,y)=x^{2}+y^{2}$。求总质量。
Use the standard circle parametrisation: $\mathbf{r}(t)=(3\cos t,\,3\sin t)$ for $t\in\bigl[0,\tfrac{\pi}{2}\bigr]$. (M1)采用标准圆参数化：$\mathbf{r}(t)=(3\cos t,\,3\sin t)$，$t\in\bigl[0,\tfrac{\pi}{2}\bigr]$。(M1)
Then $\mathbf{r}'(t)=(-3\sin t,\,3\cos t)$, so $|\mathbf{r}'(t)|=\sqrt{9\sin^{2}t+9\cos^{2}t}=3$. Hence $ds=3\,dt$. (A1)则 $\mathbf{r}'(t)=(-3\sin t,\,3\cos t)$，故 $|\mathbf{r}'(t)|=\sqrt{9\sin^{2}t+9\cos^{2}t}=3$。因此 $ds=3\,dt$。(A1)
On the curve, $x^{2}+y^{2}=9\cos^{2}t+9\sin^{2}t=9$. (M1) Therefore:在曲线上，$x^{2}+y^{2}=9\cos^{2}t+9\sin^{2}t=9$。(M1) 因此：
$$ \int_{C}\delta\,ds = \int_{0}^{\pi/2} 9 \cdot 3\,dt = 27\int_{0}^{\pi/2}dt. $$(M1 for substituting $\delta=9$ and $ds=3\,dt$, A1 for the integrand $27$)(M1 对应代入 $\delta=9$ 和 $ds=3\,dt$，A1 对应被积项 $27$)
$$ = 27\cdot\frac{\pi}{2} = \frac{27\pi}{2}. $$(A1)
The scalar line integral $\int_{C}\delta\,ds$ measures mass (or arc length weighted by density). The arc-length element $ds=|\mathbf{r}'(t)|\,dt$ is always non-negative regardless of the direction of traversal. (R1) Reversing orientation changes the sign of $d\mathbf{r}$ (the vector element) but leaves $ds=|d\mathbf{r}|$ unchanged, so the integral takes the same value. (A1)标量曲线积分 $\int_{C}\delta\,ds$ 度量质量（或弧长按密度加权）。弧长元素 $ds=|\mathbf{r}'(t)|\,dt$ 始终非负，与遍历方向无关。(R1) 反转方向会改变 $d\mathbf{r}$（向量元素）的符号，但 $ds=|d\mathbf{r}|$ 保持不变，因此积分值不变。(A1)
$\mathbf{F}(x,y)=y\,\mathbf{i}+x^{2}\,\mathbf{j}$; piecewise path $C=C_1\cup C_2$ from $(0,0)$ to $(1,0)$ to $(1,1)$. Compute the total work.$\mathbf{F}(x,y)=y\,\mathbf{i}+x^{2}\,\mathbf{j}$；分段路径 $C=C_1\cup C_2$，从 $(0,0)$ 到 $(1,0)$ 再到 $(1,1)$。求总功。
Parametrise: $\mathbf{r}(t)=(t,0)$, $t\in[0,1]$, so $d\mathbf{r}=(dt,0)$. (M1)参数化：$\mathbf{r}(t)=(t,0)$，$t\in[0,1]$，故 $d\mathbf{r}=(dt,0)$。(M1)
On $C_{1}$: $\mathbf{F}(t,0)=0\,\mathbf{i}+t^{2}\,\mathbf{j}$. Then $\mathbf{F}\cdot d\mathbf{r}=0\cdot dt+t^{2}\cdot 0=0$. (A1)在 $C_{1}$ 上：$\mathbf{F}(t,0)=0\,\mathbf{i}+t^{2}\,\mathbf{j}$。则 $\mathbf{F}\cdot d\mathbf{r}=0\cdot dt+t^{2}\cdot 0=0$。(A1)
$$ \int_{C_{1}}\mathbf{F}\cdot d\mathbf{r}=\int_{0}^{1}0\,dt=0. $$(A1)
Parametrise: $\mathbf{r}(t)=(1,t)$, $t\in[0,1]$, so $d\mathbf{r}=(0,dt)$. (M1)参数化：$\mathbf{r}(t)=(1,t)$，$t\in[0,1]$，故 $d\mathbf{r}=(0,dt)$。(M1)
On $C_{2}$: $\mathbf{F}(1,t)=t\,\mathbf{i}+1\,\mathbf{j}$. Then $\mathbf{F}\cdot d\mathbf{r}=t\cdot 0+1\cdot dt=dt$. (A1)在 $C_{2}$ 上：$\mathbf{F}(1,t)=t\,\mathbf{i}+1\,\mathbf{j}$。则 $\mathbf{F}\cdot d\mathbf{r}=t\cdot 0+1\cdot dt=dt$。(A1)
$$ \int_{C_{2}}\mathbf{F}\cdot d\mathbf{r}=\int_{0}^{1}dt=1. $$(A1)
By additivity of line integrals over piecewise paths: (A1)由曲线积分对分段路径的可加性：(A1)
$$ \int_{C}\mathbf{F}\cdot d\mathbf{r}=0+1=1. $$(A1)
$\mathbf{F}(x,y)=(2xy+e^{x})\,\mathbf{i}+(x^{2}+3y^{2})\,\mathbf{j}$. (a) Test conservatism; (b) find $f$ with $\nabla f=\mathbf{F}$; (c) use FTLI to evaluate from $(0,0)$ to $(1,2)$.$\mathbf{F}(x,y)=(2xy+e^{x})\,\mathbf{i}+(x^{2}+3y^{2})\,\mathbf{j}$。(a) 判别保守性；(b) 求 $f$ 使 $\nabla f=\mathbf{F}$；(c) 用曲线积分基本定理计算从 $(0,0)$ 到 $(1,2)$ 的积分。
Here $P=2xy+e^{x}$ and $Q=x^{2}+3y^{2}$. Compute (M1):此处 $P=2xy+e^{x}$，$Q=x^{2}+3y^{2}$。计算 (M1)：
$$ \frac{\partial P}{\partial y}=2x, \qquad \frac{\partial Q}{\partial x}=2x. $$Since $\partial P/\partial y=\partial Q/\partial x=2x$ everywhere on $\mathbb{R}^{2}$ (a simply connected domain) and all partials are continuous, $\mathbf{F}$ is conservative. (A1) The equality of mixed partials is sufficient on a simply connected domain by the converse theorem. (R1)由于 $\partial P/\partial y=\partial Q/\partial x=2x$ 在 $\mathbb{R}^{2}$（单连通区域）上处处成立，且所有偏导数连续，故 $\mathbf{F}$ 为保守场。(A1) 在单连通区域上，混合偏导数相等由逆定理即为充分条件。(R1)
We need $f$ with $f_{x}=2xy+e^{x}$ and $f_{y}=x^{2}+3y^{2}$. Integrate $f_{x}$ with respect to $x$: (M1)需要求 $f$ 使得 $f_{x}=2xy+e^{x}$，$f_{y}=x^{2}+3y^{2}$。对 $x$ 积分 $f_{x}$：(M1)
$$ f(x,y)=\int(2xy+e^{x})\,dx = x^{2}y+e^{x}+g(y), $$where $g(y)$ is an arbitrary function of $y$ alone. (A1)其中 $g(y)$ 为仅关于 $y$ 的任意函数。(A1)
Now differentiate with respect to $y$ and match to $Q$: (M1)再对 $y$ 求导并与 $Q$ 比较：(M1)
$$ f_{y}=x^{2}+g'(y)=x^{2}+3y^{2}. $$Thus $g'(y)=3y^{2}$, so $g(y)=y^{3}+C$. (A1)故 $g'(y)=3y^{2}$，即 $g(y)=y^{3}+C$。(A1)
The potential function is $f(x,y)=x^{2}y+e^{x}+y^{3}$ (taking $C=0$). Verification: $\nabla f=(2xy+e^{x},\,x^{2}+3y^{2})=\mathbf{F}$. (A1)势函数为 $f(x,y)=x^{2}y+e^{x}+y^{3}$（取 $C=0$）。验证：$\nabla f=(2xy+e^{x},\,x^{2}+3y^{2})=\mathbf{F}$。(A1)
Since $\mathbf{F}=\nabla f$ and $C$ is any smooth curve from $(0,0)$ to $(1,2)$: (M1)由于 $\mathbf{F}=\nabla f$，且 $C$ 为从 $(0,0)$ 到 $(1,2)$ 的任意光滑曲线：(M1)
$$ \int_{C}\mathbf{F}\cdot d\mathbf{r}=f(1,2)-f(0,0). $$Evaluate: (A1) $f(1,2)=(1)^{2}(2)+e^{1}+(2)^{3}=2+e+8=e+10$.计算：(A1) $f(1,2)=(1)^{2}(2)+e^{1}+(2)^{3}=2+e+8=e+10$。
$f(0,0)=(0)^{2}(0)+e^{0}+(0)^{3}=0+1+0=1$.
Therefore $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}=(e+10)-1=e+9$. (A1)因此 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}=(e+10)-1=e+9$。(A1)
$\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ with continuous partial derivatives on a simply connected open region $D$. (a) Define path independence; (b) prove conservative implies path-independent via FTLI; (c) explain the role of simple connectivity.$\mathbf{F}=P\,\mathbf{i}+Q\,\mathbf{j}$ 在单连通开区域 $D$ 上具有连续偏导数。(a) 定义路径无关性；(b) 利用曲线积分基本定理证明保守场蕴含路径无关性；(c) 解释单连通性的作用。
The line integral $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ is path independent on $D$ if, for every pair of points $A,B\in D$, the value of the integral is the same for all piecewise smooth curves $C$ in $D$ with initial point $A$ and terminal point $B$. (A1)曲线积分 $\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}$ 在 $D$ 上路径无关，是指对 $D$ 中任意一对点 $A,B$，所有以 $A$ 为起点、$B$ 为终点的分段光滑曲线 $C$ 的积分值均相同。(A1)
Equivalently, $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}=0$ for every closed piecewise smooth curve $C$ in $D$. (A1)等价地，$D$ 中每条封闭分段光滑曲线 $C$ 均有 $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}=0$。(A1)
Suppose $\mathbf{F}=\nabla f$ for some differentiable $f$ on $D$, and let $C$ be any piecewise smooth curve in $D$ from $A$ to $B$. Let $\mathbf{r}(t)$ parametrise $C$ with $\mathbf{r}(a)=A$ and $\mathbf{r}(b)=B$. (M1)设 $D$ 上存在可微函数 $f$ 使得 $\mathbf{F}=\nabla f$，设 $C$ 为 $D$ 中从 $A$ 到 $B$ 的任意分段光滑曲线，$\mathbf{r}(t)$ 为其参数化，满足 $\mathbf{r}(a)=A$，$\mathbf{r}(b)=B$。(M1)
By the chain rule and $\mathbf{F}=\nabla f$:由链式法则和 $\mathbf{F}=\nabla f$：
$$ \int_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{a}^{b}\nabla f(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt=\int_{a}^{b}\frac{d}{dt}f(\mathbf{r}(t))\,dt=f(\mathbf{r}(b))-f(\mathbf{r}(a))=f(B)-f(A). $$(A1) This value depends only on the endpoints $A$ and $B$, not on the path $C$. (R1) Hence the integral is path independent.(A1) 此值仅取决于端点 $A$ 和 $B$，与路径 $C$ 无关。(R1) 故积分路径无关。
On a simply connected domain, every closed curve can be contracted to a point without leaving $D$, so every loop bounds a region entirely inside $D$. (M1)在单连通区域上，每条封闭曲线都可以在不离开 $D$ 的情况下收缩到一点，因此每个回路所围的区域完全在 $D$ 内部。(M1)
The circulation form of Green's Theorem then gives $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}=\iint_{R}\!\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$. If $\partial P/\partial y=\partial Q/\partial x$, the double integral is zero for every such $R$. (A1)格林定理的环流形式给出 $\displaystyle\oint_{C}\mathbf{F}\cdot d\mathbf{r}=\iint_{R}\!\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$。若 $\partial P/\partial y=\partial Q/\partial x$，则对每个这样的区域 $R$，二重积分为零。(A1)
On a non-simply-connected domain (such as the punctured plane $\mathbb{R}^{2}\setminus\{0\}$), a loop may encircle a hole not in $D$, so Green's Theorem cannot be applied, and $\partial P/\partial y=\partial Q/\partial x$ no longer guarantees conservatism. (R1)在非单连通区域（如去掉原点的平面 $\mathbb{R}^{2}\setminus\{0\}$）上，回路可能围绕不属于 $D$ 的空洞，格林定理无法应用，$\partial P/\partial y=\partial Q/\partial x$ 不再保证场为保守场。(R1)
State Green's Theorem (circulation form); derive $A=\tfrac{1}{2}\oint_C x\,dy-y\,dx$; use it to find the area of the ellipse $x^{2}/a^{2}+y^{2}/b^{2}=1$.陈述格林定理（环流形式）；推导 $A=\tfrac{1}{2}\oint_C x\,dy-y\,dx$；用它求椭圆 $x^{2}/a^{2}+y^{2}/b^{2}=1$ 的面积。
Let $D$ be a simply connected region in $\mathbb{R}^{2}$ with positively oriented (counterclockwise) boundary $C$. If $P$ and $Q$ have continuous first partial derivatives on an open set containing $D$, then: (A1)设 $D$ 为 $\mathbb{R}^{2}$ 中的单连通区域，其边界 $C$ 为正向（逆时针）定向。若 $P$ 和 $Q$ 在包含 $D$ 的某开集上具有连续的一阶偏导数，则：(A1)
$$ \oint_{C} P\,dx + Q\,dy = \iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA. $$(A1)
We want $\partial Q/\partial x-\partial P/\partial y=1$. Three valid choices are: (M1)需要 $\partial Q/\partial x-\partial P/\partial y=1$。三种有效选取为：(M1)
(A1 for all three) Green's Theorem gives $A=\iint_D 1\,dA=\oint_C Q\,dy-\oint_C P\,dx$ for any such choice. Using Choice 3: (M1)（三项均正确得 A1）格林定理给出 $A=\iint_D 1\,dA=\oint_C Q\,dy-\oint_C P\,dx$，对任意选取均成立。使用选取 3：(M1)
$$ A = \oint_{C}\frac{x}{2}\,dy-\frac{-y}{2}\,dx = \frac{1}{2}\oint_{C}x\,dy-y\,dx. $$(A1) The same formula results from Choices 1 and 2 by linearity. (A1)(A1) 由线性性，选取 1 和 2 得到相同公式。(A1)
Parametrise the boundary: $x=a\cos t$, $y=b\sin t$, $t\in[0,2\pi]$. Then $dx=-a\sin t\,dt$ and $dy=b\cos t\,dt$. (M1)对边界参数化：$x=a\cos t$，$y=b\sin t$，$t\in[0,2\pi]$。则 $dx=-a\sin t\,dt$，$dy=b\cos t\,dt$。(M1)
$$ A = \frac{1}{2}\int_{0}^{2\pi}\!\bigl[(a\cos t)(b\cos t)-(b\sin t)(-a\sin t)\bigr]dt = \frac{1}{2}\int_{0}^{2\pi}\!\bigl[ab\cos^{2}t+ab\sin^{2}t\bigr]dt. $$ $$ = \frac{ab}{2}\int_{0}^{2\pi}dt = \frac{ab}{2}\cdot 2\pi = \pi ab. $$(A1) The area of the ellipse is $\pi ab$. (A1)(A1) 椭圆面积为 $\pi ab$。(A1)
Vortex field $\mathbf{F}=\dfrac{-y}{x^{2}+y^{2}}\,\mathbf{i}+\dfrac{x}{x^{2}+y^{2}}\,\mathbf{j}$ on $\mathbb{R}^{2}\setminus\{(0,0)\}$. (a) Verify $\partial P/\partial y=\partial Q/\partial x$; (b) compute $\oint_{C}\mathbf{F}\cdot d\mathbf{r}$ on the unit circle; (c) resolve the apparent contradiction.涡旋场 $\mathbf{F}=\dfrac{-y}{x^{2}+y^{2}}\,\mathbf{i}+\dfrac{x}{x^{2}+y^{2}}\,\mathbf{j}$ 定义在 $\mathbb{R}^{2}\setminus\{(0,0)\}$ 上。(a) 验证 $\partial P/\partial y=\partial Q/\partial x$；(b) 在单位圆上计算 $\oint_{C}\mathbf{F}\cdot d\mathbf{r}$；(c) 解释表观矛盾。
Let $r^{2}=x^{2}+y^{2}$. Then $P=-y/r^{2}$ and $Q=x/r^{2}$. Using the quotient rule: (M1)设 $r^{2}=x^{2}+y^{2}$。则 $P=-y/r^{2}$，$Q=x/r^{2}$。用商法则：(M1)
$$ \frac{\partial P}{\partial y}=\frac{-r^{2}+y\cdot 2y}{r^{4}}=\frac{-r^{2}+2y^{2}}{r^{4}}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}. $$ $$ \frac{\partial Q}{\partial x}=\frac{r^{2}-x\cdot 2x}{r^{4}}=\frac{r^{2}-2x^{2}}{r^{4}}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}. $$(A1 for each partial) Thus $\partial P/\partial y=\partial Q/\partial x$ on all of $\mathbb{R}^{2}\setminus\{(0,0)\}$. (A1)（各偏导数各得 A1）故 $\partial P/\partial y=\partial Q/\partial x$ 在 $\mathbb{R}^{2}\setminus\{(0,0)\}$ 上处处成立。(A1)
Parametrise: $x=\cos t$, $y=\sin t$, $t\in[0,2\pi]$. Then $dx=-\sin t\,dt$, $dy=\cos t\,dt$, and $x^{2}+y^{2}=1$. (M1)参数化：$x=\cos t$，$y=\sin t$，$t\in[0,2\pi]$。则 $dx=-\sin t\,dt$，$dy=\cos t\,dt$，$x^{2}+y^{2}=1$。(M1)
$$ \oint_{C}\mathbf{F}\cdot d\mathbf{r}=\int_{0}^{2\pi}\left[(-\sin t)(-\sin t\,dt)+(\cos t)(\cos t\,dt)\right]=\int_{0}^{2\pi}(\sin^{2}t+\cos^{2}t)\,dt. $$ $$ =\int_{0}^{2\pi}1\,dt=2\pi. $$(A1) The circulation around the unit circle is $2\pi\ne 0$. (A1)(A1) 单位圆上的环流量为 $2\pi\ne 0$。(A1)
The theorem that $\partial P/\partial y=\partial Q/\partial x$ implies conservatism requires the domain to be simply connected. (R1) The domain $D=\mathbb{R}^{2}\setminus\{(0,0)\}$ has a hole at the origin: the unit circle cannot be contracted to a point within $D$, so $D$ is not simply connected. Green's Theorem cannot be applied to the region enclosed by the unit circle because that region contains the singularity $(0,0)\notin D$. Therefore the equal-mixed-partials condition is necessary but not sufficient here, and $\mathbf{F}$ is not conservative: no single-valued potential function exists on all of $D$. (A1)$\partial P/\partial y=\partial Q/\partial x$ 蕴含保守性的定理要求区域单连通。(R1) 区域 $D=\mathbb{R}^{2}\setminus\{(0,0)\}$ 在原点有空洞：单位圆无法在 $D$ 内收缩到一点，故 $D$ 非单连通。格林定理无法应用于单位圆所围区域，因为该区域包含奇点 $(0,0)\notin D$。因此混合偏导数相等在此仅为必要条件而非充分条件，$\mathbf{F}$ 非保守场：在整个 $D$ 上不存在单值势函数。(A1)
$\mathbf{G}(x,y)=2xy^{3}\,\mathbf{i}+3x^{2}y^{2}\,\mathbf{j}$; curve $C$: $y=\sin(\pi x/2)$ from $(0,0)$ to $(1,1)$. Show $\mathbf{G}$ is conservative, use FTLI, and verify by direct parametrisation.$\mathbf{G}(x,y)=2xy^{3}\,\mathbf{i}+3x^{2}y^{2}\,\mathbf{j}$；曲线 $C$：$y=\sin(\pi x/2)$，从 $(0,0)$ 到 $(1,1)$。证明 $\mathbf{G}$ 为保守场，用曲线积分基本定理计算，并用直接参数化验证。
Here $P=2xy^{3}$ and $Q=3x^{2}y^{2}$. Check: (M1)此处 $P=2xy^{3}$，$Q=3x^{2}y^{2}$。验证：(M1)
$$ \frac{\partial P}{\partial y}=6xy^{2}, \qquad \frac{\partial Q}{\partial x}=6xy^{2}. $$Equal, so $\mathbf{G}$ is conservative on $\mathbb{R}^{2}$. (A1)相等，故 $\mathbf{G}$ 在 $\mathbb{R}^{2}$ 上为保守场。(A1)
Find $f$: integrate $f_{x}=2xy^{3}$ with respect to $x$: $f=x^{2}y^{3}+g(y)$. Differentiate with respect to $y$: $f_{y}=3x^{2}y^{2}+g'(y)=3x^{2}y^{2}$, so $g'(y)=0$ and $g$ is constant. (M1) Thus $f(x,y)=x^{2}y^{3}$. Verify: $\nabla f=(2xy^{3},3x^{2}y^{2})=\mathbf{G}$. (A1)求 $f$：对 $x$ 积分 $f_{x}=2xy^{3}$：$f=x^{2}y^{3}+g(y)$。对 $y$ 求导：$f_{y}=3x^{2}y^{2}+g'(y)=3x^{2}y^{2}$，故 $g'(y)=0$，$g$ 为常数。(M1) 因此 $f(x,y)=x^{2}y^{3}$。验证：$\nabla f=(2xy^{3},3x^{2}y^{2})=\mathbf{G}$。(A1)
Since $\mathbf{G}=\nabla f$: (M1)由于 $\mathbf{G}=\nabla f$：(M1)
$$ \int_{C}\mathbf{G}\cdot d\mathbf{r}=f(1,1)-f(0,0)=(1)^{2}(1)^{3}-(0)^{2}(0)^{3}=1-0=1. $$(A1)
Use $x=t$, $y=\sin(\pi t/2)$, $t\in[0,1]$. Then $dx=dt$ and $dy=\frac{\pi}{2}\cos(\pi t/2)\,dt$. (M1)用 $x=t$，$y=\sin(\pi t/2)$，$t\in[0,1]$。则 $dx=dt$，$dy=\frac{\pi}{2}\cos(\pi t/2)\,dt$。(M1)
$$ \int_{C}\mathbf{G}\cdot d\mathbf{r}=\int_{0}^{1}\!\left[2t\sin^{3}\!\left(\frac{\pi t}{2}\right)\cdot 1 + 3t^{2}\sin^{2}\!\left(\frac{\pi t}{2}\right)\cdot\frac{\pi}{2}\cos\!\left(\frac{\pi t}{2}\right)\right]dt. $$Recognise this as $\dfrac{d}{dt}\!\left[t^{2}\sin^{3}\!\left(\frac{\pi t}{2}\right)\right]$ by the product and chain rules: (M1)由积法则和链式法则识别为 $\dfrac{d}{dt}\!\left[t^{2}\sin^{3}\!\left(\frac{\pi t}{2}\right)\right]$：(M1)
$$ \frac{d}{dt}\!\left[t^{2}\sin^{3}\!\left(\frac{\pi t}{2}\right)\right]=2t\sin^{3}\!\left(\frac{\pi t}{2}\right)+t^{2}\cdot 3\sin^{2}\!\left(\frac{\pi t}{2}\right)\cdot\frac{\pi}{2}\cos\!\left(\frac{\pi t}{2}\right). $$(A1) This matches the integrand exactly. Therefore:(A1) 这与被积函数完全吻合。因此：
$$ \int_{0}^{1}\frac{d}{dt}\!\left[t^{2}\sin^{3}\!\left(\frac{\pi t}{2}\right)\right]dt = \Bigl[t^{2}\sin^{3}\!\left(\frac{\pi t}{2}\right)\Bigr]_{0}^{1}=(1)^{2}\sin^{3}\!\left(\frac{\pi}{2}\right)-(0)=(1)(1)=1. $$(A1) Both methods agree: $\displaystyle\int_C\mathbf{G}\cdot d\mathbf{r}=1$.(A1) 两种方法一致：$\displaystyle\int_C\mathbf{G}\cdot d\mathbf{r}=1$。
$\mathbf{H}(x,y)=(x^{3}-y^{3})\,\mathbf{i}+(x^{3}+y^{3})\,\mathbf{j}$; $C$ = boundary of $D$ between $y=x^{2}$ and $y=x$, $0\le x\le 1$, counterclockwise. Apply Green's Theorem and evaluate.$\mathbf{H}(x,y)=(x^{3}-y^{3})\,\mathbf{i}+(x^{3}+y^{3})\,\mathbf{j}$；$C$ 为 $y=x^{2}$ 和 $y=x$（$0\le x\le 1$）之间区域 $D$ 的边界，逆时针方向。应用格林定理并计算。
Here $P=x^{3}-y^{3}$ and $Q=x^{3}+y^{3}$. Verify the hypotheses: $P$ and $Q$ have continuous first partial derivatives everywhere on $\mathbb{R}^{2}$, and $D$ is a simply connected region. (M1) Compute the curl integrand:此处 $P=x^{3}-y^{3}$，$Q=x^{3}+y^{3}$。验证假设条件：$P$ 和 $Q$ 在 $\mathbb{R}^{2}$ 上处处具有连续的一阶偏导数，且 $D$ 为单连通区域。(M1) 计算旋度被积项：
$$ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=3x^{2}-(-3y^{2})=3x^{2}+3y^{2}. $$(A1) Green's Theorem gives: (M1)(A1) 格林定理给出：(M1)
$$ \oint_{C}\mathbf{H}\cdot d\mathbf{r}=\iint_{D}(3x^{2}+3y^{2})\,dA. $$(A1)
The region $D$ is bounded below by $y=x^{2}$ and above by $y=x$ for $x\in[0,1]$. (M1)区域 $D$ 在 $x\in[0,1]$ 上，下界为 $y=x^{2}$，上界为 $y=x$。(M1)
$$ \iint_{D}(3x^{2}+3y^{2})\,dA=\int_{0}^{1}\int_{x^{2}}^{x}(3x^{2}+3y^{2})\,dy\,dx. $$Evaluate the inner integral: (M1)计算内层积分：(M1)
$$ \int_{x^{2}}^{x}(3x^{2}+3y^{2})\,dy=\left[3x^{2}y+y^{3}\right]_{y=x^{2}}^{y=x}=\bigl(3x^{3}+x^{3}\bigr)-\bigl(3x^{4}+x^{6}\bigr)=4x^{3}-3x^{4}-x^{6}. $$(A1) Integrate over $x$:(A1) 对 $x$ 积分：
$$ \int_{0}^{1}(4x^{3}-3x^{4}-x^{6})\,dx=\left[x^{4}-\frac{3x^{5}}{5}-\frac{x^{7}}{7}\right]_{0}^{1}=1-\frac{3}{5}-\frac{1}{7}=\frac{35}{35}-\frac{21}{35}-\frac{5}{35}=\frac{9}{35}. $$(A1)
$\displaystyle\oint_{C}\mathbf{H}\cdot d\mathbf{r}=\dfrac{9}{35}$. (A1) [Cross-check: $1-\tfrac{3}{5}-\tfrac{1}{7}=\tfrac{35-21-5}{35}=\tfrac{9}{35}$.]$\displaystyle\oint_{C}\mathbf{H}\cdot d\mathbf{r}=\dfrac{9}{35}$。(A1) [验算：$1-\tfrac{3}{5}-\tfrac{1}{7}=\tfrac{35-21-5}{35}=\tfrac{9}{35}$。]
Direct computation would require parametrising two curves separately: along $y=x$ from $(0,0)$ to $(1,1)$ with $P=x^3-x^3=0$ and $Q=2x^3$, and along $y=x^2$ reversed. While manageable here, the real power of Green's Theorem appears when $\partial Q/\partial x-\partial P/\partial y$ is simpler than the boundary parametrisations, reducing the problem from multiple contour integrals to one iterated double integral. (A1)直接计算需分别对两段曲线参数化：沿 $y=x$ 从 $(0,0)$ 到 $(1,1)$（$P=x^3-x^3=0$，$Q=2x^3$），以及反向沿 $y=x^2$。虽然本题尚可处理，但格林定理真正的优势在于当 $\partial Q/\partial x-\partial P/\partial y$ 比边界参数化更简单时，将多个围道积分化为一个迭代二重积分。(A1)
$\mathbf{K}(x,y)=(x+y^{2})\,\mathbf{i}+(x^{2}-y)\,\mathbf{j}$; triangular $D$ with vertices $(0,0)$, $(2,0)$, $(0,4)$. (a) Compute outward flux via flux-form Green's Theorem; (b) verify area $=4$ via $A=\tfrac{1}{2}\oint x\,dy-y\,dx$.$\mathbf{K}(x,y)=(x+y^{2})\,\mathbf{i}+(x^{2}-y)\,\mathbf{j}$；顶点为 $(0,0)$、$(2,0)$、$(0,4)$ 的三角形区域 $D$。(a) 用通量形格林定理计算向外通量；(b) 用 $A=\tfrac{1}{2}\oint x\,dy-y\,dx$ 验证面积 $=4$。
The flux (outward) form of Green's Theorem states: $\displaystyle\oint_{C}\mathbf{K}\cdot\hat{\mathbf{n}}\,ds=\iint_{D}\!\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA$. (M1)格林定理的（向外）通量形式为：$\displaystyle\oint_{C}\mathbf{K}\cdot\hat{\mathbf{n}}\,ds=\iint_{D}\!\left(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}\right)dA$。(M1)
Here $P=x+y^{2}$ and $Q=x^{2}-y$. Compute the divergence: (A1)此处 $P=x+y^{2}$，$Q=x^{2}-y$。计算散度：(A1)
$$ \frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}=1+(-1)=0. $$Since the divergence is identically zero on $D$: (M1)由于散度在 $D$ 上恒为零：(M1)
$$ \oint_{C}\mathbf{K}\cdot\hat{\mathbf{n}}\,ds=\iint_{D}0\,dA=0. $$(A1) The outward flux is $0$. Note this holds regardless of the shape of $D$. (A1)(A1) 向外通量为 $0$。注意这与 $D$ 的形状无关。(A1)
Traverse $C$ counterclockwise through the three edges: $(0,0)\to(2,0)$, then $(2,0)\to(0,4)$, then $(0,4)\to(0,0)$. (M1)逆时针遍历 $C$ 的三条边：$(0,0)\to(2,0)$，再 $(2,0)\to(0,4)$，再 $(0,4)\to(0,0)$。(M1)
Edge 1: $(0,0)\to(2,0)$: $\mathbf{r}(t)=(2t,0)$, $t\in[0,1]$, $dx=2\,dt$, $dy=0$. Contribution: $\tfrac{1}{2}\int_0^1[2t\cdot 0-0\cdot 2]\,dt=0$.边 1：$(0,0)\to(2,0)$：$\mathbf{r}(t)=(2t,0)$，$t\in[0,1]$，$dx=2\,dt$，$dy=0$。贡献：$\tfrac{1}{2}\int_0^1[2t\cdot 0-0\cdot 2]\,dt=0$。
Edge 2: $(2,0)\to(0,4)$: parametrise as $x=2-2t$, $y=4t$, $t\in[0,1]$, $dx=-2\,dt$, $dy=4\,dt$. Contribution:边 2：$(2,0)\to(0,4)$：参数化为 $x=2-2t$，$y=4t$，$t\in[0,1]$，$dx=-2\,dt$，$dy=4\,dt$。贡献：
$$ \frac{1}{2}\int_{0}^{1}\bigl[(2-2t)(4\,dt)-(4t)(-2\,dt)\bigr]=\frac{1}{2}\int_{0}^{1}(8-8t+8t)\,dt=\frac{1}{2}\int_{0}^{1}8\,dt=4. $$Edge 3: $(0,4)\to(0,0)$: $x=0$ throughout, $dx=0$. Contribution: $\tfrac{1}{2}\int[(0)dy-(y)(0)]=0$.边 3：$(0,4)\to(0,0)$：全程 $x=0$，$dx=0$。贡献：$\tfrac{1}{2}\int[(0)dy-(y)(0)]=0$。
(A1) Total: $A=0+4+0=4$. (A1) The area of the triangle with base $2$ and height $4$ is indeed $\tfrac{1}{2}(2)(4)=4$, confirming the formula.(A1) 总计：$A=0+4+0=4$。(A1) 底为 $2$、高为 $4$ 的三角形面积确为 $\tfrac{1}{2}(2)(4)=4$，验证了公式。
For each field, state whether it is conservative and, if so, evaluate $\int_C \mathbf{F}\cdot d\mathbf{r}$ from $(1,0)$ to $(0,1)$: (a) $\mathbf{F}_1=2x\,\mathbf{i}+2y\,\mathbf{j}$; (b) $\mathbf{F}_2=y\,\mathbf{i}+x\,\mathbf{j}$; (c) $\mathbf{F}_3=y^{2}\,\mathbf{i}+2xy\,\mathbf{j}$; (d) $\mathbf{F}_4=e^x\sin y\,\mathbf{i}+e^x\cos y\,\mathbf{j}$.对每个向量场，判断是否为保守场，若是则计算从 $(1,0)$ 到 $(0,1)$ 的 $\int_C \mathbf{F}\cdot d\mathbf{r}$：(a) $\mathbf{F}_1=2x\,\mathbf{i}+2y\,\mathbf{j}$；(b) $\mathbf{F}_2=y\,\mathbf{i}+x\,\mathbf{j}$；(c) $\mathbf{F}_3=y^{2}\,\mathbf{i}+2xy\,\mathbf{j}$；(d) $\mathbf{F}_4=e^x\sin y\,\mathbf{i}+e^x\cos y\,\mathbf{j}$。
$\partial P/\partial y=0$, $\partial Q/\partial x=0$. Equal, so conservative. (M1) Potential: $f=x^{2}+y^{2}$. By FTLI:$\partial P/\partial y=0$，$\partial Q/\partial x=0$。相等，故为保守场。(M1) 势函数：$f=x^{2}+y^{2}$。由曲线积分基本定理：
$$ \int_C\mathbf{F}_1\cdot d\mathbf{r}=f(0,1)-f(1,0)=(0+1)-(1+0)=0. $$(A1)
$\partial P/\partial y=1$, $\partial Q/\partial x=1$. Equal, so conservative. (M1) Potential: $f=xy$. By FTLI:$\partial P/\partial y=1$，$\partial Q/\partial x=1$。相等，故为保守场。(M1) 势函数：$f=xy$。由曲线积分基本定理：
$$ \int_C\mathbf{F}_2\cdot d\mathbf{r}=f(0,1)-f(1,0)=(0\cdot 1)-(1\cdot 0)=0. $$(A1)
$\partial P/\partial y=2y$, $\partial Q/\partial x=2y$. Equal everywhere, so conservative on $\mathbb{R}^2$. (M1) Potential: $f_x=y^2$ gives $f=xy^2+g(y)$; then $f_y=2xy+g'(y)=2xy$ so $g'=0$ and $f=xy^2$. By FTLI:$\partial P/\partial y=2y$，$\partial Q/\partial x=2y$。处处相等，故在 $\mathbb{R}^2$ 上为保守场。(M1) 势函数：$f_x=y^2$ 给出 $f=xy^2+g(y)$；则 $f_y=2xy+g'(y)=2xy$，故 $g'=0$，$f=xy^2$。由曲线积分基本定理：
$$ \int_C\mathbf{F}_3\cdot d\mathbf{r}=f(0,1)-f(1,0)=(0\cdot 1)-(1\cdot 0)=0. $$(A1)
$\partial P/\partial y=e^{x}\cos y$, $\partial Q/\partial x=e^{x}\cos y$. Equal everywhere, so conservative. (M1) Potential: $f_x=e^x\sin y$ gives $f=e^x\sin y+g(y)$; then $f_y=e^x\cos y+g'(y)=e^x\cos y$ so $g'=0$ and $f=e^x\sin y$. By FTLI:$\partial P/\partial y=e^{x}\cos y$，$\partial Q/\partial x=e^{x}\cos y$。处处相等，故为保守场。(M1) 势函数：$f_x=e^x\sin y$ 给出 $f=e^x\sin y+g(y)$；则 $f_y=e^x\cos y+g'(y)=e^x\cos y$，故 $g'=0$，$f=e^x\sin y$。由曲线积分基本定理：
$$ \int_C\mathbf{F}_4\cdot d\mathbf{r}=f(0,1)-f(1,0)=e^{0}\sin 1-e^{1}\sin 0=\sin 1-0=\sin 1. $$(A1)