Finding Flux of Vector Field F Through Cube S

[asi123]

Homework Statement

I need to find the flux of the vector field F through S (in the pic), when S represent the edges of a cube.
My question is, how do I find N (normal)? Do I need to split the curb and to find the flux through each face?

Homework Equations

The Attempt at a Solution

 

[asi123]

I think I found a way using Gauss' law, is this right?

 

[HallsofIvy]

Yes, Gauss' law is the way to go.

To answer your original question, and possibly as a check, doing the surface integral directly, the surface consists of the six faces of the cube and you would need to do each face separately. The normal vector is just the unit vector in the direction perpendicular to the face- in the direction of a coordinate axis.

For example, the (external) normal vector to the face defined by x= 1 is just \vec{i} and the integral over that face would just be the integral of the x component, with x= 1, dydz with y and z limits of integration 0 and 1.

The (external) normal vector to the face defined by x= 0 would be -\vec{i}

More generally, the "vector differential of surface area" of a surface given by the parametric vector equation \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k} is the cross product \vec{r}_u\times\vec{r}_v times dudv. The sign of course depends on the order of the cross product- you choose that to give the vector in the right direction for the problem.

Although in this simple cube problem we can find the normal vectors directly, the face x= 1 is given by the vector equation \vec{i}+ y\vec{j}+ z\vec{k} and we can use y and z themselves as the two parameters. Then
\vec{r}_y\times\vec{r}_z= \vec{i}\times\vec{k}= \vec{i}
as I said before. Of course since the cube itself lies between 0 and 1, the "outward pointing" or external normal must point in the positive x direction at x= 1 and in the negative x direction at x= 0.

 

[asi123]

Yes, Gauss' law is the way to go.

To answer your original question, and possibly as a check, doing the surface integral directly, the surface consists of the six faces of the cube and you would need to do each face separately. The normal vector is just the unit vector in the direction perpendicular to the face- in the direction of a coordinate axis.

For example, the (external) normal vector to the face defined by x= 1 is just \vec{i} and the integral over that face would just be the integral of the x component, with x= 1, dydz with y and z limits of integration 0 and 1.

The (external) normal vector to the face defined by x= 0 would be -\vec{i}

More generally, the "vector differential of surface area" of a surface given by the parametric vector equation \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k} is the cross product \vec{r}_u\times\vec{r}_v times dudv. The sign of course depends on the order of the cross product- you choose that to give the vector in the right direction for the problem.

Although in this simple cube problem we can find the normal vectors directly, the face x= 1 is given by the vector equation \vec{i}+ y\vec{j}+ z\vec{k} and we can use y and z themselves as the two parameters. Then
\vec{r}_y\times\vec{r}_z= \vec{i}\times\vec{k}= \vec{i}
as I said before. Of course since the cube itself lies between 0 and 1, the "outward pointing" or external normal must point in the positive x direction at x= 1 and in the negative x direction at x= 0.

So I can't represent the entire surface of the cube using parametric vector equation \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}?

And another thing, in this equation, what's u and what's v? is it like x and y?

10x.

 

[Defennder]

I'm pretty sure you must parametrise each individual cube face, although if the cube is oriented parallel to the coordinate axes it becomes pretty easy. u,v are just parameters. For example the vector equation of a line is \vec{OP_0} + t\vec{v} where t is the parameter. You need 2 parameters to represent a surface, just 1 for a line, and 3 for a volume.

 

[HallsofIvy]

So I can't represent the entire surface of the cube using parametric vector equation \vec{r}(u,v)= x(u,v)\vec{i}+ y(u,v)\vec{j}+ z(u,v)\vec{k}?lk

Not using differentiable functions, you can't. That is signaled by the sharp edges on the cube. Our ways of writing formulas have developed so that only smooth curves are surfaces can be represented by a single formula. Typically, you have "corners" or "edges" you need to break the problem into pieces.

And another thing, in this equation, what's u and what's v? is it like x and y?

10x.

Well, yes, "like" x and y. A surface is, by definition, two dimensional. That means that every point on a (smooth) surface can be represented as some function of two parameters. I'd be very surprised if you got this far without having met parametric equations for at least curves (which depend on one parameter because a curve is one dimensional). I was using u and v to represent whatever the parameters are. Of course, if you can write z as a function of x and y for the surface then you can use x and y themselves as the parameters. For example, the parabolid z= x²+ y[/sup]2[/sup] can be written with parametric equations x= u, y= v, z= u²+ v[/sup]2[/sup], so that a vector equation is \vec{r}(u,v)= x\vec{i}+ y\vec{v}+ z\vec{k}=u\vec{i}+ v\vec{v}+ (u^2+ v^2)\vec{k} or just use the letters x and y instead of u and v:\vec{r}(x,y)= x\vec{i}+ y\vec{v}+ (x^2+ y^2)\vec{k}

On the other hand, we cannot do that with a sphere, because z cannot be written as a single valued function of x and y (nor can y be written as a function of x and z nor x as a function of y and z). What we can do in that situation is use the spherical coordinates for 3-space with \rho set equal to the radius of the sphere: x= R cos(\theta) sin(\phi), y= R sin(\theta) sin(\phi), z= R cos(\phi) where the parameters are now \theta and \phi.

In your problem, you could represent the side of the cube where y= 1 by using x and z, which can take on any values between 0 and 1, as parameters. You could write the vector equation of that side as \vec{r}(x,z)= x\vec{i}+ \vec{j}+ z\vec{k}[/tex].

 

[asi123]

Not using differentiable functions, you can't. That is signaled by the sharp edges on the cube. Our ways of writing formulas have developed so that only smooth curves are surfaces can be represented by a single formula. Typically, you have "corners" or "edges" you need to break the problem into pieces.


Well, yes, "like" x and y. A surface is, by definition, two dimensional. That means that every point on a (smooth) surface can be represented as some function of two parameters. I'd be very surprised if you got this far without having met parametric equations for at least curves (which depend on one parameter because a curve is one dimensional). I was using u and v to represent whatever the parameters are. Of course, if you can write z as a function of x and y for the surface then you can use x and y themselves as the parameters. For example, the parabolid z= x²+ y[/sup]2[/sup] can be written with parametric equations x= u, y= v, z= u²+ v[/sup]2[/sup], so that a vector equation is \vec{r}(u,v)= x\vec{i}+ y\vec{v}+ z\vec{k}=u\vec{i}+ v\vec{v}+ (u^2+ v^2)\vec{k} or just use the letters x and y instead of u and v:\vec{r}(x,y)= x\vec{i}+ y\vec{v}+ (x^2+ y^2)\vec{k}

On the other hand, we cannot do that with a sphere, because z cannot be written as a single valued function of x and y (nor can y be written as a function of x and z nor x as a function of y and z). What we can do in that situation is use the spherical coordinates for 3-space with \rho set equal to the radius of the sphere: x= R cos(\theta) sin(\phi), y= R sin(\theta) sin(\phi), z= R cos(\phi) where the parameters are now \theta and \phi.

In your problem, you could represent the side of the cube where y= 1 by using x and z, which can take on any values between 0 and 1, as parameters. You could write the vector equation of that side as \vec{r}(x,z)= x\vec{i}+ \vec{j}+ z\vec{k}[/tex].

<br /> <br /> thanks.