Unit C7: Vector Fields, Line Integrals, and Green's Theorem单元 C7：向量场、线积分与格林定理

Describe $\mathbf{F}(x,y) = \langle x, y\rangle$.描述 $\mathbf{F}(x,y) = \langle x, y\rangle$。

At the point $(x,y)$ the vector is $\langle x,y\rangle$, which is the position vector itself. So every arrow points directly away from the origin, and its length is $\sqrt{x^2+y^2}$, the distance from the origin.在点 $(x,y)$ 处的向量是 $\langle x,y\rangle$，正是位置向量本身。于是每个箭头都直接背离原点，其长度为 $\sqrt{x^2+y^2}$，即到原点的距离。

$$ |\mathbf{F}(x,y)| = \sqrt{x^2+y^2}. $$

The field is radial and grows linearly with distance: short arrows near the origin, long arrows far away. This is the gradient field of $f(x,y) = \tfrac12(x^2+y^2)$, since $\nabla f = \langle x, y\rangle$.该场是径向的，且随距离线性增长：原点附近箭头短，远处箭头长。它是 $f(x,y) = \tfrac12(x^2+y^2)$ 的梯度场，因为 $\nabla f = \langle x, y\rangle$。

Describe $\mathbf{F}(x,y) = \langle -y, x\rangle$.描述 $\mathbf{F}(x,y) = \langle -y, x\rangle$。

The vector $\langle -y, x\rangle$ is perpendicular to the position vector $\langle x, y\rangle$, since their dot product is $(-y)(x) + (x)(y) = 0$. Its length is again $\sqrt{x^2+y^2}$.向量 $\langle -y, x\rangle$ 垂直于位置向量 $\langle x, y\rangle$，因为二者的标量积为 $(-y)(x) + (x)(y) = 0$。其长度同样是 $\sqrt{x^2+y^2}$。

So the arrows circle the origin counterclockwise with speed proportional to the radius. This is the model velocity field of a rigid rotation. As we will see in Section 6, this field is not a gradient field.于是箭头绕原点逆时针环行，速度与半径成正比。这是刚性转动的标准速度场模型。正如第 6 节将看到的，这个场不是梯度场。

Show that the inverse-square field $\mathbf{F} = -\dfrac{\langle x, y, z\rangle}{(x^2+y^2+z^2)^{3/2}}$ is the gradient of $f = \dfrac{1}{\sqrt{x^2+y^2+z^2}}$ away from the origin.证明：在原点之外，平方反比场 $\mathbf{F} = -\dfrac{\langle x, y, z\rangle}{(x^2+y^2+z^2)^{3/2}}$ 是 $f = \dfrac{1}{\sqrt{x^2+y^2+z^2}}$ 的梯度。

Write $r = \sqrt{x^2+y^2+z^2}$, so $f = r^{-1}$. Differentiate using $\dfrac{\partial r}{\partial x} = \dfrac{x}{r}$:记 $r = \sqrt{x^2+y^2+z^2}$，则 $f = r^{-1}$。利用 $\dfrac{\partial r}{\partial x} = \dfrac{x}{r}$ 求导：

$$ \frac{\partial f}{\partial x} = -r^{-2}\cdot\frac{\partial r}{\partial x} = -r^{-2}\cdot\frac{x}{r} = -\frac{x}{r^3}. $$

By symmetry the $y$ and $z$ partials are $-y/r^3$ and $-z/r^3$, so由对称性，关于 $y$ 和 $z$ 的偏导分别为 $-y/r^3$ 与 $-z/r^3$，于是

$$ \nabla f = -\frac{\langle x, y, z\rangle}{r^3} = \mathbf{F}. $$

This is exactly the Newtonian gravitational and Coulomb electrostatic field. Because it is a gradient field, the work it does between two points depends only on the radial distances of those points, a fact we exploit in Sections 4 and 5.这正是牛顿引力场与库仑静电场。因为它是梯度场，它在两点之间所做的功只依赖于这两点的径向距离，这一事实我们将在第 4、5 节加以利用。

Evaluate $\displaystyle\int_C 2x\,ds$ where $C$ is the parabola $y = x^2$ from $(0,0)$ to $(1,1)$.求 $\displaystyle\int_C 2x\,ds$，其中 $C$ 是抛物线 $y = x^2$ 从 $(0,0)$ 到 $(1,1)$ 的一段。

Parametrize by $x = t$, $y = t^2$, $0 \le t \le 1$. Then $\mathbf{r}'(t) = \langle 1, 2t\rangle$ and $|\mathbf{r}'(t)| = \sqrt{1 + 4t^2}$.参数化为 $x = t$，$y = t^2$，$0 \le t \le 1$。则 $\mathbf{r}'(t) = \langle 1, 2t\rangle$，且 $|\mathbf{r}'(t)| = \sqrt{1 + 4t^2}$。

$$ \int_C 2x\,ds = \int_0^1 2t\,\sqrt{1+4t^2}\,dt. $$

Let $u = 1 + 4t^2$, so $du = 8t\,dt$ and $2t\,dt = \tfrac14\,du$. As $t$ runs $0 \to 1$, $u$ runs $1 \to 5$.令 $u = 1 + 4t^2$，则 $du = 8t\,dt$，$2t\,dt = \tfrac14\,du$。当 $t$ 从 $0$ 到 $1$ 时，$u$ 从 $1$ 到 $5$。

$$ \int_0^1 2t\sqrt{1+4t^2}\,dt = \frac14\int_1^5 \sqrt{u}\,du = \frac14\cdot\frac{2}{3}\,u^{3/2}\Big|_1^5 = \frac{1}{6}\left(5^{3/2} - 1\right). $$

A wire follows the helix $\mathbf{r}(t) = \langle \cos t,\ \sin t,\ t\rangle$, $0 \le t \le 2\pi$, with linear density $\rho(x,y,z) = z$. Find its mass.一根金属丝沿螺旋线 $\mathbf{r}(t) = \langle \cos t,\ \sin t,\ t\rangle$ 分布，$0 \le t \le 2\pi$，线密度为 $\rho(x,y,z) = z$。求其质量。

The velocity is $\mathbf{r}'(t) = \langle -\sin t,\ \cos t,\ 1\rangle$, so the speed is constant:速度为 $\mathbf{r}'(t) = \langle -\sin t,\ \cos t,\ 1\rangle$，故速率是常数：

$$ |\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t + 1} = \sqrt{2}. $$

On the curve $z = t$, so $\rho = t$ and在曲线上 $z = t$，故 $\rho = t$，于是

$$ \text{mass} = \int_C z\,ds = \int_0^{2\pi} t\,\sqrt{2}\,dt = \sqrt{2}\cdot\frac{t^2}{2}\Big|_0^{2\pi} = \sqrt{2}\,\frac{(2\pi)^2}{2} = 2\sqrt{2}\,\pi^2. $$

The constant speed of a helix is what makes its arc-length integrals so clean: $ds = \sqrt{2}\,dt$ throughout.螺旋线的速率恒定，正是它的弧长积分如此简洁的原因：处处 $ds = \sqrt{2}\,dt$。

Evaluate $\displaystyle\int_C x\,ds$ where $C$ goes from $(0,0)$ to $(1,0)$ along the $x$-axis, then from $(1,0)$ to $(1,1)$ straight up.求 $\displaystyle\int_C x\,ds$，其中 $C$ 先沿 $x$ 轴从 $(0,0)$ 到 $(1,0)$，再竖直向上从 $(1,0)$ 到 $(1,1)$。

A line integral over a piecewise curve is the sum of the integrals over each smooth piece. On $C_1$ use $\mathbf{r}(t) = \langle t, 0\rangle$, $0 \le t \le 1$, so $ds = dt$ and the integrand is $x = t$:分段曲线上的线积分是各光滑段上积分之和。在 $C_1$ 上取 $\mathbf{r}(t) = \langle t, 0\rangle$，$0 \le t \le 1$，故 $ds = dt$，被积量为 $x = t$：

$$ \int_{C_1} x\,ds = \int_0^1 t\,dt = \tfrac12. $$

On $C_2$ use $\mathbf{r}(t) = \langle 1, t\rangle$, $0 \le t \le 1$, so again $ds = dt$ but now $x = 1$ throughout:在 $C_2$ 上取 $\mathbf{r}(t) = \langle 1, t\rangle$，$0 \le t \le 1$，故仍有 $ds = dt$，但此时处处 $x = 1$：

$$ \int_{C_2} x\,ds = \int_0^1 1\,dt = 1. $$

Adding the pieces gives $\int_C x\,ds = \tfrac12 + 1 = \tfrac32$.两段相加得 $\int_C x\,ds = \tfrac12 + 1 = \tfrac32$。

Claim: $\int_C f\,ds$ does not depend on how $C$ is parametrized. Suppose $\mathbf{r}(t)$, $a \le t \le b$, and a second parametrization $\mathbf{R}(\tau) = \mathbf{r}(\phi(\tau))$, $\alpha \le \tau \le \beta$, where $\phi$ is a smooth increasing change of variable with $\phi(\alpha) = a$ and $\phi(\beta) = b$.断言：$\int_C f\,ds$ 不依赖 $C$ 的参数化方式。设有 $\mathbf{r}(t)$，$a \le t \le b$，以及第二种参数化 $\mathbf{R}(\tau) = \mathbf{r}(\phi(\tau))$，$\alpha \le \tau \le \beta$，其中 $\phi$ 是光滑的递增换元，满足 $\phi(\alpha) = a$，$\phi(\beta) = b$。

By the chain rule $\mathbf{R}'(\tau) = \mathbf{r}'(\phi(\tau))\,\phi'(\tau)$, and since $\phi' > 0$ we have $|\mathbf{R}'(\tau)| = |\mathbf{r}'(\phi(\tau))|\,\phi'(\tau)$. Substitute into the second integral:由链式法则 $\mathbf{R}'(\tau) = \mathbf{r}'(\phi(\tau))\,\phi'(\tau)$，又因 $\phi' > 0$，故 $|\mathbf{R}'(\tau)| = |\mathbf{r}'(\phi(\tau))|\,\phi'(\tau)$。代入第二个积分：

$$ \int_\alpha^\beta f\big(\mathbf{R}(\tau)\big)\,|\mathbf{R}'(\tau)|\,d\tau = \int_\alpha^\beta f\big(\mathbf{r}(\phi(\tau))\big)\,|\mathbf{r}'(\phi(\tau))|\,\phi'(\tau)\,d\tau. $$

Now set $t = \phi(\tau)$, so $dt = \phi'(\tau)\,d\tau$, and the limits become $a$ to $b$:再令 $t = \phi(\tau)$，则 $dt = \phi'(\tau)\,d\tau$，积分限变为 $a$ 到 $b$：

$$ = \int_a^b f\big(\mathbf{r}(t)\big)\,|\mathbf{r}'(t)|\,dt. $$

The two integrals are equal, so the value belongs to the curve, not the parametrization. The same substitution with a decreasing $\phi$ (a reversal) introduces $|\phi'| = -\phi'$ and swaps the limits, and the two sign changes cancel, which is why reversing direction also leaves $\int_C f\,ds$ unchanged.两个积分相等，可见其值属于曲线本身，而非参数化。若 $\phi$ 递减（即反向），同样的换元会带来 $|\phi'| = -\phi'$ 并交换积分限，这两处符号变化相互抵消，这正是反转方向也不改变 $\int_C f\,ds$ 的原因。

Compute $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle y, x\rangle$ along the segment from $(0,0)$ to $(2,3)$.对 $\mathbf{F} = \langle y, x\rangle$，沿从 $(0,0)$ 到 $(2,3)$ 的线段计算 $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$。

Parametrize $\mathbf{r}(t) = \langle 2t, 3t\rangle$, $0 \le t \le 1$, so $\mathbf{r}'(t) = \langle 2, 3\rangle$. On the path, $\mathbf{F} = \langle 3t, 2t\rangle$.参数化 $\mathbf{r}(t) = \langle 2t, 3t\rangle$，$0 \le t \le 1$，则 $\mathbf{r}'(t) = \langle 2, 3\rangle$。在路径上 $\mathbf{F} = \langle 3t, 2t\rangle$。

$$ \mathbf{F}\cdot\mathbf{r}'(t) = (3t)(2) + (2t)(3) = 12t. $$ $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 12t\,dt = 6. $$

Compute $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle -y, x\rangle$ around the upper half of the unit circle from $(1,0)$ to $(-1,0)$.对 $\mathbf{F} = \langle -y, x\rangle$，沿单位圆上半部分从 $(1,0)$ 到 $(-1,0)$ 计算 $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$。

Use $\mathbf{r}(t) = \langle \cos t, \sin t\rangle$, $0 \le t \le \pi$, so $\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$. On the circle $\mathbf{F} = \langle -\sin t, \cos t\rangle$.取 $\mathbf{r}(t) = \langle \cos t, \sin t\rangle$，$0 \le t \le \pi$，则 $\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$。在圆上 $\mathbf{F} = \langle -\sin t, \cos t\rangle$。

$$ \mathbf{F}\cdot\mathbf{r}'(t) = \sin^2 t + \cos^2 t = 1, \qquad \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^\pi 1\,dt = \pi. $$

Traversing the same arc backward, from $(-1,0)$ to $(1,0)$, gives $-\pi$. The magnitude is fixed by the path; the sign is fixed by the orientation.把同一段弧反向走，从 $(-1,0)$ 到 $(1,0)$，得 $-\pi$。大小由路径决定，符号由定向（orientation）决定。

Compute $\displaystyle\int_C y\,dx$ for $\mathbf{F} = \langle y, 0\rangle$ along two different routes from $(0,0)$ to $(1,1)$.对 $\mathbf{F} = \langle y, 0\rangle$，沿从 $(0,0)$ 到 $(1,1)$ 的两条不同路线计算 $\displaystyle\int_C y\,dx$。

Route 1, the straight segment.路线 1，直线段。 Use $\mathbf{r}(t) = \langle t, t\rangle$, $0 \le t \le 1$, so $dx = dt$ and $y = t$:取 $\mathbf{r}(t) = \langle t, t\rangle$，$0 \le t \le 1$，故 $dx = dt$，$y = t$：

$$ \int_{C_1} y\,dx = \int_0^1 t\,dt = \tfrac12. $$

Route 2, along the parabola路线 2，沿抛物线 $y = x^2$. Use $\mathbf{r}(t) = \langle t, t^2\rangle$, so $dx = dt$ and $y = t^2$:$y = x^2$。取 $\mathbf{r}(t) = \langle t, t^2\rangle$，故 $dx = dt$，$y = t^2$：

$$ \int_{C_2} y\,dx = \int_0^1 t^2\,dt = \tfrac13. $$

The two answers differ, $\tfrac12 \ne \tfrac13$, so the work done by $\mathbf{F} = \langle y, 0\rangle$ depends on the path, not just the endpoints. This field is therefore not conservative, a point Section 5 makes precise: indeed $P_y = 1 \ne 0 = Q_x$.两个答案不同，$\tfrac12 \ne \tfrac13$，故 $\mathbf{F} = \langle y, 0\rangle$ 所做的功依赖路径，而不只是端点。因此该场不是保守场，这一点第 5 节会精确说明：事实上 $P_y = 1 \ne 0 = Q_x$。

Let $\mathbf{F} = \nabla f$ and parametrize $C$ by $\mathbf{r}(t)$, $a \le t \le b$. Substitute into the work integral:设 $\mathbf{F} = \nabla f$，把 $C$ 参数化为 $\mathbf{r}(t)$，$a \le t \le b$，代入功积分：

$$ \int_C \nabla f\cdot d\mathbf{r} = \int_a^b \nabla f\big(\mathbf{r}(t)\big)\cdot \mathbf{r}'(t)\,dt. $$

By the multivariable chain rule, the integrand is exactly the $t$-derivative of the composition $g(t) = f(\mathbf{r}(t))$:由多元链式法则，被积量恰是复合函数 $g(t) = f(\mathbf{r}(t))$ 对 $t$ 的导数：

$$ \frac{d}{dt}\,f\big(\mathbf{r}(t)\big) = \nabla f\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t) = g'(t). $$

Now apply the single-variable Fundamental Theorem of Calculus to $g$:再对 $g$ 应用一元微积分基本定理：

$$ \int_a^b g'(t)\,dt = g(b) - g(a) = f\big(\mathbf{r}(b)\big) - f\big(\mathbf{r}(a)\big). $$

This is the claimed result. The route from $A$ to $B$ never enters the answer.这就是所断言的结论。从 $A$ 到 $B$ 的具体路线从未进入答案。

Compute $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle 2xy, x^2\rangle$ along any path from $(0,0)$ to $(3,1)$.对 $\mathbf{F} = \langle 2xy, x^2\rangle$，沿从 $(0,0)$ 到 $(3,1)$ 的任意路径计算 $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$。

First recognize a potential: we need $f$ with $f_x = 2xy$ and $f_y = x^2$. Integrating $f_x$ in $x$ gives $f = x^2 y + h(y)$; then $f_y = x^2 + h'(y) = x^2$ forces $h'(y) = 0$, so $f = x^2 y$.先找势函数：需要 $f$ 满足 $f_x = 2xy$ 与 $f_y = x^2$。把 $f_x$ 对 $x$ 积分得 $f = x^2 y + h(y)$；再由 $f_y = x^2 + h'(y) = x^2$ 推出 $h'(y) = 0$，故 $f = x^2 y$。

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = f(3,1) - f(0,0) = (9)(1) - 0 = 9. $$

For $\mathbf{F} = \langle 2xy, x^2\rangle$ with potential $f = x^2 y$, verify the answer of Worked Example 4.1 by integrating directly along the straight segment from $(0,0)$ to $(3,1)$.对带势函数 $f = x^2 y$ 的 $\mathbf{F} = \langle 2xy, x^2\rangle$，沿从 $(0,0)$ 到 $(3,1)$ 的直线段直接积分，以验证例题 4.1 的答案。

Parametrize $\mathbf{r}(t) = \langle 3t, t\rangle$, $0 \le t \le 1$, so $\mathbf{r}'(t) = \langle 3, 1\rangle$. On the path $\mathbf{F} = \langle 2(3t)(t),\ (3t)^2\rangle = \langle 6t^2,\ 9t^2\rangle$.参数化 $\mathbf{r}(t) = \langle 3t, t\rangle$，$0 \le t \le 1$，则 $\mathbf{r}'(t) = \langle 3, 1\rangle$。在路径上 $\mathbf{F} = \langle 2(3t)(t),\ (3t)^2\rangle = \langle 6t^2,\ 9t^2\rangle$。

$$ \mathbf{F}\cdot\mathbf{r}'(t) = (6t^2)(3) + (9t^2)(1) = 27t^2. $$ $$ \int_C \mathbf{F}\cdot d\mathbf{r} = \int_0^1 27t^2\,dt = 27\cdot\frac{t^3}{3}\Big|_0^1 = 9. $$

The direct computation agrees with the potential method. The Fundamental Theorem saves the parametrization work entirely once a potential is in hand, which is why recognizing a gradient field is so valuable.直接计算与势函数方法一致。一旦手握势函数，基本定理就完全省去了参数化的功夫，这正是识别梯度场如此宝贵的原因。

Compute $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle yz,\ xz,\ xy\rangle$ from $(1,1,1)$ to $(2,3,4)$.对 $\mathbf{F} = \langle yz,\ xz,\ xy\rangle$，从 $(1,1,1)$ 到 $(2,3,4)$ 计算 $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$。

Look for $f$ with $f_x = yz$. Integrating in $x$ gives $f = xyz + h(y,z)$. Then $f_y = xz + h_y = xz$ forces $h_y = 0$, and $f_z = xy + h_z = xy$ forces $h_z = 0$, so $h$ is constant and $f = xyz$.寻找满足 $f_x = yz$ 的 $f$。对 $x$ 积分得 $f = xyz + h(y,z)$。再由 $f_y = xz + h_y = xz$ 推出 $h_y = 0$，由 $f_z = xy + h_z = xy$ 推出 $h_z = 0$，故 $h$ 为常数，$f = xyz$。

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = f(2,3,4) - f(1,1,1) = (2)(3)(4) - (1)(1)(1) = 24 - 1 = 23. $$

No path was specified, and none is needed: the field is a gradient, so the answer is fixed by the endpoints.题目未给路径，也无需给定：该场是梯度场，故答案由端点确定。

Is $\mathbf{F} = \langle 3 + 2xy,\ x^2 - 3y^2\rangle$ conservative? If so, find $f$.$\mathbf{F} = \langle 3 + 2xy,\ x^2 - 3y^2\rangle$ 是保守场吗？若是，求 $f$。

Test: $P = 3 + 2xy$ gives $P_y = 2x$; $Q = x^2 - 3y^2$ gives $Q_x = 2x$. They agree on all of the plane, which is simply connected, so $\mathbf{F}$ is conservative.判别：$P = 3 + 2xy$ 给出 $P_y = 2x$；$Q = x^2 - 3y^2$ 给出 $Q_x = 2x$。二者在整个平面上相等，而平面单连通，故 $\mathbf{F}$ 保守。

Integrate $P$ in $x$: $f = 3x + x^2 y + h(y)$. Then $f_y = x^2 + h'(y)$ must equal $Q = x^2 - 3y^2$, so $h'(y) = -3y^2$ and $h(y) = -y^3$.把 $P$ 对 $x$ 积分：$f = 3x + x^2 y + h(y)$。再令 $f_y = x^2 + h'(y)$ 等于 $Q = x^2 - 3y^2$，得 $h'(y) = -3y^2$，$h(y) = -y^3$。

$$ f(x,y) = 3x + x^2 y - y^3 + \text{constant}. $$

Is $\mathbf{F} = \langle x^2 y,\ xy^2\rangle$ conservative on the plane?$\mathbf{F} = \langle x^2 y,\ xy^2\rangle$ 在平面上是保守场吗？

Compute the mixed partials. With $P = x^2 y$ we get $P_y = x^2$. With $Q = xy^2$ we get $Q_x = y^2$.计算混合偏导。由 $P = x^2 y$ 得 $P_y = x^2$；由 $Q = xy^2$ 得 $Q_x = y^2$。

$$ P_y = x^2, \qquad Q_x = y^2. $$

These are not equal as functions (they agree only on the lines $y = \pm x$), so $P_y \ne Q_x$ and $\mathbf{F}$ is not conservative. No potential exists, and the work it does will in general depend on the path. There is nothing to integrate to find $f$, so we stop as soon as the test fails.作为函数它们并不相等（仅在直线 $y = \pm x$ 上相等），故 $P_y \ne Q_x$，$\mathbf{F}$ 不是保守场。势函数不存在，它做的功一般依赖路径。既无 $f$ 可求，判别一失败便就此停手。

Evaluate $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F} = \langle e^y,\ x e^y\rangle$ from $(0,0)$ to $(2,\ln 3)$.对 $\mathbf{F} = \langle e^y,\ x e^y\rangle$，从 $(0,0)$ 到 $(2,\ln 3)$ 计算 $\displaystyle\int_C \mathbf{F}\cdot d\mathbf{r}$。

Test first: $P = e^y$ gives $P_y = e^y$; $Q = x e^y$ gives $Q_x = e^y$. They agree on the whole plane, which is simply connected, so $\mathbf{F}$ is conservative.先判别：$P = e^y$ 给出 $P_y = e^y$；$Q = x e^y$ 给出 $Q_x = e^y$。二者在整个平面上相等，而平面单连通，故 $\mathbf{F}$ 保守。

Find $f$: integrate $P = e^y$ in $x$ to get $f = x e^y + h(y)$. Then $f_y = x e^y + h'(y)$ must equal $Q = x e^y$, so $h'(y) = 0$ and $f = x e^y$.求 $f$：把 $P = e^y$ 对 $x$ 积分得 $f = x e^y + h(y)$。再令 $f_y = x e^y + h'(y)$ 等于 $Q = x e^y$，得 $h'(y) = 0$，$f = x e^y$。

$$ \int_C \mathbf{F}\cdot d\mathbf{r} = f(2, \ln 3) - f(0,0) = 2e^{\ln 3} - 0 = 2\cdot 3 = 6. $$

Consider the vortex field on the plane minus the origin:考虑挖去原点的平面上的涡旋场：

$$ \mathbf{F} = \left\langle \frac{-y}{x^2+y^2},\ \frac{x}{x^2+y^2}\right\rangle. $$

A direct computation gives $P_y = Q_x = \dfrac{y^2 - x^2}{(x^2+y^2)^2}$, so the component test is satisfied everywhere the field is defined.直接计算得 $P_y = Q_x = \dfrac{y^2 - x^2}{(x^2+y^2)^2}$，故分量判别在场有定义之处处处满足。

Yet integrate around the unit circle $\mathbf{r}(t) = \langle\cos t, \sin t\rangle$, $0 \le t \le 2\pi$. On the circle $\mathbf{F} = \langle -\sin t, \cos t\rangle$ and $\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$, so $\mathbf{F}\cdot\mathbf{r}' = 1$ and然而沿单位圆 $\mathbf{r}(t) = \langle\cos t, \sin t\rangle$，$0 \le t \le 2\pi$ 积分。在圆上 $\mathbf{F} = \langle -\sin t, \cos t\rangle$，$\mathbf{r}'(t) = \langle -\sin t, \cos t\rangle$，故 $\mathbf{F}\cdot\mathbf{r}' = 1$，于是

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} 1\,dt = 2\pi \ne 0. $$

A nonzero loop integral means $\mathbf{F}$ is not conservative, even though $P_y = Q_x$. The obstruction is the hole at the origin: the domain is not simply connected, so the test alone cannot certify a potential.回路积分非零意味着 $\mathbf{F}$ 不是保守场，尽管 $P_y = Q_x$。障碍就在原点处的洞：定义域不单连通，单凭判别无法确证势函数存在。

Evaluate $\displaystyle\oint_C (y^2)\,dx + (3xy)\,dy$ where $C$ is the boundary of the unit disk $x^2 + y^2 \le 1$, oriented counterclockwise.求 $\displaystyle\oint_C (y^2)\,dx + (3xy)\,dy$，其中 $C$ 是单位圆盘 $x^2 + y^2 \le 1$ 的边界，逆时针定向。

Here $P = y^2$ and $Q = 3xy$, so $Q_x = 3y$ and $P_y = 2y$, giving $Q_x - P_y = y$.这里 $P = y^2$，$Q = 3xy$，故 $Q_x = 3y$，$P_y = 2y$，得 $Q_x - P_y = y$。

$$ \oint_C P\,dx + Q\,dy = \iint_D y\,dA. $$

The region $D$ is symmetric about the $x$-axis and $y$ is an odd function of $y$, so the integral vanishes:区域 $D$ 关于 $x$ 轴对称，而 $y$ 是 $y$ 的奇函数，故积分为零：

$$ \iint_D y\,dA = 0. $$

Find the area enclosed by the ellipse $\mathbf{r}(t) = \langle a\cos t,\ b\sin t\rangle$, $0 \le t \le 2\pi$.求椭圆 $\mathbf{r}(t) = \langle a\cos t,\ b\sin t\rangle$（$0 \le t \le 2\pi$）所围的面积。

Use the area formula. With $x = a\cos t$, $y = b\sin t$, we get $dx = -a\sin t\,dt$ and $dy = b\cos t\,dt$, so使用面积公式。取 $x = a\cos t$，$y = b\sin t$，得 $dx = -a\sin t\,dt$，$dy = b\cos t\,dt$，于是

$$ x\,dy - y\,dx = (a\cos t)(b\cos t)\,dt - (b\sin t)(-a\sin t)\,dt = ab(\cos^2 t + \sin^2 t)\,dt = ab\,dt. $$ $$ \text{Area} = \frac12\int_0^{2\pi} ab\,dt = \frac12\,ab\,(2\pi) = \pi ab. $$

Evaluate $\displaystyle\oint_C (x^4 - y)\,dx + (x + y^4)\,dy$, where $C$ is the boundary of the square $0 \le x \le 1$, $0 \le y \le 1$, counterclockwise.求 $\displaystyle\oint_C (x^4 - y)\,dx + (x + y^4)\,dy$，其中 $C$ 是正方形 $0 \le x \le 1$、$0 \le y \le 1$ 的边界，逆时针。

Direct parametrization would mean four separate segment integrals. Green's Theorem replaces all of them with one double integral. Here $P = x^4 - y$ and $Q = x + y^4$, so直接参数化意味着四段分别积分。格林定理把它们全部换成一个二重积分。这里 $P = x^4 - y$，$Q = x + y^4$，故

$$ Q_x - P_y = (1) - (-1) = 2. $$

The integrand is the constant $2$, so the loop integral is just $2$ times the area of the square:被积量是常数 $2$，故回路积分就是正方形面积的 $2$ 倍：

$$ \oint_C P\,dx + Q\,dy = \iint_D 2\,dA = 2\cdot\text{Area}(D) = 2\cdot 1 = 2. $$

The high powers $x^4$ and $y^4$ never matter, because they sit in the components whose relevant partials cancel. Recognizing this in advance is the whole art of applying Green's Theorem.高次项 $x^4$ 与 $y^4$ 始终无关紧要，因为它们恰好落在那些相关偏导会抵消的分量里。事先看出这一点，正是运用格林定理的全部技巧所在。

Evaluate $\displaystyle\oint_C (-y^3)\,dx + (x^3)\,dy$, where $C$ is the circle $x^2 + y^2 = 4$, counterclockwise.求 $\displaystyle\oint_C (-y^3)\,dx + (x^3)\,dy$，其中 $C$ 是圆 $x^2 + y^2 = 4$，逆时针。

With $P = -y^3$ and $Q = x^3$ we get $Q_x - P_y = 3x^2 - (-3y^2) = 3(x^2+y^2)$. Green's Theorem turns the loop into取 $P = -y^3$，$Q = x^3$，得 $Q_x - P_y = 3x^2 - (-3y^2) = 3(x^2+y^2)$。格林定理把回路化为

$$ \oint_C P\,dx + Q\,dy = \iint_D 3(x^2+y^2)\,dA. $$

The disk of radius $2$ calls for polar coordinates, $x^2+y^2 = r^2$ and $dA = r\,dr\,d\theta$:半径为 $2$ 的圆盘宜用极坐标，$x^2+y^2 = r^2$，$dA = r\,dr\,d\theta$：

$$ \iint_D 3r^2\,dA = \int_0^{2\pi}\!\!\int_0^2 3r^2\cdot r\,dr\,d\theta = \int_0^{2\pi}\!\!\int_0^2 3r^3\,dr\,d\theta = \int_0^{2\pi} \frac{3r^4}{4}\Big|_0^2 d\theta = \int_0^{2\pi} 12\,d\theta = 24\pi. $$

Prove the special case $\oint_C P\,dx = -\iint_D P_y\,dA$ for a region $D$ that is vertically simple: $a \le x \le b$, $g_1(x) \le y \le g_2(x)$.对竖直方向简单的区域 $D$（$a \le x \le b$，$g_1(x) \le y \le g_2(x)$），证明特例 $\oint_C P\,dx = -\iint_D P_y\,dA$。

Evaluate the right side by integrating $P_y$ in $y$ first, using the single-variable Fundamental Theorem:先对 $y$ 积分 $P_y$，用一元基本定理算出右端：

$$ \iint_D P_y\,dA = \int_a^b \big[\,P(x, g_2(x)) - P(x, g_1(x))\,\big]\,dx. $$

Now trace the boundary $C$ counterclockwise. The lower curve $y = g_1(x)$ runs left to right and contributes $\int_a^b P(x, g_1)\,dx$; the upper curve $y = g_2(x)$ runs right to left and contributes $-\int_a^b P(x, g_2)\,dx$. The vertical sides have $dx = 0$ and contribute nothing.再逆时针描出边界 $C$。下方曲线 $y = g_1(x)$ 从左到右，贡献 $\int_a^b P(x, g_1)\,dx$；上方曲线 $y = g_2(x)$ 从右到左，贡献 $-\int_a^b P(x, g_2)\,dx$。竖直两侧 $dx = 0$，无贡献。

$$ \oint_C P\,dx = \int_a^b P(x, g_1)\,dx - \int_a^b P(x, g_2)\,dx = -\iint_D P_y\,dA. $$

An identical argument across horizontally simple regions gives $\oint_C Q\,dy = \iint_D Q_x\,dA$. Adding the two identities yields Green's Theorem; general regions are handled by cutting them into simple pieces, whose interior boundaries cancel in pairs.对水平方向简单的区域作同样论证得 $\oint_C Q\,dy = \iint_D Q_x\,dA$。两式相加即得格林定理；一般区域则切成若干简单片处理，其内部边界成对抵消。

Compute the scalar curl and the divergence of $\mathbf{F} = \langle x^2 - y,\ x + y^2\rangle$.计算 $\mathbf{F} = \langle x^2 - y,\ x + y^2\rangle$ 的标量旋度与散度。

Here $P = x^2 - y$ and $Q = x + y^2$.这里 $P = x^2 - y$，$Q = x + y^2$。

$$ \text{curl} = Q_x - P_y = (1) - (-1) = 2, \qquad \text{div} = P_x + Q_y = (2x) + (2y) = 2x + 2y. $$

The curl is constant, so the field has uniform swirl. The divergence varies with position: it is positive in the first quadrant (a source) and negative in the third (a sink).旋度是常数，故场具有均匀的旋转。散度随位置变化：在第一象限为正（源），在第三象限为负（汇）。

Find the outward flux of $\mathbf{F} = \langle x, y\rangle$ across the circle $x^2 + y^2 = 9$ using the flux form of Green's Theorem.用格林定理的通量形式，求 $\mathbf{F} = \langle x, y\rangle$ 穿过圆 $x^2 + y^2 = 9$ 的向外通量。

The divergence is $P_x + Q_y = 1 + 1 = 2$, a constant. The flux form converts the boundary integral into a double integral of the divergence over the enclosed disk $D$:散度为 $P_x + Q_y = 1 + 1 = 2$，是常数。通量形式把边界积分化为散度在所围圆盘 $D$ 上的二重积分：

$$ \oint_C \mathbf{F}\cdot\mathbf{n}\,ds = \iint_D (P_x + Q_y)\,dA = \iint_D 2\,dA = 2\cdot\text{Area}(D). $$

The disk has radius $3$, so its area is $9\pi$ and the flux is $18\pi$. This matches intuition: the radial field $\langle x, y\rangle$ flows uniformly outward, so its net outflow grows with the enclosed area.圆盘半径为 $3$，故面积为 $9\pi$，通量为 $18\pi$。这与直觉相符：径向场 $\langle x, y\rangle$ 均匀向外流动，故其净外流随所围面积增大。

The area formula.面积公式。 Apply Green's Theorem with $P = -\tfrac{y}{2}$ and $Q = \tfrac{x}{2}$. Then对 $P = -\tfrac{y}{2}$、$Q = \tfrac{x}{2}$ 应用格林定理。则

$$ Q_x - P_y = \frac12 - \left(-\frac12\right) = 1, $$

so the double integral collapses to the area itself:于是二重积分坍缩为面积本身：

$$ \frac12\oint_C x\,dy - y\,dx = \iint_D 1\,dA = \text{Area}(D). $$

Any choice with $Q_x - P_y = 1$ works; the symmetric choice above is simply the most memorable.任何使 $Q_x - P_y = 1$ 的选取都行；上面这种对称选取只是最便于记忆而已。

A region with a hole.带洞的区域。 Green's Theorem extends to a region $D$ between two curves, an outer boundary $C_{\text{out}}$ counterclockwise and an inner boundary $C_{\text{in}}$ clockwise, by cutting $D$ along a bridge and observing the bridge is traversed twice in opposite directions, so it cancels. The result is格林定理可推广到夹在两条曲线之间的区域 $D$：外边界 $C_{\text{out}}$ 逆时针、内边界 $C_{\text{in}}$ 顺时针。沿一条"桥"把 $D$ 切开，注意这座桥被反向走了两次而相互抵消。结果是

$$ \oint_{C_{\text{out}}} P\,dx + Q\,dy \;-\; \oint_{C_{\text{in}}} P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dA, $$

with both boundaries oriented counterclockwise. For the vortex field of Section 5, $Q_x - P_y = 0$ on the annulus that avoids the origin, so the two loop integrals must be equal. That is why every loop encircling the origin once gives the same value $2\pi$: the integral cannot be deformed across the puncture, but it is unchanged by any deformation that does not cross it. This is the seed of the homotopy invariance that runs through the rest of vector calculus.其中两条边界都按逆时针定向。对第 5 节的涡旋场，在避开原点的圆环上 $Q_x - P_y = 0$，故两个回路积分必相等。这正是每条绕原点一圈的回路都给出同一值 $2\pi$ 的原因：积分无法越过那个穿孔变形，但任何不跨越它的变形都不改变积分值。这便是贯穿向量微积分其余部分的同伦不变性的雏形。