I'm currently working on a practice problem for my Calculus III class:
"Evaluate the line integral along the negatively-oriented closed curve C, where C is the boundary of the triangle with vertices $(0,0)$, $(0,1)$, and $(1,0)$:" $$\oint_{c}^{}(e^{2x+y}dx) + (e^{-y}dy)$$
I know that the first order of business here is to break the integral into two (due to the addition) & find suitable parametric functions for the integrands, but I'm at a loss as to how to do that for exponentials such as these.
Many thanks in advance for any help you might be able to offer!
Given the orientation is negative, we must traverse the curve in a clockwise fashion when viewed from above : i.e, you must first go from $(0,0)$ to $(0,1)$ vertically upwards along $\ x = 0$, then proceed towards $(1,0)$ along $\ y + \ x = 1$ and then move left towards $(0,0)$ along $\ y = 0$
Break the curve C into three parts $\ C_1, \ C_2$ and $\ C_3$, each representing the three paths described above, respectively.
For $\ C_1$, it can be parametrized as $\ r(t) = 0 \ i + t \ j$, where 0 <= $\ t$ <= 1. Note that this implies $\ x = 0$ and $\ y = t $.
The reason for this is obvious - if you travel along the $\ y$-axis, $\ x$ remains 0 throughout, while $\ y$ is directly proportional to $\ t$.
The formula for the line integral of a given vector field $\ F$ along a given curve $\ r(t)$ is:
$$ \int \ F \cdot \ r'(t) dt $$
Notice from the given expression that $\ F = e^{2x+y} \ i + e^{-y} \ j $ which, along $\ r(t)$, further equals $\ e^{t} \ i + e^{-t} \ j$.
Now, computing the line integral, we get:
$$ I_1 = \int_0^1 (e^{t} \ i + e^{-t}\ j) \cdot ( 0 \ i + 1 \ j) dt $$
$$ = \int_0^1 e^{-t} dt = \frac{e-1}{e} $$
Similarly, $\ C_2$ can be parameterized as : $\ x(t) = t, y(t) = (1-t)$ for $\ 0 \le t \le 1$ and $\ C_3$ can be parametrized as : $\ x(t) = t, \ y(t) = 0 $ for $\ 0 \le t \le 1 $.
The rest is merely a matter of computation.