Independence of Path and Fundamental Theorem of Calculus?

Question

a) Show that the given line integral is independent of path. How would you show this? Does this require assigning $C_1$ and $C_2$ two the two legs of the line?

b) Then, evaluate the line integral I by finding a potential function f and applying the Fundamental Theorem of Line Integrals. $$I= \int_{(0, 0)}^{(1, 2)} (x+y dx) + (x-y)dy$$

For b) my final answer was 5/2, but I'm not quite sure. Is this correct? If not, can you please show me how.

Thank you in advance!

@VincentPfenninger thanks for checking that. How would I show that the line given for A is independent of path? — Paul Barrett, Jul 04 '13 at 23:14
@VincentPfenninger It's the brief description right before the question a) — Paul Barrett, Jul 04 '13 at 23:21
Path independence follows from the fundamental theorem of line integrals since the vector field has a potential and is thus conservative. — Vincent Pfenninger, Jul 04 '13 at 23:32
@VincentPfenninger great, thank you very much for your help! — Paul Barrett, Jul 04 '13 at 23:34
Another way to show that the integral is path independent (without finding the potential function in advance) is to show that the curl of the vector field is zero, which guarantees the existence of a suitable potential function. — Ben Grossmann, Jul 04 '13 at 23:44

score 1 · Answer 1 · answered Jul 05 '13 at 02:33

A line integral $ \int_{C} \vec F\cdot d \vec r$ is path independent if $\vec F$ is a conservative field. $\vec F$ is conservative iff the domain of $\vec F$ is simply connected (in $\mathbb{R}^2$ this usually means that $\vec F$ has no poles) and $\nabla \times \vec F = \vec 0$.

It's easy enough to see that the domain of $\vec F$ is simply connected here and a quick calculation shows that $\nabla \times \vec F = (0, 0, 1-1) = \vec 0$, so $\vec F$ is conservative. Hence $\vec F = \nabla \phi$ for some scalar function $\phi$. Hence $x + y = \frac {\partial \phi}{\partial x}$, so $\phi = \frac{x^2}{2} + yx + f(y) $. Differentiating this with respect to y, we find $ \frac {\partial \phi}{\partial y} = x + f'(y)$, comparing this to our original integral, we see that $f'(y) = -y$, so $f(y) = -\frac{y^2}{2}$ and $\phi = \frac{x^2}{2} - \frac{y^2}{2} + yx$.

So $\int_{(0, 0)}^{(1, 2)} (x+y dx) + (x-y)dy = \phi|_{0,0}^{1,2} = \frac{1}{2} - \frac{4}{2} + (2)(1) = \frac{1}{2}$.

Independence of Path and Fundamental Theorem of Calculus?

1 Answers1