Surface integrals of vector fields

[cjc0117]

The integral for calculating the flux of a vector field through a surface S with parametrization r(u,v) can be written as:

$\int\int_{D}F\bullet(r_{u}\times r_{v})dA$

But what's to stop one from multiplying the normal vector $r_{u}\times r_{v}$ by a scalar, which would result in a different flux? When you're asked to find the flux of a vector field through a surface, how do you know which scalar to use?

 

[Dick]

dA is du*dv, right? That's your formula. It doesn't say you can multiply the normal vector in the formula by an arbitrary number. Why would you do that?

 

[cjc0117]

I got confused because I was trying to solve a problem where you had to use Stoke's theorem to find the flux of the vector field <2x,3y,z³> through the curve of intersection of the paraboloid z=x²+y² and the plane 2x+4y+z=1, when C is traversed conterclockwise when viewed from the origin. So I found the curl to be <0,0,5>, the normal to the plane <2,4,1> and the region D traced out by the curve C to be a circle of radius sqrt(6). The flux I think is just -<0,0,5>*<2,4,1>π(sqrt(6))²=-30π. But then I started thinking, since any normal can be used to find the scalar equation of a plane, why can't the normal <2,4,1> be multiplied by, say, 2 to get <4,8,2>. It'll still be normal to the plane, right? Then the flux would be -60π. But now that I think of it more, that doesn't make sense, since the normal vector was derived from the scalar equation of the plane, which dictates that the normal vector be <2,4,1>.

 

[Dick]

I got confused because I was trying to solve a problem where you had to use Stoke's theorem to find the flux of the vector field <2x,3y,z³> through the curve of intersection of the paraboloid z=x²+y² and the plane 2x+4y+z=1, when C is traversed conterclockwise when viewed from the origin. So I found the curl to be <0,0,5>, the normal to the plane <2,4,1> and the region D traced out by the curve C to be a circle of radius sqrt(6). The flux I think is just -<0,0,5>*<2,4,1>π(sqrt(6))²=-30π. But then I started thinking, since any normal can be used to find the scalar equation of a plane, why can't the normal <2,4,1> be multiplied by, say, 2 to get <4,8,2>. It'll still be normal to the plane, right? Then the flux would be -60π. But now that I think of it more, that doesn't make sense, since the normal vector was derived from the scalar equation of the plane, which dictates that the normal vector be <2,4,1>.

Your formula uses a specific normal vector and a specific surface measure dA=du*dv. Sometimes you do it with a unit normal vector and a different notion of surface measure. The two are interrelated. You can't change one without changing the other.

 

[cjc0117]

Thank you. So it can't be changed just by virtue of the definition of the differential dS=(r_u x r_v)dudv, which relies on a unique parameterization r(u,v) (or at least unique within the scope of a single coordinate system)? I really don't understand why this was troubling me; I must have been very tired last night.