Question

Good evening. :) I need a little help understanding the paramiterization of variables along a rectangular path for setting up a line integral. Image attached.

Answer 1

I've left it generic so that I'll be able to work through it on my own.

Answer 2

Lecture 20 shows a similar problem
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-20-path-independence/

Answer 3

Great, thanks. I'll take a look right now.

Answer 4

if we let F= <3y, 2x> and dr = <dx,dy>
then
\[ \int F \cdot dr = \int 3y \ dx + 2x \ dy \]
starting at 0,0 and moving to a,0
x = 0 to a
y = 0 and so dy = 0
using this info in the integral
\[ \int 3 \cdot 0 \ dx + 2x \cdot 0 = 0 \]
i.e with y=0 and dy=0 both terms drop out and we get 0

Answer 5

Now let's integrate from (a,0) to (a,b)
x= a (constant) so dx = 0
y = 0 to b
the integral is
\[ \int_0^b 3 y \cdot 0 + 2a\ dy \]
i.e. the first term with dx=0 drops out. x is a constant value =a
thus the line integral becomes an integral over dy
\[ 2a \int_0^b dy = 2a \ y\bigg|_0^b = 2ab
\]

Answer 6

the other paths are treated similarly. The only thing is be careful of the order of the start and end values of the integral (we go "backwards")

Answer 7

So in your first response, am I interpreting correctly if I look at r as a vector and the right-hand side of the equation is the same equation broken into it's x and y components?

Answer 8

yes, though I defined dr = <dx,dy>

Answer 9

Lecture 19 (the one preceding the one I just posted) introduces these ideas

Answer 10

Ok, great. I think that's enough to go on. I usually try to intuit my way through things until I'm just short of an aneurysm. It's probably time to surrender my pride. :) Thank you for your help.

Answer 11

If nothing else, Denis Auroux is an excellent lecturer (French accent not withstanding), and generally very clear.

Answer 12

I speak French (though not as well as he speaks English), so it should be fine. Thanks again.

Answer 13

Got it! I'll do a bunch more, of course.