How do you remember the order for the cross product in this surface integral?

[flyingpig]

Homework Statement

[tex]\iint \mathbf{F} \cdot \mathbf{dS} = \iint \mathbf{F} \cdot |\mathbf{r_u} \times \mathbf{r_v} | \mathbf{dA} = \iint \mathbf{F} \cdot |\mathbf{r_u} \times \mathbf{r_v} | \mathbf{\hat{n}}dA = \iint \mathbf{F} \cdot |\mathbf{r_u} \times \mathbf{r_v} | \frac{\mathbf{r_u} \times \mathbf{r_v} }{|\mathbf{r_u} \times \mathbf{r_v} |} dA[/tex]

In the last equation, notice how we no longer have an absolute value bar up there? Like the bottom and top cancels out for the absolute value cross products because the order doesn't matter, but what happens if you accidentally you reverse the terms in the cross product? How do you know you are wrong?

 

[HallsofIvy]

You have to have more information. Specifically, you need to know how the surface is to be oriented. The orientation of a surface determines the direction of its normal vector. For example, if your surface is the paraboloid z= x^2+ y^2, then r= <x, y, x^2+ y^2> so r_x= <1, 0, 2x> and r_y= <0, 1, 2y>. The two orders for their cross product will give -2xi- 2yj+ k which is oriented upward and 2xi+ 2yj- k which is oriented downward (look at the sign on k). Similarly, if the surface were a sphere, the orientation would be "inward" or "outward".

You can think of the integral of a vector over an oriented surface as giving the total flow of the of a fluid having the vector field as velocity field through the surface. But that flow can be positive or negative depending upon whether the flow in the direction of or opposite the direction of the normal to the surface, which has to be given.

 

[LCKurtz]

I always taught my classes that the correct formula was

[tex]\pm\iint_{(u,v)} \vec F \cdot \vec R_u \times \vec R_v\ dudv[/tex]

where the sign must be chosen to make the direction of the cross product agree with the orientation of the surface.

 

[flyingpig]

You have to have more information. Specifically, you need to know how the surface is to be oriented. The orientation of a surface determines the direction of its normal vector. For example, if your surface is the paraboloid z= x^2+ y^2, then r= <x, y, x^2+ y^2> so r_x= <1, 0, 2x> and r_y= <0, 1, 2y>. The two orders for their cross product will give -2xi- 2yj+ k which is oriented upward and 2xi+ 2yj- k which is oriented downward (look at the sign on k). Similarly, if the surface were a sphere, the orientation would be "inward" or "outward".

You can think of the integral of a vector over an oriented surface as giving the total flow of the of a fluid having the vector field as velocity field through the surface. But that flow can be positive or negative depending upon whether the flow in the direction of or opposite the direction of the normal to the surface, which has to be given.

What do you mean look at k? What if the surface was directed at the y-axs and you can't imagine what the surface looks like?

 

[HallsofIvy]

What do you mean look at k? What if the surface was directed at the y-axs and you can't imagine what the surface looks like?

I said "The two orders for their cross product will give -2xi- 2yj+ k which is oriented upward and 2xi+ 2yj- k which is oriented downward (look at the sign on k)". The first is oriented "upward" because the k component is positive and the second is oriented downward because the k component is negative.

I not sure what you mean by "directed at the y axis" but (orientable) surface has two sides and so two orientations, depending upon which way the normal vector is pointed. Whatever you call the orientation- upward/downward, left/right, forward/backward, inward downward, the two choices for sign in the cross product (order of multiplication) is determined by the orientation.