Compute the flux from left to right across a curve

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Homework Statement



Compute the flux of [itex]\overrightarrow{F}(x,y) = (-y,x)[/itex] from left to right across the curve that is the image of the path [itex]\overrightarrow{\gamma} : [0, \pi /2] \rightarrow \mathbb{R}^2[/itex], [itex]t \mapsto (t\cos(t), t\sin(t))[/itex].

A (2-space) graph was actually given, and the problem referenced "the curve" as given in polar coordinates by [itex]r=\theta[/itex] and [itex]0\leq \pi/2[/itex], so the above parameterization was my doing.

Homework Equations



Flux is given by [itex]\int\int_{S}{\overrightarrow{F} \cdot d\overrightarrow{S}} = \int\int_{D}{\overrightarrow{F} \cdot \overrightarrow{n} dA} = \int\int_{D}{- F_1 g_x - F_2 g_y + F_3} dA[/itex], where [itex]\overrightarrow{F}[/itex] is a vector field defined on a surface [itex]S[/itex] given by [itex]z = g(x,y)[/itex], oriented by a unit normal [itex]\overrightarrow{n}[/itex].

The Attempt at a Solution



First, I am thrown off by the "from left to right" requirement--I don't know what that means. The idea I have is that it has something to do with the direction of the normal.

Also, I have only really dealt with (3-space) surfaces.

Second, I am used to computing the normal as the cross product of the parametrized surface of [itex]S[/itex] differentiated with respect to each of the two variables parameterizing it. Here, I only have one variable. Would I change my parameterization to have two variable but only use one?

Thanks.
 

Answers and Replies

  • #2
Dick
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Homework Statement



Compute the flux of [itex]\overrightarrow{F}(x,y) = (-y,x)[/itex] from left to right across the curve that is the image of the path [itex]\overrightarrow{\gamma} : [0, \pi /2] \rightarrow \mathbb{R}^2[/itex], [itex]t \mapsto (t\cos(t), t\sin(t))[/itex].

A (2-space) graph was actually given, and the problem referenced "the curve" as given in polar coordinates by [itex]r=\theta[/itex] and [itex]0\leq \pi/2[/itex], so the above parameterization was my doing.

Homework Equations



Flux is given by [itex]\int\int_{S}{\overrightarrow{F} \cdot d\overrightarrow{S}} = \int\int_{D}{\overrightarrow{F} \cdot \overrightarrow{n} dA} = \int\int_{D}{- F_1 g_x - F_2 g_y + F_3} dA[/itex], where [itex]\overrightarrow{F}[/itex] is a vector field defined on a surface [itex]S[/itex] given by [itex]z = g(x,y)[/itex], oriented by a unit normal [itex]\overrightarrow{n}[/itex].

The Attempt at a Solution



First, I am thrown off by the "from left to right" requirement--I don't know what that means. The idea I have is that it has something to do with the direction of the normal.

Also, I have only really dealt with (3-space) surfaces.

Second, I am used to computing the normal as the cross product of the parametrized surface of [itex]S[/itex] differentiated with respect to each of the two variables parameterizing it. Here, I only have one variable. Would I change my parameterization to have two variable but only use one?

Thanks.

Yes, it's the choice of which normal to use. If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve. So use the normal pointing away from the origin. And if you take the derivative with respect to your parameter you'll get a tangent vector. In 2d finding the normal to that is easy. Just rotate it 90 degrees.
 
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Yes, it's the choice of which normal to use. If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve. So use the normal pointing away from the origin. And if you take the derivative with respect to your parameter you'll get a tangent vector. In 2d finding the normal to that is easy. Just rotate it 90 degrees.

Thanks for the response.

"If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve."

I don't understand what you are saying here. Could you please explain more explicitly? I am thinking you are talking about drawing the vector field of [itex]\overrightarrow{F}[/itex] over the graph of the curve. If that is the case, I am having trouble seeing that the vector field flows in the direction of increase of the curve.
 
  • #4
Dick
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Thanks for the response.

"If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve."

I don't understand what you are saying here. Could you please explain more explicitly? I am thinking you are talking about drawing the vector field of [itex]\overrightarrow{F}[/itex] over the graph of the curve. If that is the case, I am having trouble seeing that the vector field flows in the direction of increase of the curve.

No, I'm not paying any attention to F. Your curve goes out from the origin and then back to the y-axis in the first quadrant. You have two choices of normal, it can be rotated right or left from the tangent. If they say flux from left to right, I think you should pick the normal to the right.
 
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  • #5
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No, I'm not paying any attention to F. Your curve goes out from the origin and then back to the y-axis in the first quadrant. You have two choices of normal, it can be rotated right or left from the tangent. If they say flux from left to right, I think you should pick the normal to the right.

Thanks for the reply.

Alright... I think I need to work on my intuition for this stuff to better understand your help.

Anyway, I think I've got something decent, and I'm done with it. For anyone looking at this thread for help, I found this link helpful:

http://web.mit.edu/jorloff/www/18.02a-esg-iap/topic40.pdf

Thanks again Dick.
 
  • #6
Dick
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Thanks for the reply.

Alright... I think I need to work on my intuition for this stuff to better understand your help.

Anyway, I think I've got something decent, and I'm done with it. For anyone looking at this thread for help, I found this link helpful:

http://web.mit.edu/jorloff/www/18.02a-esg-iap/topic40.pdf

Thanks again Dick.

That's great! But let me try and explain again. If they give you a closed curve and say compute the flux they usually mean outward going flux. So you should pick the outward normal. So if they give you a curve and describe the direction (here left to right) you should pick the normal pointed outward from that direction. So that's rightwards.
 
  • #7
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That's great! But let me try and explain again. If they give you a closed curve and say compute the flux they usually mean outward going flux. So you should pick the outward normal. So if they give you a curve and describe the direction (here left to right) you should pick the normal pointed outward from that direction. So that's rightwards.

Gotcha; that makes a bit more sense now. Thanks for the generalization!
 

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