스토크스 정리

Considering that there are ten line segments in $C$, it is probably not very efficient to use direct evaluation. Two other possible methods come to mind. If $\mathbf{F}$ is conservative, we can use $\mathbf{F}$'s potential. Even if $\mathbf{F}$ is not conservative, it may be possible to efficiently use Stokes' (or Green's) theorem. So let's compute

$$\begin{aligned} { \boldsymbol{\nabla} }\times\mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} &\hat{\boldsymbol{\jmath}} &\hat{\mathbf{k}} \\ \tfrac{\partial}{\partial x} & \tfrac{\partial}{\partial y} & \tfrac{\partial}{\partial z} \\ y & 2x-10 & 0 \end{matrix} \right] =\hat{\mathbf{k}} \end{aligned}$$

As ${ \boldsymbol{\nabla} }\times\mathbf{F}\ne\mathbf{0}$, the vector field $\mathbf{F}$ is not conservative. As ${ \boldsymbol{\nabla} }\times\mathbf{F}\ne\mathbf{0}$ is very simple, it looks like Stokes' theorem could provide an efficient way to compute the integral. The left figure below contains a sketch of $C$.

The curve $C$ is not closed, and so is not the boundary of a surface, so we cannot apply Stokes' theorem directly. But we can easily come up with a surface whose boundary contains $C$. Let $R$ be the shaded region in the figure on the right above. The boundary $\partial R$ of $R$ consists of two parts — $C$ and the line segment $L$. The normal of $R$ for $-\hat{\mathbf{k}}$ (since $\partial R$ is oriented clockwise). So Stokes' theorem gives

$$\begin{aligned} \int_C \mathbf{F} \cdot\mathrm{d}\mathbf{r} + \int_L \mathbf{F} \cdot\mathrm{d}\mathbf{r} = \iint _R { \boldsymbol{\nabla} }\times\mathbf{F} \cdot(-\hat{\mathbf{k}})\ \mathrm{d}S = \iint _R (\hat{\mathbf{k}}) \cdot(-\hat{\mathbf{k}})\ \mathrm{d}S = -\text{Area}(R) \end{aligned}$$

$R$ is the union of $5$ triangles, each of height $1$ and base $1$. So

$$\text{Area}(R) = 5\times \frac{1}{2}\times 1\times 1 =\frac{5}{2}$$

If we denote by $-L$ the line segment from $(0,0)$ to $(5,5)$, we can parametrize $-L$ by $\mathbf{r}(t) = t(5,5)$, $0\le t\le 1$ and

$$\begin{aligned} \int_{-L} \mathbf{F} \cdot\mathrm{d}\mathbf{r} &=\int_0^1 \overbrace{\big(5t\,\hat{\boldsymbol{\imath}} + (10t-10)\,\hat{\boldsymbol{\jmath}}\big)}^{\mathbf{F}(\mathbf{r}(t))} \cdot\overbrace{(5\,\hat{\boldsymbol{\imath}}+5\,\hat{\boldsymbol{\jmath}})}^{\mathbf{r}'(t)}\ \mathrm{d}t =\int_0^1 5\big(15t-10\big) \mathrm{d}t =25\left(\frac{3}{2}-2\right) \\ &=-\frac{25}{2} \end{aligned}$$

All together

$$\begin{aligned} \int_C \mathbf{F} \cdot\mathrm{d}\mathbf{r} = -\text{Area}(R) - \int_L \mathbf{F} \cdot\mathrm{d}\mathbf{r} = -\text{Area}(R) + \int_{-L} \mathbf{F} \cdot\mathrm{d}\mathbf{r} =-\frac{5}{2}-\frac{25}{2} =-15 \end{aligned}$$