I really need some help.

1. Apr 29, 2005

ArnfinnS

hi..i really need some help. we have that alpha(x,y,x) = 1/r , where r^= x^2 + y^2 + z^2

and where the fector field F = nabla alpha (gradient of alpha)

i calculated F to be :
(-xx^ + yy^+zz^)/r^3

but iam supposed to calculate the flux integral
I_s F*n dS
but i dont know how to find the normal vector n. can anyone help me with that?
(you dont need to calculate it, i just need hints of how i can find it.)

PS ..s is a area without the point (0,0,0)

2. Apr 29, 2005

Hippo

One needs two parameters to define a surface in 3-dimensions.

Often, people use

$$z = f(x, y)$$

or, perhaps,

$$r = f(\theta, \phi)$$

Regardless of whether such functions can be provided, any point on the surface may be represented as a vector in the space where the surface exists.

At each such point one may specify two additional vectors, those tangent to the surface at the point.

Obviously, a cross product of those vectors at a point will be normal to the surface at that point.

You may wonder, but will there not be two cross products, each the negation of the other? Yes, that's true, and the choice of the normal vector is known as the orientation, "positive" or "negative" for a given surface.

Of course, the above holds only for orientable surfaces, that is, those with two distinct sides (e.g. It wouldn't hold for a
Mobius Strip.).

So...

Let us define every point on a surface S as

$$\vec S = \left< f(u,v), g(u, v), h(u, v) \right>$$

where u and v are the necessary parameters.

The tangent vectors would be

$$\vec S _{u} = \left< f_{u}, g_{u}, h_{u} \right>$$

and

$$\vec S _{v} = \left< f_{v}, g_{v}, h_{v} \right>$$

where the subscripts denote partial differentiation.

Thus, the normal vectors would be

$$\vec S _{u} \times \vec S _{v}$$

and

$$\vec S _{v} \times \vec S _{u}$$

[EDIT: The above information may not be sufficient, but I find it difficult to provide more, seeing as you didn't really specify your surface of integration. I'd like to be more helpful, if you'd be more specific.]

Last edited by a moderator: Apr 29, 2005
3. Apr 30, 2005

ArnfinnS

thanks for helping.

thanks for help.
well here is more about my problems :

we have Greens Formula I (Pdx + Qdy)=II_omega(dQ/dx - dP/dy)dxdy
= I_(boundary of omega) F * dr
where the scalar product is supposed to be that it is possible to orthogonal project P on the tangent to Boundary of omega (can anyone in this world explain to me what this means????!!!!!)

Futher : From the Linear algebra it is supposed to be known that (Tx)*Ty = X*y where T is a orthogonal transformation on R^3? ( what is an orthogonal transformation????)

We should see futher that the vector filed F and F^ is ortogonal where F=(P,Q) and F^=(-Q,R) Iam supposed from this information to interpret the integral :

I_(boundary omega) (-Qdx + Pdy) = I(_boundary omega) F*dr
geometrically.

Hint : if ds is the length of boundary omega so is dr = t^ds where t^is the unit tangent to boundary omega. If F Rotate the angle pi/2 , we have to rotate t^with the same angle for that the value of scalar product shouldnt change.

Could someone help and PLEASE show me how i can do this? its really important for me to get this problem solved...and understand the idea behind it...exam is around soon ...:(