Question

example
find by using greens theorem and parametrization

Answer 1

@Zepdrix

Answer 2

What does the thing at the bottom of the contour say?
Gamma ... if you use Greene's Theorem? 0_o

Sec I gotta go read Paul's Notes a sec x'D

Answer 3

Oh it has to be a closed loop in order to apply this theorem, right?

Answer 4

yes thats gamma for greens theorem
but she said if u were to parametrize it u dont need gamma

Answer 5

Hmm :\ thinking...

Answer 6

I don't see any values labeled on the graph.
How high does c1 go? Does it end at y=1 or something?

Answer 7

Do you have points for the c's?

Answer 8

oh wait there are points leh me upload the one she wrote on the board

Answer 9

k lemm figure this out :x

Answer 10

And the final answer is supposed to be -2?
Darn, I came up with -4 ;c

Back to the drawing board.

Answer 11

hmm yeh

Answer 12

Ohh I made an Algebra mistake... I see it now.
Ok lemme fix my mistake and see if that tidies things up ;p
Then we'll go over it.

Answer 13

okie;p

Answer 14

Hmm that's even more wrong ;c lol
Thinkinggg

Answer 15

We could get some boundaries for the region if we label the sides like that:

0 <= x <= 2
0 <= y <= x+2

I'm not sure how, or why, we would parameterize that.. hmm

Answer 16

yeah

Answer 17

Nah :(

There is an **unlimited** number of ways to param this (\(\gamma\)) path integral

** Neat 'n' simple one:

\(x(t) = 2-t, y(t) = 0 \implies \mathbf r_{\gamma}(t) = <2-t,0>_{t \in [0,2]}\)

\(\mathbf r'_{\gamma}(t) = <-1,0>\)

\(\int_{\gamma} \mathbf F \cdot d \mathbf r = \int\limits_{0}^{2} <(2 - t) + (2 - t) \cdot 0, 0> \cdot <-1,0> dt = -2\)

** More Contrived, yet Simple One:

\( \mathbf r_{\gamma}(t) = <2- 2 \sin t,0>_{t \in [0,\frac{\pi}{2}]}\)

\(\mathbf r'_{\gamma}(t) = <- 2 \cos t,0>\)

\(\int_{\gamma} \mathbf F \cdot d \mathbf r = \int\limits_{0}^{\frac{\pi}{2}} <(2 - 2 \sin t) + (2 - 2 \sin t) \cdot 0, 0> \cdot <- 2 \cos t,0> dt\)

\(= \int\limits_{0}^{\frac{\pi}{2}} - 4 \cos t + 2 \sin 2t ~ d t = -2\)

** Obvious One:

\(x(t) = t, y(t) = 0 \implies \mathbf r_{\gamma}(t) = <t,0>_{t \in [2,0]}\)

Do that yourself, that's maybe the best place to start if you are lost; but careful as the interval is \(t = 2 \to 0\).

Thinking of parameter t as a proxy for time is so helpful as physics is where all this stuff originally came from, and that intuition drives a real practical understanding of what it really adds up to

ASIDE:

Green's Theorem: the CW path you have drawn/been given will give you the negative answer. Green's (and Stokes') and the whole of vector calculus are driven by the right hand/thumb rule....etc...so the algebra itself is asymmetric. If you do all the other line integrals, you need to watch out for the -ve sign you will inevitably get from Green if you work it correctly