Particle traveling on a Helix curve and work against gravity A particle moves around Helix curve $x = a\cos t, y=b\sin t, z = ct$ when $0 \le t \le 2\pi $
How do I calculate the work what the particle does against gravity $\mathbf F(x,y,z)=-mg\,\mathbf k$
I know that the work is $-\int_C \mathbf F\cdot \mathbf {dr}$
 A: You already have the curve in parametric form... If a curve $C$ has a parametrization $r: \mathbb{R}\to \mathbb{R}^3$ given by $r(t), t \in [a,b]$, then
$$
\int_C F \cdot dr = \int_a^b F(r(t))\cdot r'(t) dt
$$
In this case you have $r(t) = (a \cos t, b \sin t, ct)$, $r'(t) = (-a \sin t, b \cos t, c)$, $k = (0,0,1)$. So,
$$
-\int_C F \cdot dr= -\int_0^{2 \pi} (0,0,-mg)\cdot r'(t)dt = \int_0^{2 \pi} mgc dt = 2 \pi mgc
$$
[edit] Corrected $F(r(t))$.
A: Since the gravity is vertical, the work expression simplifies greatly due to $\mathbf k \cdot d\mathbf r = dz$,
$$-\int_C \mathbf F\cdot d\mathbf {r}=\int_z mgdz= mgc\int_0^{2\pi}dt= 2\pi mgc$$
A: Since all work done against gravity is converted into gravitational potential energy, the work done is the same as the change in potential energy, which you should be able to determine without integrating anything. At least that is how I would solve a problem like that in real life. 
Of course to do this you must use some general results in mathematical physics that depend on integration. And perhaps this exercise is set as a concrete demonstration of those results. Is it? This seems like an academic exercise, so context is everything. 
A: First note that with
$\mathbf r = \mathbf r(t), \tag 1$
that is, with $\mathbf r$ a function of $t$, we have
$d\mathbf r = \dot{\mathbf r}(t) \; dt; \tag 2$
and with $x(t)$, $y(t)$, $z(t)$ as given:
$x(t) = a\cos t, \tag 3$
$y(t) = b\sin t, \tag 3$
$z(t) = ct, \tag 4$
where
$0 \le t \le  2\pi, \tag 5$
we may write
$\mathbf r(t) = a\cos t \; \mathbf i + b\sin t \; \mathbf j + ct \; \mathbf k, \tag 6$
whence (2) becomes
$d\mathbf r = \dot{\mathbf r}(t) \; dt = -a\sin t \; dt \; \mathbf i + b\cos t \; dt \; \mathbf j + cdt \; \mathbf k,  \tag 7$
so with the force
$\mathbf F = -mg \mathbf k, \tag 8$
we have
$\mathbf F \cdot d\mathbf r  = -mg \mathbf k \cdot (  -a\sin t \; dt \; \mathbf i + b\cos t \; dt \; \mathbf j + cdt \; \mathbf k) = -mgc \; dt , \tag 9$
since
$\mathbf i \cdot \mathbf i = \mathbf j \cdot \mathbf j = \mathbf k \cdot \mathbf k = 1, \tag{10}$
and
$\mathbf i \cdot \mathbf j = \mathbf i \cdot \mathbf k = \mathbf j \cdot \mathbf k = 0; \tag{11}$
we may now compute
$-\displaystyle \int_C \mathbf F \cdot d\mathbf r =  -\int_0^{2\pi} \mathbf F \cdot d\mathbf r = -\int_0^{2\pi} ( -mgc )\; dt = 2\pi mgc.  \tag{12} $
