Calculus problem #2069

$\textbf{r}(t)=t\,\textbf{i}+sint\,\textbf{j}+cost\,\textbf{k}$ .

So $\;\:\mathrm{d}\textbf{r}=\mathrm{d}t\,\textbf{i}+cost\,\mathrm{d}t\textbf{j}-sint\,\mathrm{d}t\,\textbf{k}$

Now $F\cdot\,\mathrm{d}\textbf{r}=cost\,\mathrm{d}t+sint\,cost\,\mathrm{d}t+t\,sint\,\mathrm{d}t$ .

The given integral equals $\int_0^\pi\,cost\,\mathrm{d}t+sint\,cost\,\mathrm{d}t+t\,sint\,\mathrm{d}t=\pi=\boxed{3.142}$ .

NOTE: The first two integrands integrate to zero, which can be easily realized from their respective graphs.

From the definition of line integral of a vector field:

$N = \int_C \textbf{F} \cdot d\textbf{r} = \int_0^\pi \textbf{F}(\gamma (t)) \cdot \gamma ' (t)dt$

Since

$\gamma (t) = (t, \sin t, \cos t)$ , $\gamma ' (t) = (1, \cos t, -\sin t)$ , $\textbf{F}(\gamma (t)) = (z(t), y(t), -x(t)) = (\cos t, \sin t, -t)$

It follows that

$F(\gamma(t)) \cdot \gamma ' (t) = (\cos t, \sin t, -t) \cdot (1, \cos t, -\sin t) = \cos t + \sin t \cos t + t\sin t$

Hence

$N = \int_0^\pi (\cos t + \sin t \cos t + t\sin t)dt = (\sin t - \frac{\cos 2t}{4} + \sin t - t \cos t )\Bigg|_0^\pi$

$=-\pi \cos (-\pi) = \pi \approx \boxed{3.142}$

$\int_C \mathbf{F} \cdot d \mathbf{r} = \int_0^{\pi} \cos{t} \cdot \frac{d}{dt} t + \sin{t} \cdot \frac{d}{dt} \sin{t} - t \cdot \frac{d}{dt} \cos{t} = \boxed{\pi}$