선적분

(a) The curve is $\mathbf{r}(\theta) = x(\theta)\,\hat{\boldsymbol{\imath}}+ y(\theta)\,\hat{\boldsymbol{\jmath}}$ with $x(\theta)=r(\theta)\cos\theta$, $y(\theta)=r(\theta)\sin\theta$ and $\theta_1\le \theta\le \theta_2$. On this curve

$$\begin{aligned} \mathbf{v}(\theta) =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}(\theta)&= x'(\theta)\hat{\boldsymbol{\imath}}+ y'(\theta)\hat{\boldsymbol{\jmath}} =\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hat{\boldsymbol{\imath}}+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hat{\boldsymbol{\jmath}}\\ \implies \frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\\ &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{aligned}$$

Hence

$$\begin{aligned} \int_\mathcal{C} f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta \end{aligned}$$

(b) In this case $f(x,y)=1$, $r(\theta)=1+\cos\theta$, $\theta_1=0$ and $\theta_2=2\pi$,

$$\begin{aligned} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\mathrm{d}\theta =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\mathrm{d}\theta\\ &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\mathrm{d}\theta =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\mathrm{d}\theta =4\int_0^{\pi} \cos\frac{\theta}{2}\ \mathrm{d}\theta =8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{aligned}$$

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.

Since $\mathbf{F}$ is conservative, and $A$ and $B$ start and end at the same points, by path-independence $\int_B \mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_A \mathbf{F}\cdot\mathrm{d}\mathbf{r}=5.$

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 $\mathbf{r}(t_1)=(0,-2,3)$,

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

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