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 line integral is $\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r}$ with $\mathbf{F} = (ye^x + \sin y)\,\hat{\boldsymbol{\imath}} + (e^x + \sin y + x \cos y)\,\hat{\boldsymbol{\jmath}}$. We are to show that it is independent of path. That is the case if and only if $\mathbf{F}$ is conservative. So let's look for a potential $\varphi$ for $\mathbf{F}$. That is, let's look for a function $\varphi$ that obeys

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y) &= ye^x + \sin y \\ \frac{\partial \varphi}{\partial y}(x,y) &= e^x + \sin y + x \cos y \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y) = ye^x + x\sin y + f(y)$$

Substituting this into the second equation gives

$$e^x + x\cos y + f'(y) = e^x + \sin y + x \cos y\qquad\text{or}\quad f'(y) = \sin y$$

which integrates to

$$f(y) = -\cos y + C$$

So $\mathbf{F}$ is indeed conservative with one potential being $\varphi(x,y) = ye^x + x\sin y -\cos y$ and the line integral is

$$\begin{aligned} \int_C (ye^x + \sin y)\,\mathrm{d}x + (e^x + \sin y + x \cos y)\,\mathrm{d}y &=\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} =\varphi(x,y)\Big|^{(0,\pi/2)}_{(1,0)} \\ &=\Big[ye^x + x\sin y -\cos y\Big]^{(0,\pi/2)}_{(1,0)} \\ &=1+\frac{\pi}{2} \end{aligned}$$