### Alternate notation for vector line integrals

In the introduction to line integrals of vector fields, we denoted line integrals \begin{align} \dlint = \dplint, \label{original_notation} \end{align} for a curve $C$ parametrized by $\dllp(t)$ for $t \in [a,b]$. Sometimes, we use an alternative notation for these line integrals, where we essential “multiply out” the dot products on both sides of the equation.

Let's assume we are in three dimensions, though this notation can also be used for two dimensions. We'll denote the components of the vector field as $\dlvf = (\dlvfc_1, \dlvfc_2, \dlvfc_3)$. We can think of the $d\lis$ as being a vector of infinitesimals in all three coordinates and denote it $d\lis = (dx,dy,dz)$. With this notational convention, we can rewrite the left hand side of equation \eqref{original_notation} so that it becomes \begin{align*} \dlint= \int_\dlc \dlvfc_1 dx +\dlvfc_2 dy + \dlvfc_3 dz. \end{align*}

We can write the components of $\dllp$ as $\dllp(t) = (x(t),y(t),z(t))$ and its derivative as \begin{align*} \dllp'(t) = \left( \diff{x}{t}, \diff{y}{t}, \diff{z}{t}\right). \end{align*} Multiplying out the dot product on the right hand side of equation \eqref{original_notation}, we can rewwrite it as \begin{align*} \dplint = \int_a^b \left(\dlvfc_1(\dllp(t)) \diff{x}{t} + \dlvfc_2(\dllp(t)) \diff{y}{t} + \dlvfc_3(\dllp(t)) \diff{z}{t}\right)dt. \end{align*}

Frequently, line integral problems are presented in this notation. In this case, one can compute them directly using the formula \begin{align} \int_\dlc \dlvfc_1 dx + \dlvfc_2 dy + \dlvfc_3 dz= \int_a^b \left(\dlvfc_1(\dllp(t)) \diff{x}{t} + \dlvfc_2(\dllp(t)) \diff{y}{t} + \dlvfc_3(\dllp(t)) \diff{z}{t}\right)dt. \label{alternative_formula} \end{align}

#### Example

Evaluate \begin{align*} \int_\dlc y\,dx + (x+y) dy + dz \end{align*} where thee curve $\dlc$ is parameterized by $\dllp(t)=(t, 1-t, 1)$, $0 \le t \le 1$.

**Solution**: If we write $\dllp(t) = (t,1-t,1) = (x(t),y(t),z(t))$, Then
$\diff{x}{t} = 1$, $\diff{y}{t} = -1$, and $\diff{z}{t} =0$. Since
$\dlvfc_1(x,y,z) = y$, $\dlvfc_2(x,y,z) = x+y$, and
$\dlvfc_3(x,y,z) =1$, we can use the formula of
equation \eqref{alternative_formula}
to calculate the integral.
\begin{align*}
&\int_\dlc y\, dx + (x+y) dy\\
&= \int_0^1 \left(y(t) \diff{x}{t} + (x(t)+y(t))\diff{y}{t}
+ \diff{z}{t} \right)dt\\
&= \int_0^1 \left[(1-t)(1) + (t + (1-t)) (-1)\right] dt\\
&= \int_0^1 [(1 - t) + (-1)] dt\\
&= \int_0^1 (-t)dt = \left. - \frac{t^2}{2} \right|_0^1 = -\frac{1}{2}
\end{align*}

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