How do you parametrize a face of a cube?

[flyingpig]

Homework Statement

Evaluate the surface integral. For closed surfaces, use a positive orientation.

[tex]\mathbf{F} = <x,2y,3z>[/tex]

S is a cube with vertices [tex]<\pm1, \pm1,\pm1>[/tex]

Solution

[PLAIN]http://img34.imageshack.us/img34/9730/unledjfx.jpg

The Attempt at a Solution

I understand that my book broke the surface into 6 faces, but how did they parametrize the faces? I don't even understand how they got the partial derivatives?

How come the partial derivatives at the unit vectors? I am so confused

 

[tiny-tim]

hi flyingpig!

I understand that my book broke the surface into 6 faces, but how did they parametrize the faces?

eg on the top face, dS is (0, 0, dxdy), and they integrate both x and y from -1 to 1

I don't even understand how they got the partial derivatives?

How come the partial derivatives at the unit vectors?

what partial derivatives?

 

[flyingpig]

r_y x r_z

In the very beginning

 

[tiny-tim]

oh, no they're not partial derivatives, they're a weird way (which i don't completely understand ) of finding the direction of the unit normal

 

[flyingpig]

I know I am suppose to "try it first", but I really don't know how to set it up, could you show me how a regular person would work this problem? Please don't use that solution, I've staring blankly at this chapter (surface integrals) for a few days now

 

[HallsofIvy]

oh, no they're not partial derivatives, they're a weird way (which i don't completely understand ) of finding the direction of the unit normal

That's not a weird way- that's my favorite way. If a surface is defined by parametric equations x= f(u,v), y= g(u,v), z= h(u,v) or, equivalently, the vector equation f(u,v)= f(u,v)i+ g(u,v)j+ h(u,v)k then the two derivatives f_ui+ g_uj+ h_uk and f_vi+ g_vj+ h_vk are tangent vectors whose lengths reflect how u and v measure the surface. Their cross product is a vector perpendicular to the surface whose length gives the area measurement of the surface.

This is a consequence of the fact that if a parallelogram has two sides given by vectors u and b, then the area of the parallelogram is |u x v|.

 

[HallsofIvy]

One face of the cube with, say, vertices (1, 1, 1), (-1, 1, 1), (-1, -1, 1), (1, -1, 1), is the face defined by z= 1 (recall that a plane is defined by 3 points and any 3 of those points satisfy z= 1). It should be obvious that the "differential of area" on that plane (which is parallel to the xy-plane) is dxdy.

Of course, the position vector of any point on that plane is r(x,y)= xi+ yj+ k. The two partial derivatives of that vector (and they are partial derivatives) are r_x= i r_y= j. The cross product is, of course, k (or, changing the order of multiplication, -k).

The "vector differential of area" is k dx dy (or -k dxdy depending on the orientation of the surface) and the "scalar differential of area" is |k|dxdy= dxdy.

 

[flyingpig]

Is the picture in the solutions even right? The whole cube is shifted into the first octant...

 

[flyingpig]

Okay I remember how to find the planes now, but if I have to do each plane (6 of them) separately, isn't that going to be a lot of work?

normal vector dot <x-x0,y-y0,z-z0>

 

[LCKurtz]

Is the picture in the solutions even right? The whole cube is shifted into the first octant...

No, it isn't shifted. Look carefully at the axes.

 

[LCKurtz]

Okay I remember how to find the planes now, but if I have to do each plane (6 of them) separately, isn't that going to be a lot of work?

It might seem like a lot of work until you learn it. Once you understand, by practicing, you would likely be able to work the problem faster than you could post your questions.

For example, once you understand the basics, it is immediately obvious that, for example, on the left face the natural parameters are x and z, the area element is dxdz and the outward normal is -j.

One uncertainty I see in the statement of the problem is "for closed surfaces use positive orientation". Does that mean outward? Up? Right? Or what.

 

[Dick]

oh, no they're not partial derivatives, they're a weird way (which i don't completely understand ) of finding the direction of the unit normal

They really are partial derivatives. I think sides might be labelled incorrectly in the diagram, it's hard to tell. But if it really is a y-z side then a general vector on say x=1, has the representation r=<1,y,z>. That makes r_y (partial derivative) of r <0,1,0> and r_z=<0,0,1>. Those are both tangent vectors. So the normal is proportional to <0,1,0>x<0,0,1>. You have to orient it by eye. This is more useful in problems where the normal vector isn't kind of obvious.

 

[tiny-tim]

That's not a weird way- that's my favorite way. If a surface is defined by parametric equations x= f(u,v), y= g(u,v), z= h(u,v) …

ohhh!

but it seems like a sledgehammer to crack a nut in this case …

it parametrises the top face by x = 1x + 0y, y = 0x + 1y,

so ∂(xi + yj)/∂x = i, ∂(xi + yj)/∂y = j, giving us the normal as i x j = k

 

[HallsofIvy]

In this very simple case, yes, it is.

 

[flyingpig]

I still don't understand how could the normal vector just happened to be the unit vectors i,j,k.

 

[LCKurtz]

I still don't understand how could the normal vector just happened to be the unit vectors i,j,k.

The sides of the cube are parallel to the coordinate planes. If you draw a unit vector perpendicular to, for example, the xy plane, it will be k or -k won't it?