- #1
FrogPad
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This question is throwing me for a loop.
Q: If u = x^2 in the square S = \{ -1<x,y<1\}, verify the divergence theorem when \vec w = \Nabla u:
\int\int_S div\,grad\,u\,dx\,dy = \int_C \hat n \cdot grad\,u\,ds
If a different u satisfies Laplace's equation in S, what is the net flow through C?
----
So the first thing I should do is show that each side of the expression match:
\vec w = \Nabla u = (2x,0)
(1) \int\int_S div\,grad\,u\,dx\,dy = \int_{-1}^{1} \int_{-1}^{1} \Nabla \cdot \vec w \, dx \, dy = 2 \int_{-1}^{1} \int_{-1}^{1} \,dx\,dy = 8
(2) \int_C \hat n \cdot \nabla u \,ds = \sum\limits_{i = 1}^4 \int_C \hat n_i \cdot \vec w(\vec r_i) ds
\vec w = (2x,0)
ds = \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} dt = \frac{d}{dt}(2(\vec r_{xi}(t)))dt
A parametric line is of the form:
\vec r(t) = \vec r_0 - t\vec r_0 + t\vec r_1
Since we have \vec w \cdot \hat n_i only the x-component of the path is used. And to trace the square we have the parametric paths:
r_1(t) = (2t-1,---)
r_2(t) = (1,---)
r_3(t) = (-2t+1,---)
r_4(t) = (-1,---)
The normal vectors for each path would be:
\hat n_1 = (0,1)
\hat n_2 = (1,0)
\hat n_3 = (0,1)
\hat n_4 = (-1,0)
Evaluating:
\int_0^1 (1,0)\cdot (2(1),0)2(1)dt + \int_0^1 (-1,0) \cdot (2(-1),0) 2(-1)dt = 4 - 4
Now I'm doing something wrong. But to be honest, I feel so lost on this problem it's not funny.
Some of my concerns are as follows:
* I'm not doing this question correctly AT ALL.
* \vec w is conservative. So the fundamental theorem of vector calculus should hold. So going from the start point to the end point which is the same point will yield 0? Or is this when you are only traveling in a vector field.
* This part of the question:
If a different u satisfies Laplace's equation in S, what is the net flow through C?
What is C? Is C the path traced around the square?* Ahhh! :(
If anyone could shed some light on this problem for me, that'd be awesome. Thanks.
Q: If u = x^2 in the square S = \{ -1<x,y<1\}, verify the divergence theorem when \vec w = \Nabla u:
\int\int_S div\,grad\,u\,dx\,dy = \int_C \hat n \cdot grad\,u\,ds
If a different u satisfies Laplace's equation in S, what is the net flow through C?
----
So the first thing I should do is show that each side of the expression match:
\vec w = \Nabla u = (2x,0)
(1) \int\int_S div\,grad\,u\,dx\,dy = \int_{-1}^{1} \int_{-1}^{1} \Nabla \cdot \vec w \, dx \, dy = 2 \int_{-1}^{1} \int_{-1}^{1} \,dx\,dy = 8
(2) \int_C \hat n \cdot \nabla u \,ds = \sum\limits_{i = 1}^4 \int_C \hat n_i \cdot \vec w(\vec r_i) ds
\vec w = (2x,0)
ds = \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} dt = \frac{d}{dt}(2(\vec r_{xi}(t)))dt
A parametric line is of the form:
\vec r(t) = \vec r_0 - t\vec r_0 + t\vec r_1
Since we have \vec w \cdot \hat n_i only the x-component of the path is used. And to trace the square we have the parametric paths:
r_1(t) = (2t-1,---)
r_2(t) = (1,---)
r_3(t) = (-2t+1,---)
r_4(t) = (-1,---)
The normal vectors for each path would be:
\hat n_1 = (0,1)
\hat n_2 = (1,0)
\hat n_3 = (0,1)
\hat n_4 = (-1,0)
Evaluating:
\int_0^1 (1,0)\cdot (2(1),0)2(1)dt + \int_0^1 (-1,0) \cdot (2(-1),0) 2(-1)dt = 4 - 4
Now I'm doing something wrong. But to be honest, I feel so lost on this problem it's not funny.
Some of my concerns are as follows:
* I'm not doing this question correctly AT ALL.
* \vec w is conservative. So the fundamental theorem of vector calculus should hold. So going from the start point to the end point which is the same point will yield 0? Or is this when you are only traveling in a vector field.
* This part of the question:
If a different u satisfies Laplace's equation in S, what is the net flow through C?
What is C? Is C the path traced around the square?* Ahhh! :(
If anyone could shed some light on this problem for me, that'd be awesome. Thanks.
Last edited: