선적분

The square has four sides, each of which is a line segment.

• On the first side, $y=0$ and $\mathrm{d}y=0$. That is, we may parametrize the first side by $\mathbf{r}(x)=x\,\hat{\boldsymbol{\imath}}$ with $0\le x\le 1$.
• On the second side, $x=1$ and $\mathrm{d}x=0$. We may parametrize the second side by $\mathbf{r}(y)=\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}$ with $0\le y\le 1$.
• On the third side, $y=1$ and $\mathrm{d}y=0$. We may parametrize the third side by $\mathbf{r}(x)=x\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$ with $x$ running from $1$ to $0$.
• On the final side, $x=0$ and $\mathrm{d}x=0$. We may parametrize the fourth side by $\mathbf{r}(y)=y\,\hat{\boldsymbol{\jmath}}$ with $y$ running from $1$ to $0$.

So

$$\begin{aligned} \int_\mathcal{C} x^2y^2\,\mathrm{d}x+x^3y\,\mathrm{d}y &=\int_0^1 x^2\times 0^2\,\mathrm{d}x +\int_0^1 1^3\times y\,\mathrm{d}y +\int_1^0 x^2\times 1^2\,\mathrm{d}x +\int_1^0 0^3\times y\,\mathrm{d}y \\ &=\frac{1}{2}-\frac{1}{3} =\frac{1}{6} \end{aligned}$$

Since $\mathbf{F}$ is conservative, $\int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r}=0$ over any closed curve $C$. The given curve is closed, so the integral is simply zero.

The point here is that $\mathbf{F}$ is conservative, as $\mathbf{F}=\nabla\phi$ with

$$\phi=\frac{x^2}{2}+yx-yz+\frac{z^2}{2}$$

So, for all paths from $\mathbf{r}(t_0)=(1,0,-1)$ to $\mathbf{r}(t_1)=(0,-2,3)$,

$$\begin{aligned} \int_\mathcal{C}\mathbf{F}\cdot \mathrm{d}\mathbf{r} =\phi\big(\mathbf{r}(t_1)\big)-\phi\big(\mathbf{r}(t_0)\big) &=\phi(0, -2, 3)-\phi(1,0,-1) \\ &=\left[0+0+6+\frac{9}{2}\right]-\left[\frac{1}{2}+0-0+\frac{1}{2}\right] \\ &=9\frac{1}{2} \end{aligned}$$

.