# Thread: Using symmetry for curve integrals...

1. ## Using symmetry for curve integrals...

Problem: Use symmetry considerations to find the following. (Picture of curves attached.)

(a) Let C be the polygonal curve shown in Diagram (a).

Compute $\oint (e^{x^2} - 2xy)dx + (2xy - x^2)dy$.

(b) Let C be the curve shown in Diagram (b); you might visualize as a racetrack with two semicircular ends.

Compute $\oint (4x^3y - 3y^2)dx + (x^4 + e^{\sin(y)})dy$.

For (a), my method was to split the curve in five curves, $C_1 \to C_5$, where
$C_1$ goes from (-1,1) to (3,1),
$C_2$ goes from (3,1) to (4,2),
$C_3$ goes from (4,2) to (3,3),
$C_4$ goes from (3,3) to (-1,3), and
$C_5$ goes from (-1,3) to (-1,1).

Then I parametrized each of the curves as follows (where $0 \leq t \leq 1$ for each curve):
$C_1: x = 4t - 1, y = 1, dx = 4dt, dy = 0$,
$C_2: x = t + 3, y = t + 1, dx = dt, dy = dt$,
$C_3: x = 4 - t, y = t + 2, dx = -dt, dy = dt$,
$C_4: x = 3 - 4t, y = 3, dx = -4dt, dy = 0$, and
$C_5: x = -1, y = 3 - 2t, dx = 0, dy = -2dt$.

Then the next step would be to split the integral into five integrals, one for each curve, and add them together. But the problem is the $e^{x^2}$ term that hinders my integration attempts after the parametrization, and I can't evaluate the integral.

I assume that there must be a better approach to doing this problem since I haven't actually used symmetry as it asks. But I'm not exactly sure how to apply symmetry to curve integrals.

2. I'm pretty sure you just want to simplify the region by cutting in in half over its axis of symmetry and then double the result.

It's been awhile since I took Calculus III though so I could be wrong.

3. $\oint{e^{x^2}}dx\:=\:0$ - due to symmetry
$\int_{-1}^{4}{e^{x^2}}dx\:+\int_{4}^{-1}{e^{x^2}}dx\:=\:0$

4. So if I use that concept in looking at my first four curves (since dx = 0 in the fifth curve), I can apply it as follows:

$\int_{-1}^3 {(e^{x^2})dx} + \int_{3}^4 {(e^{x^2})dx} + \int_{4}^3 {(e^{x^2})dx} + \int_{3}^{-1} {(e^{x^2})dx}$

$= \left( \int_{-1}^3 {(e^{x^2})dx} + \int_{3}^{-1} {(e^{x^2})dx} \right)$ $+ \left(\int_{3}^4 {(e^{x^2})dx} + \int_{4}^3 {(e^{x^2})dx} \right)$

$= \left(\int_{-1}^3 {(e^{x^2})dx} - \int_{-1}^{3} {(e^{x^2})dx} \right)$ $+ \left(\int_{3}^4 {(e^{x^2})dx} - \int_{3}^4 {(e^{x^2})dx} \right) = 0$

So the problem then can be solved as I had originally set it up, with the only difference being that I need to now solve the closed integral:

$\oint{(-2xy)dx + (2xy - x^2)dy}$

5. Could I use the exact same logic to also show that $\oint{(-2xy)dx} = 0$ by symmetry to simplify the integral further?
6. Upon further reflection, I figured that it can't work with respect to $-2xy$ since the integral has a y in it and I am only integrating with respect to x for symmetrical purposes. So then y would be in the final answer, which isn't good.