Because the gradient transforms a scalar function into a vector function, not the other way around like you have it here.
Just a quick question about notation.
I was given the vector field
F = r + grad(1/bar(r)) where r= (x)i+(y)j+(z)k.
grad is just written as the upside down delta (gradient) and the bar I wrote in the above equation looks like an absolute value around just the r (although I don't think it is absolute value).
What would be a simplification of this vector field so that I can solve the rest of the problem?
I want to find its flux across the surface of a sphere.
I don't know. I specifically have the problem:
F= r + F= r + 1/double bar(r)
The first part is easy enough, but what is 1/double bar (r) if r=xi+yj+zk?
The double bar above is literally just two bars one each side surrounding r.
This is my problem word for word:
Consider the vector field
F= r + grad(1/double bar(r)).
Compute the flux (double integral F ndS) of F across the surface of the sphere x^2+y^2+z^2=a^2 where a>0. ndS is the vector element of surface with n the unit normal which here is assumed to point away from the enclosed volume.
If I break it into two parts, I get F=r and F= grad(1/double bar(r))
F=r just has flux 4*pi*a^3 since F=xi+yj+zk and F*N=a where * is the dot product. Note that N=(x/a)i+(y/a)j+(z/a)k. I know this just by looking at it, but how would I set up this double integral to get 4*pi*a^3?
Also, I don't get the second part. What is grad(1/double bar(r))?
To find the gradient you need to find the partial derivatives, which are all symettric.
How do I find the flux of this mess then?
I can simplify it by using a^2=x^2+y^2+z^2 and use polar coordinates, but how do I do this?
I get that grad(1/magnitude(r))=
[(-x)/a^(3/2)]i + [(-y)/a^(3/2)]j + [(-z)/a^(3/2)]k
I know that the unit normal N=(x/a)i + (y/a)j + (z/a)k.
Thus the dot product of grad(1/magnitude(r)) and the unit normal=
(-x^2)/a^(5/2) + (-y^2)/a^(5/2) + (-z^2)/a^(5/2) =
Now how do I integrate this to get the flux of the surface?
integral from 0 to 4pi, integral from 0 to a^2 of (-a^(-3/2))dr dtheta?