Help in understanding how to evaluate line integrals. I'm attempting to teach myself some vector calculus before starting university next month in hope of getting my head around some of the concepts as I can foresee this being a weak topic for me.
I have been 'learning' from some online lecture notes related to my course. The notes talk about line integrals but as far as I understand say little on how to evaluate them and only gives one quick example in the form below that I didn't find terribly useful. As a result I'm not entirely sure how to evaluate the line integral below and so I would ask that someone answer the below question, but if possible perhaps give more detail than would usually be necessary, talking through each step with a specific emphasis on the difference between evaluating (i) and (ii), thank you.

Evaluate explicitly the line integral $\int(y$ $dx+x$ $dy+dz)$ along
  (i) the straight path from the origin to $x=y=z=1$ and (ii) the
  parabolic path given parametrically by $x = t,y = t,z = t^2$ from $t=0$ to
  $t=1$.

Any help is appreciated.
Thank you.
 A: The trick mainly consists of parameterizing the curve $C$ in some parameter $t \in [0,1]$ and then you integrate
$$
\int_C f(x,y,z) = \int_0^1 f(x(t), y(t), z(t))
       \sqrt{|x'(t)|^2 + |y'(t)|^2 + |z'(t)|^2}dt.
$$
Let's do the first one together. The parameterization is obvious $x=y=z=t$ with $t \in [0,1]$, so $dx=dy=dz=dt$ and the integral becomes
$$
\int_C(ydx + xdy + dz) = \int_0^1 (tdt + tdt + dt) = \int_0^1 (2t+1)dt = \left. t^2 + t \right|_0^1 = 2.
$$
Please do the second one yourself.
A: You have to make a substitution with the parameter. I'll denote paths under the generic name $\gamma$.
For the first  integral, the parametrisation of the straight line is clearly $\;\begin{cases}x=t\\y=t\\z=t\end{cases}\enspace(0\le t\le 1)$, so $\;\mathrm d\mkern1mux=\mathrm d\mkern1mut=\mathrm d\mkern1muy=\mathrm d\mkern1muz $, and
$$\int_\gamma(y\,\mathrm d\mkern1mu x+x\,\mathrm d\mkern1mu y+\mathrm d\mkern1muz)=
\int_0^1(t\,\mathrm d\mkern1mu t+t\,\mathrm d\mkern1mu t+\mathrm d\mkern1mu t)=\int_0^1(2t+1)\,\mathrm d\mkern1mu t= t^2+t\,\bigg\vert_0^1=2.$$
A: The path $C$ is parametrized as $\mathbf r(t) = t\mathbf x + t\mathbf y + t\mathbf z$ for $t\in[0,1]$.  Then you use the line integral formula
$$\int_{C} \mathbf f\cdot d\mathbf r = \int_{t_0}^{t_1} \mathbf f(\mathbf r)\cdot \mathbf r'(t)\ dt$$
In this case $\mathbf f(\mathbf r)\cdot \mathbf r'(t) = (t\mathbf x + t\mathbf y + 1\mathbf z)\cdot(1\mathbf x + 1\mathbf y + 1\mathbf z) = 2t+1$.
Thus your line integral reduces to the Riemann integral
$$\int_0^1 [2t+1]\ dt$$
which is pretty easy to evaluate.
