# Line Integral Around a Triangle

Let $R$ be the interior of the triangle with vertices $(0,0), (4,2),$ and $(0,2)$. Let $C$ be the boundary of $R$, oriented counterclockwise. Now evaluate the integral below.

$$\int_C(y+e^\sqrt{x}) dx + (xe^{y^2}) dy$$

I know this has to be parametrized somehow, but I'm not sure where to start. Could someone show me how to set up the integral so it can be evaluated?

Thanks.

-
I think you can use Green's theorem to avoid those dreadful parametrizations and then apply Fubini's theorem. – el.Salvador Jul 22 '14 at 17:39
So would $(y+e^{\sqrt{x}})\textbf{i}+(xe^{y^2})\textbf{j}$ be the vector field? I haven't learned Fubini's Theorem yet, so is there another way? @Pringoooals – user7000 Jul 22 '14 at 17:49
It turns out to be a bad idea, because $\frac{\partial}{\partial x} \left ( xe^{y^2} \right ) - \frac{\partial}{\partial y} \left ( y + e^{\sqrt{x}} \right ) = e^{y^2} - 1$ and $e^{y^2}$ doesn't have an antiderivative in elementary terms. Too bad; you'll have to do a bunch of calculations. – el.Salvador Jul 22 '14 at 17:59
@Pringooals, that's what I tried the first time :/ Is there another way? Thanks – user7000 Jul 22 '14 at 17:59

As Pringoooals suggested, use Green's Theorem. The key is to integrate $x$ first, then $y$.

$\displaystyle\int_C(y+e^\sqrt{x}) dx + (xe^{y^2}) dy = \iint_{\Omega}\left[\dfrac{\partial}{\partial x}(xe^{y^2}) - \dfrac{\partial}{\partial x}(y+e^{\sqrt{x}})\right]\,dx\,dy = \iint_{\Omega}(e^{y^2}-1)\,dx\,dy = \int_{0}^{2}\int_{0}^{2y}(e^{y^2}-1)\,dx\,dy = \int_{0}^{2}2y(e^{y^2}-1)\,dy = \left[e^{y^2}-y^2\right]_{0}^{2} = e^4-5$.

-

Equivalently, you may go from the following way:

$$A: x=0,~ 0\le y\le 2\longrightarrow\int_A=0$$ $$B: y=2,~0\le x\le4\longrightarrow \int_B=\int_0^4(2+e^{\sqrt x})dx$$ $$C:=0\le x\le4,~y=\frac{x}2\longrightarrow \int_C=\int_0^4\left(\frac{x}2+e^{\sqrt x}\right)dx+\frac{1}2\int_0^4(xe^{x^2/4})dx$$

-

You don't really need to setup any parametrization first. You should first simplify the integral and then try either $x$ or $y$ as your parametrization unless there are other obvious choices.

1. $\displaystyle\;\int_C y dx$ is just negative of the area of $R$ and hence $= -\frac12 (2 \times 4) = -4$.
2. $e^{\sqrt{x}} dx$ is a total differential, so its integral over any closed curve, eg. $C$, will be zero.

3. For $\displaystyle\;\int_C x e^{y^2} dy$,

• the contribution from the line segment $(0,2) \to (0,0)$ is zero because $x = 0$ there.
• the contribution from the line segment $(4,2) \to (0,2)$ is zero because $dy = 0$ there.

The leaves us with the integral along the line segment $(0,0) \to (4,2)$. Since $x = 2y$ on that line segment, we get $$\int_C x e^{y^2} dy = \int_0^2 2y e^{y^2} dy = \left[ e^{y^2} \right]_0^2 = e^4 - 1$$

Combine all these, we find the integral is simply $e^4 - 5$.

As you can see, you don't really need to worry about the parametrization. Once you simplify your integral and identify the piece you need to deal with, the choice of parametrization is obvious.

To pick a parametrization, the main criterion is having one that make your life easier. Until you analyze your integrand, you usually don't know what is the best choice. The best course of action is delay the decision until you have a better picture what you are dealing with.

-

Parametrize each segment separately:

1)Segment from $(0,0)$ to $(4,2)$: $\gamma(t)=(4,2)t$ for $0<t<1$

2)From $(4,2)$ to $(0,2)$:$\gamma(t)=(4,2)+t((0,2)-(4,2))$ for $0<t<1$

3)From $(0,2)$ to $(0,0)$:$\gamma(t)=(0,2)+t((0,0)-(0,2))$ for $0<t<1$.

-