선적분

You may have noticed that the orientation of (i.e. the direction of motion along) the curve $C$ was not specified in the question. That's because the value of the integral is independent of the orientation of $C$:

• At each point of $C$, $\mathbf{F}(x,y) = xy^2 \,\hat{\boldsymbol{\imath}} + ye^x \,\hat{\boldsymbol{\jmath}}$, independent of the orientation.
• At each point of $C$, $\hat{\mathbf{n}}$ is the outward pointing normal, independent of the orientation.
• $ds>0$, independent of the orientation.

Here is a sketch of the rectangle $R$.

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We have arbitrarily chosen to traverse $C$ in the counterclockwise direction. The boundary of $R$ boundary consists of four line segments.

• $L_1$ from $(0,-1)$ to $(3,-1)$, with $\hat{\mathbf{n}} = -\hat{\boldsymbol{\jmath}}$
• $L_2$ from $(3,-1)$ to $(3,1)$, with $\hat{\mathbf{n}} = \hat{\boldsymbol{\imath}}$
• $L_3$ from $(3,1)$ to $(0,1)$, with $\hat{\mathbf{n}} = \hat{\boldsymbol{\jmath}}$
• $L_4$ from $(0,1)$ to $(0,-1)$, with $\hat{\mathbf{n}} = -\hat{\boldsymbol{\imath}}$

So

$$\begin{aligned} \int_C \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}s &= \int_{L_1} \mathbf{F}\cdot(-\hat{\boldsymbol{\jmath}})\,\mathrm{d}s +\int_{L_2} \mathbf{F}\cdot \hat{\boldsymbol{\imath}}\,\mathrm{d}s +\int_{L_3} \mathbf{F}\cdot(\hat{\boldsymbol{\jmath}})\,\mathrm{d}s +\int_{L_4} \mathbf{F}\cdot(-\hat{\boldsymbol{\imath}})\,\mathrm{d}s \\ &= \int_0^3 -\overbrace{(-1)}^{y}e^x\,\mathrm{d}x +\int_{-1}^1 \overbrace{(3)}^{x}y^2\,\mathrm{d}y +\int^0_3 \overbrace{(1)}^{y}e^x\,\overbrace{(-\mathrm{d}x)}^{\mathrm{d}s} +\int^{-1}_1 \overbrace{(0)}^{x}y^2\,\overbrace{(-\mathrm{d}y)}^{\mathrm{d}s} \\ &=\big[e^3-1\big]+\big[1^3-(-1)^3\big] + \big[e^3-1\big]+0 \\ &=2e^3 \end{aligned}$$

The trickiest part of this computation is getting $\mathrm{d}s$ correct on $L_3$ and $L_4$ (remembering that $\mathrm{d}s$ is the arc length traveled and so is positive, while $\mathrm{d}x<0$ on $L_3$ and $\mathrm{d}y<0$ on $L_4$). To make a more detailed computation of $\int_{L_3} \mathbf{F}\cdot(\hat{\boldsymbol{\jmath}})\,\mathrm{d}s$, parametrize $L_3$ by

$$\mathbf{r}(t) = (3,1) + t\big\{(0,1)-(3,1)\big\} = \big(3-3t,1\big)\qquad 0\le t\le 1$$

so that $\mathbf{r}(0) = (3,1)$ is the initial point of $L_3$ and $\mathbf{r}(1) = (0,1)$ is the final point of $L_3$. Then

$$\mathbf{r}'(t) = (-3,0)\qquad \frac{\mathrm{d}s}{\mathrm{d}t}(t) = |\mathbf{r}'(t)| = 3$$

and

$$\begin{aligned} \int_{L_3} \mathbf{F}\cdot \hat{\boldsymbol{\jmath}}\,\mathrm{d}s =\int_0^1 \mathbf{F}\big(\mathbf{r}(t)\big)\cdot \hat{\boldsymbol{\jmath}}\,\frac{\mathrm{d}s}{\mathrm{d}t}(t)\,\mathrm{d}t =\int_0^1\overbrace{e^{3-3t}}^{y(t)e^{x(t)}}\, \overbrace{3}^{\frac{\mathrm{d}s}{\mathrm{d}t}(t)}\,\mathrm{d}t =-e^{3-3t}\Big|_0^1 =e^3-1 \end{aligned}$$

As a check, we'll now verify that we get the same answer for $\int_{L_3} \mathbf{F}\cdot(\hat{\boldsymbol{\jmath}})\,\mathrm{d}s$, when we parametrize $L_3$ by

$$\mathbf{r}(t) = (0,1) + t\big\{(3,1)-(0,1)\big\} = \big(3t,1\big)\qquad 0\le t\le 1$$

so that, this time, $\mathbf{r}(0) = (0,1)$ is the initial point of $L_3$ and $\mathbf{r}(1) = (3,1)$ is the final point of $L_3$. This time

$$\mathbf{r}'(t) = (3,0)\qquad \frac{\mathrm{d}s}{\mathrm{d}t}(t) = |\mathbf{r}'(t)| = 3$$

and

$$\begin{aligned} \int_{L_3} \mathbf{F}\cdot \hat{\boldsymbol{\jmath}}\,\mathrm{d}s =\int_0^1 \mathbf{F}\big(\mathbf{r}(t)\big)\cdot \hat{\boldsymbol{\jmath}}\,\frac{\mathrm{d}s}{\mathrm{d}t}(t)\,\mathrm{d}t =\int_0^1\overbrace{e^{3t}}^{y(t)e^{x(t)}}\, \overbrace{3}^{\frac{\mathrm{d}s}{\mathrm{d}t}(t)}\,\mathrm{d}t =e^{3t}\Big|_0^1 =e^3-1 \end{aligned}$$

So both orientations of $L_3$ give the same value of $\int_{L_3} \mathbf{F}\cdot(\hat{\boldsymbol{\jmath}})\,\mathrm{d}s$.