그린 정리

Solution 1 (direct evaluation): Here is a sketch of $C$.

The square consists of four line segments.

• The bottom line segment may be parametrized $\mathbf{r}(x)=(x,0),\ 0\le x\le 3$. So the line integral along this segment is

$$\int_0^3 \mathbf{F}(\mathbf{r}(x))\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}x}\ \mathrm{d}x =\int_0^3 (0, 0)\cdot(1, 0)\ \mathrm{d}x =0$$

• The second line segment may be parametrized $\mathbf{r}(y)=(3,y),\ 0\le y\le 3$. So the line integral along this segment is

$$\int_0^3 \mathbf{F}(\mathbf{r}(y))\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}y}\ \mathrm{d}y =\int_0^3 (9y^2, 6y)\cdot(0, 1)\ \mathrm{d}y =\int_0^3 6y\ \mathrm{d}y =27$$

• The third line segment may be parametrized $\mathbf{r}(t)=(3-t,3),\ 0\le t\le 3$. So the line integral along this segment is

$$\int_0^3 \mathbf{F}(\mathbf{r}(t))\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\ \mathrm{d}t =\int_0^3 \big(9(3-t)^2\,,\, 6(3-t)\big)\cdot(-1\,,\,0)\ \mathrm{d}t =-\int_0^3 9(3-t)^2\ \mathrm{d}t =-81$$

• The final line segment may be parametrized $\mathbf{r}(t)=(0,3-t),\ 0\le t\le 3$. So the line integral along this segment is

$$\int_0^3 \mathbf{F}(\mathbf{r}(t))\cdot\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\ \mathrm{d}t =\int_0^3 (0, 0)\cdot(0, -1)\ \mathrm{d}t =0$$

The full line integral is

$$\oint_C\mathbf{F}\cdot d\mathbf{r}=0+27-81+0=-54$$

Solution 2 (Green's theorem): We apply Green's Theorem.

$$\begin{aligned} \oint_C x^2y^2\,\mathrm{d}x+2xy\,\mathrm{d}y &=\int_0^3\mathrm{d}x\int_0^3 \mathrm{d}y\ \left[\frac{\partial }{\partial x}(2xy)-\frac{\partial }{\partial y}(x^2y^2)\right]\\ &=\int_0^3\mathrm{d}x\int_0^3 \mathrm{d}y\ \big[2y-2x^2y\big]\\ &=\int_0^3\mathrm{d}x\ \big[9-9x^2\big]\\ &=27-9\frac{3^3}{3}=-54 \end{aligned}$$

Call the trapezoid $T$.

By Green's theorem,

$$\begin{aligned} &\oint_C (x\sin y^2 -y^2)\,\mathrm{d}x+(x^2y\cos y^2+3x)\,\mathrm{d}y \\ &=\iint _T\left\{\frac{\partial }{\partial x}(x^2y\cos y^2+3x) -\frac{\partial }{\partial y} (x\sin y^2 -y^2)\right\}\,\mathrm{d}x\,\mathrm{d}y\\ &=\iint _T\big(2xy\cos y^2+3-2xy\cos y^2+2y\big)\,\mathrm{d}x\,\mathrm{d}y\\ &=\iint _T\big(3+2y\big)\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$$

The integral $\iint _T(2y)\,\mathrm{d}x\,\mathrm{d}y$ vanishes because $2y$ changes sign under $y\rightarrow-y$ while the domain of integration is invariant under $y\rightarrow -y$. The integral $\iint _T 3\,\mathrm{d}x\,\mathrm{d}y$ is $3$ times the area of the trapezoid, which is its width (1) times the average of its heights $(\tfrac{1}{2}[2+4])=3$. So

$$\oint_C (x\sin y^2 -y^2)\,\mathrm{d}x+(x^2y\cos y^2+3x)\,\mathrm{d}y =3\times 1\times 3 = 9$$

The integral that would be used for direct evaluation looks very complicated. So let's try Green's theorem. The curve $C$ is the boundary of the triangle

$$T = \big\{ \ (x,y)\ \big|\ 0\le x\le 1, 0\le y\le 2x\ \big\}$$

So

$$\begin{aligned} \int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} &= \int_C \big\{\big(y^2 - e^{-y^2} + \sin x\big)\mathrm{d}x +\big(2xye^{-y^2} + x\big)\mathrm{d}y\big\} \\ &= \iint _T \big\{ \tfrac{\partial}{\partial x}\big(2xye^{-y^2} + x\big) -\tfrac{\partial}{\partial y}\big(y^2 - e^{-y^2} + \sin x\big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \iint _T \big\{ \big(2ye^{-y^2} + 1\big) - \big(2y +2y e^{-y^2} \big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \int_0^1\mathrm{d}x\int_0^{2x}\mathrm{d}y\ \big\{1-2y\big\} \\ &= \int_0^1\mathrm{d}x\ \big\{2x-4x^2\big\} \\ &=1-\frac{4}{3} = -\frac{1}{3} \end{aligned}$$

Direct evaluation will lead to three integrals, one for each side of the triangle. The integral from $(0,0)$ and $(1,-2)$ and the integral from $(1,2)$ to $(0,0)$ will each contain six (nonconstant) terms. This does not look very efficient. So let's try Green's theorem. Denote by $T$, the triangle

$$T = \big\{ \ (x,y)\ \big|\ 0\le x\le 1,\ -2x\le y\le 2x\ \big\} {-50pt}[40pt][40pt]{  }$$

It has boundary $\partial T=C$, oriented counterclockwise as desired. So, by Green's theorem,

$$\begin{aligned} \int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} &= \int_{\partial T} \big\{\big(\tfrac{3}{2}y^2 + e^{-y} +\sin x\big)\mathrm{d}x +\big(\tfrac{1}{2}x^2+x-x e^{-y}\big)\mathrm{d}y\big\} \\ &= \iint _T \big\{ \tfrac{\partial}{\partial x}\big(\tfrac{1}{2}x^2+x-x e^{-y}\big) -\tfrac{\partial}{\partial y}\big(\tfrac{3}{2}y^2 + e^{-y} +\sin x\big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \iint _T \big\{ \big(x + 1-e^{-y}\big) - \big(3y - e^{-y} \big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \iint _T \big\{ x -3y + 1 \big\}\mathrm{d}x\mathrm{d}y \end{aligned}$$

Now

$$\begin{aligned} \iint _T \mathrm{d}x\mathrm{d}y &=\text{Area}(T) = \frac{1}{2}(4)(1) = 2 \\ \iint _T y\ \mathrm{d}x\mathrm{d}y &= 0 \qquad\text{since $y$ is odd under $y\rightarrow-y$} \\ \iint _T x\ \mathrm{d}x\mathrm{d}y &=\int_0^1\mathrm{d}x\int_{-2x}^{2x} \mathrm{d}y\ x =\int_0^1 4x^2\ \mathrm{d}x =\frac{4}{3} \end{aligned}$$

So

$$\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} = \frac{4}{3}-3\times 0 +2 =\frac{10}{3}$$