- #1
Orson1981
- 15
- 0
Why can't I get Stoke's Theorem to Work!?
This is very similar to the problem I put up last night actually part c of the same problem. I really have put an amazing amount of time into this problem (week and a half/ 5 hrs a day) and it countinues to stump me. At this point I'm spinning mental wheels and could use some help.
we have a 3-4-5 right triangle lying on the x-y plane with the length three side on the x-axis and the length 4 side on the y-axis. we define a field
[tex]\vec{u}=x^2\hat{x} + xy\hat{y}[/tex]
and I am trying to evaluate the integral:
[tex]\oint{\vec{u}d\vec{l}}[/tex]
I need to solve both:
[tex]\oint{\vec{u}d\vec{l}}[/tex]
and
[tex]\oint{(\nabla{}x\vec{u}) . \hat{n} dA}[/tex]
please excuse my clumsy notation, this should be del cross the vector u
The surface integral after doing del cross u I end up with
[tex]\frac {1}{2} \oint{ y \hat{k} dxdy} [/tex]
evaluated from 0 - 3 on the x-axis and 0 - 4 on the y-axis
Doing this I get 12.
the 1/2 I added into this equation is largely a guess on my part, and could be correct or not, just seemed to make some sense for a triagle who's area is 1/2 base*height
Looking at the line integral
I placed curve 1 along the y-axis from 4 - 0 then paramaterized x = 0 and dx = 0
so for curve 1 is zero
Curve 2, I paramaterized y = 0, dy = 0 that left
[tex]\int{x^2 dx}[/tex] evaluated from 0 - 3 gives me 9
Curve 3 along the hypotinous (I can't spell ... sorry) paramaterized y = 4 - 4/3 * x dy = -3/4 * dx
plugging these new values into y and dy and integrating from 3 - 0 gives me -1
So for the whole curve I have 8
So somewhere I missed something because last I checked 8 =! 12, or if you ignore my guess of 1/2 in the surface integral 8 =! 24
so I'm only off by a factor of 3 somewhere, though it could be coincidence that's a pretty nice number to be off by, makes me think I'm just missing a term somewhere.
If I had to guess I would say somehow [tex] \hat{n} [/tex] is somehow equal to 1/3, or perhaps 1/6 to include my guess about 1/2 in the surface integral
Thank you for reading this long rambling thing, and thanks for any help
-Mo
Homework Statement
This is very similar to the problem I put up last night actually part c of the same problem. I really have put an amazing amount of time into this problem (week and a half/ 5 hrs a day) and it countinues to stump me. At this point I'm spinning mental wheels and could use some help.
we have a 3-4-5 right triangle lying on the x-y plane with the length three side on the x-axis and the length 4 side on the y-axis. we define a field
[tex]\vec{u}=x^2\hat{x} + xy\hat{y}[/tex]
and I am trying to evaluate the integral:
[tex]\oint{\vec{u}d\vec{l}}[/tex]
Homework Equations
I need to solve both:
[tex]\oint{\vec{u}d\vec{l}}[/tex]
and
[tex]\oint{(\nabla{}x\vec{u}) . \hat{n} dA}[/tex]
please excuse my clumsy notation, this should be del cross the vector u
The Attempt at a Solution
The surface integral after doing del cross u I end up with
[tex]\frac {1}{2} \oint{ y \hat{k} dxdy} [/tex]
evaluated from 0 - 3 on the x-axis and 0 - 4 on the y-axis
Doing this I get 12.
the 1/2 I added into this equation is largely a guess on my part, and could be correct or not, just seemed to make some sense for a triagle who's area is 1/2 base*height
Looking at the line integral
I placed curve 1 along the y-axis from 4 - 0 then paramaterized x = 0 and dx = 0
so for curve 1 is zero
Curve 2, I paramaterized y = 0, dy = 0 that left
[tex]\int{x^2 dx}[/tex] evaluated from 0 - 3 gives me 9
Curve 3 along the hypotinous (I can't spell ... sorry) paramaterized y = 4 - 4/3 * x dy = -3/4 * dx
plugging these new values into y and dy and integrating from 3 - 0 gives me -1
So for the whole curve I have 8
So somewhere I missed something because last I checked 8 =! 12, or if you ignore my guess of 1/2 in the surface integral 8 =! 24
so I'm only off by a factor of 3 somewhere, though it could be coincidence that's a pretty nice number to be off by, makes me think I'm just missing a term somewhere.
If I had to guess I would say somehow [tex] \hat{n} [/tex] is somehow equal to 1/3, or perhaps 1/6 to include my guess about 1/2 in the surface integral
Thank you for reading this long rambling thing, and thanks for any help
-Mo