선적분

(a) Parametrize $C$ by $x$. When the first component of a point on the curve is $x$, then the second component, $y$, must be $x^2$ and the third component, $z$, must be $x^3$. So

$$\begin{aligned} \mathbf{r}(x)&=x\,\hat{\boldsymbol{\imath}}+x^2\,\hat{\boldsymbol{\jmath}}+x^3\,\hat{\mathbf{k}}\qquad 0\le x\le 1\\ \mathbf{r}'(x)&=\hat{\boldsymbol{\imath}}+2x\,\hat{\boldsymbol{\jmath}}+3x^2\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}x}(x)&=\sqrt{1+4x^2+9x^4} \end{aligned}$$

and

$$\begin{aligned} \rho(x)\ \frac{\mathrm{d}s}{\mathrm{d}x}(x) &=\big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\\ \int_C\rho\ \mathrm{d}s &=\int_0^1 \big(8x+36x^3\big)\sqrt{1+4x^2+9x^4}\,\mathrm{d}x \end{aligned}$$

Substituting $u=1+4x^2+9x^4$, $\mathrm{d}u = \big(8x+36x^3\big)\,\mathrm{d}x$, $u(0)=1$, $u(1)=14$,

$$\begin{aligned} \int_C\rho\ \mathrm{d}s &=\int_1^{14} \sqrt{u}\ \mathrm{d}u =\frac{2}{3}u^{3/2}\bigg|_1^{14}\\ &=\frac{2}{3}\big[14^{3/2}-1\big]\approx 34.26 \end{aligned}$$

(b) Since $\mathbf{F}(x,y,z)=\nabla f(x,y,z)$ with $f(x,y,z)=x\sin y+yz+\tfrac{1}{2} z^2$,

$$\int_C \mathbf{F}\cdot \mathrm{d}\mathbf{r}=f(1,1,1)-f(0,0,0) =\sin 1+\frac{3}{2} \approx 2.3415$$

The potential $f$ was just guessed. Alternatively, it can be found by solving

$$\begin{aligned} \frac{\partial f}{\partial x}(x,y,z) &= \sin y \\ \frac{\partial f}{\partial y}(x,y,z) &= x\cos y+z \\ \frac{\partial f}{\partial z}(x,y,z) &= y+z \end{aligned}$$

Integrating the first of these equations gives

$$f(x,y,z) = x\sin y +g(y,z)$$

Substituting this into the second equation gives

$$x\cos y+\frac{\partial g}{\partial y}(y,z) = x\cos y+z\qquad\text{or}\qquad \frac{\partial g}{\partial y}(x,z)=z$$

which forces

$$g(y,z) = yz + h(z)$$

Finally, substituting $f(x,y,z) = x\sin y +yz + h(z)$ into the last equation gives

$$y + h'(z) =y+z \quad\text{or}\quad h'(z) = z$$

So $h(x) = \frac{z^2}{2}+C$ and hence $f(x,y,z) = x\sin y + yz + \frac{z^2}{2}+C$ for any constant $C$.