Question

How to find the closed path when using Stoke's theorem from points given?

Answer 1

For example:
(0,0,0) to (2,0,0) to (0,2,1) to (0,0,0)

Answer 2

At Z = y/2

Answer 3

The easiest thing to do is parameterize the path like this:
If you're going from point \(\vec{a}\) to point \(\vec{b}\), then
\[ \vec{r}(t) = \vec a + t\cdot \left(\vec{b} - \vec{a} \right) \]
where t runs from zero to one.

For example, along the first segment,
\[\vec{r}(t) = <0,0,0> + t\cdot<2,0,0>= <2t,0,0> \]
You can see that at t=0 you're at point \(\vec a\) and at t=1 you're at point \(\vec b\).

Answer 4

Right and at those points they would help me to determine the limits of integration of a function?

Answer 5

\[\int\limits_{C}^{} F dr = \int\limits_{}^{} _S \int\limits_{}^{} (Curl F) Nds\]

Answer 6

Remember that
\[\oint \vec f \cdot d\vec r = \int_1 \vec f \cdot d\vec{r}_1 + \int_2 \vec f \cdot \vec{r}_2 + \int_3 \vec f \cdot d\vec r_3 \]
So you need to break the integral up into three integrals, one along each segment. If you parameterize like I did, then each one of those integrals will become an integral over the variable t, which will run from 0 to 1 each time.

Answer 7

And how would the Z = y/2 come into play?

Answer 8

That does not make sense in the context of the question.

Answer 9

I have a vector \[F(x,y,z) = -3y^2 i + 4z j +6xk\]
I found the curl to be: <-4, -6, 6y >

Answer 10

to place it into stokes theorem I was given that triangular path which lies in the plane Z =y/2

Answer 11

choose ur surface and find its normal vector

Answer 12

What do you mean by choosing my surface?

Answer 13

projected onto the x-z plane

Answer 14

choose your surface as the plane passing containing the path

Answer 15

Not sure how you got the dS?

Answer 16

never seen that before ?

Answer 17

oh right the -g_x (x,y) and - g_y (x,y)

Answer 18

yes dot that with curl and setup the double integral xy plane

Answer 19

sorry the earlier formila is for ndS :

z = f(x,y)

ndS = <-f_x, -f_y, 1>dxdy

Answer 20

z = f(x,y) = y/2

ndS = <0, -1/2, 1> dxdy
right ?