Should I use $dS$ or $dr$? The question states "Integrate $f(x, y, z) = x + \sqrt y -z^2$ over the patch $C = C_1 + C_2$ and they proceed to give $C_1$ and $C_2$ as $r_1(t)$ and $r_1(t)$. I've been dealing with 2 forms of line integrals so far, integrations with respect to $dS$ where $dS = ||r'(t)||dt$ and integrals with $dr$ where $dr=r'(t)dt$ . I am fairly certain that I must use $dS$ since $f$ does not appear to be a vector, but I want to make sure.
 A: 
I've been dealing with 2 forms of line integrals so far, integrations
  with respect to $dS$ where $dS = ||r'(t)||dt$ and integrals with $dr$
  where $dr=r'(t)dt$ .

$ds$ is used for differential arc lenghts which should be a non-negative scalar.
$dr$ is a differential change of $r$ and is of the same type as $r$, e.g. scalar or vectorial.
It is used e.g. for summing up a scalar field values along a curve
$$
du = f(r) \lVert dr \rVert
$$
or e.g. for summing up scalar products of the path elements with a vector field
$$
du = f(r) \cdot dr
$$

I am fairly certain that I must use $dS$ since $f$ does not appear to
  be a vector, but I want to make sure.

The task is integration of $f$ values along some curve:
\begin{align}
I &= \int\limits_C f(r) \, \lVert dr \rVert \\
&= \int\limits_{C_1 + C_2} f(r) \, \lVert dr \rVert \\
&= \int\limits_{C_1} f(r) \, \lVert dr \rVert + 
\int\limits_{C_2} f(r) \, \lVert dr \rVert \\
&=
\int\limits_{a_1}^{b_1} f(r_1(t)) \lVert\dot{r_1}(t)\rVert \, dt +
\int\limits_{a_2}^{b_2} f(r_2(t)) \lVert\dot{r_2}(t)\rVert \, dt
\end{align}
A: I found this question while studying for my Math 13 (Multivariable Calculus) exam.  My curriculum uses Jon Rogawski & Collin Adam's FREEMAN/MacMillan Multivariable Calculus Third Edition textbook.  (I am not in any way implicitly or explicitly endorsing any textbook in this answer, but the specific textbook is important to the answer, so bear with me).  Based off of my experience with material put online by classes covering the same material at Colombia, Harvard, Duke, and other schools, my impression is that my textbook's choice in notation, though a bit convoluted, is not uncommon.
Here is the notation, with page references, from my textbook.


*

*ds = ||r'(t)||dt 


(p.922, Chapter 17)  Note that this is a scalar. Used for scalar path integrals over a curve C.


*

*dr = r'(t)dt


(p.926, Chapter 17)  Note that this is a vector.  Used for vector path integrals over a curve C.


*

*dS = ||N(u,v)||dudv


(p.953, Chapter 17.4)  Note that this is a scalar.  Used for scalar surface integrals (e.g, Flux) over a surface S with perimeter dS.  In this scenario, the perimeter dS is often referred to as D, for example, the double integral over D of dA (which results in the area of the mapped surface S).


*

*dS = N(u,v)dudv


(p.961, Chapter 17.5)  Note that this is a vector.  Used for vector surface integrals, for example, for calculating "the flux F across or through S."
The explicit delineation of definitions of ds, dr, dS, and dS in a particular textbook may seem a-topical and not addressing the specific question.  However, I believe that 1) the implicit statement that these definitions are a little bit arbitrary and may very between textbooks, and 2) the listed explanations of definitions for what is one of the two or three most commonly used textbooks for this subject, is worthy of being shared as its own answer.  I mean for this to more supplement than in any way compete with the other, more clearly explained and broadly useful math-based answer.  If this is in violation of S/M rules, comment and I will remove it.
A: The vector equation is formed from the parametric equation:
$$\vec r=\left\langle x(t),y(t),z(t)\right\rangle$$
$$\frac {d\vec r}{dt}=\left\langle\frac {dx} {dt},\frac {dy} {dt},\frac {dz} {dt}\right\rangle$$
Which can also be written as:
$$ d\vec r=\left\langle dx, dy, dz \right\rangle $$
Also a vector, but as we all know any vector can be written as a magnitude and a unit vector.
$$ d\vec r=\sqrt{(dx)^2+(dy)^2+(dz)^2}\, \hat u $$
$$ \boxed{d\vec r=ds\,\hat u}$$
Where the unit vector is:
$$ \hat u=\begin{pmatrix} \frac {dx} {\sqrt{(dx)^2+(dy)^2+(dz)^2}}\\ \frac {dy} {\sqrt{(dx)^2+(dy)^2+(dz)^2}}\\ \frac {dz} {\sqrt{(dx)^2+(dy)^2+(dz)^2}} \end{pmatrix}$$
$$ \hat u=\left\langle \frac {dx}{ds}, \frac {dy}{ds}, \frac {dz}{ds} \right\rangle $$
So there is the clear relation between $d\vec r$ and $ds$, the vector element and the arc length element.
