Scalar line integrals vs. line integrals when to calculate each?

I have read some discussion of the difference between scalar line integrals and line integrals, but I am still confused. For the problem: 1) Evaluate the scalar line integral $\int_{H}^{}$ $(x^2 + y^2 + z^2)$ds where H is the helix given by $\vec{c}^{\,}$(t) = $2cos(t)\vec{i}^{\,} + 2sin(t)\vec{j}^{\,} +2t\vec{k}^{\,}$, 0 $\leq$ t $\leq$2$\pi$. In the solution we do $\int_{0}^{2\pi}$ $4cos^2(t) + 4sin^2(t) +4t^2*\sqrt{4sin^2(t) + 4cos^2(t) + 4}$dt.

On the other hand for the problem: 2) Evaluate the line integral $\int_{C}^{}$ $(e^x\vec{i}^{\,} + xy\vec{j}^{\,})d\vec{s}^{\,}$ where C is parametrized by $\vec{r}^{\,}(t)=t\vec{i}^{\,}-t^2\vec{j}^{\,}, 0 \leq t \leq 1$. In the solution we get r'(t) = <1,-2t> and calculate $\int_{0}^{1} e^t(1) + t(-t^2)(-2t)dt$.

Why is that for the first one we use the square root of the derivative of $\vec{c}^{\,}$ in the square root and for the second we plug it in directly? Is it because the second problem is a line integral and the first is a scalar line integral? Any help is appreciated.

• Both are "line integrals", i.e., integrals whose values are independent of the parametrization used for the curve at stake (the values may depend on the "direction" of the curve). In each case: Compute what you are told to do. Various kinds of line integrals have a definite physical interpretation. – Christian Blatter Mar 18 '17 at 19:07
• So how should I know to use the first formula, taking the square root of the squared values and multiplying it by the problem? – user3293643 Mar 19 '17 at 18:35
• Compute what you are told to. In the second problem $d\vec s$ should be replaced by $d\vec r$, in so far as the variable $s$ seems to refer to arc length, so that $ds=\sqrt{|\dot r(t)|^2}\>dt$. – Christian Blatter Mar 19 '17 at 19:26
• I think I am having trouble figuring out what I am told to do. In the first one do I do arc length because it says ds rather than $d\vec{s}^{\,}$ – user3293643 Mar 19 '17 at 20:40