JJRKnights said:
1.If you have a vector field F = <-y,x>/(x^2 + y^2)
is that the same thing as saying: F = < -y/(x^2+y^2), x/(x^2 + y^2)>?
That's basically saying that the function 1/(x^2 + y^2) is a scalar.
Yes, that is correct.
2.How do you show that vector fields are not conservative for certain values?
A vector field, F, is "conservative" on a region if its integral from one point in that region to another is independent of the path. That leads to the fact that the integral around a closed path must be 0. It can be shown that that is the same as saying it has a "potential function", a scalar function, \phi such that \nabla \phi= F. In that case, if F= <f(x,y), g(x,y)>, we must have \partial \phi/\partial x= f(x, y) and \partial \phi/\partial y= g(x,y). And now the "mixed derivatives", \partial^2\phi/\partial x\partial y= f_y(x,y) and \partial^2\phi/\partial x\partial y= g_x, must be equal so we must have f_y= g_x. This last is a necessary condition for a "conservative" vector field but is a "sufficient" condition only if the function is continuous in a "simply connected" region. I mention that because in your next example the function is not continuous at (0, 0) and so not in any region containing that point.
Ex: F = <-y,x>/|r|^p show it's not conservative for p = 2.
Integrate around the unit circle: x= cos(\theta) and y= sin(\theta) so that d\vec{r}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta and the function is -y\vec{i}+ x\vec{j}= -sin(\theta)\vec{i}+ cos(\theta)\vec{j} and the integral becomes
\int_0^{2\pi} (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})(-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta= \int_0^{2\pi} (1)d\theta= 2\pi
Since that is not 0, the function is not conservative.
3. How do you find normal vectors to rounded surfaces given a vector field and the function of the rounded surface? (paraboloids, spheres)
A surface is two dimensional and so can be written in terms of two parameters: \vec{r}= f(u, v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}. The two derivative vectors, \vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k} and \vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k} lie in the tangent plane at each point. their
cross product is a normal vector to the surface and, in fact, its length gives the "differential of surface area".
For example, the sphere of radius R can be written using spherical coordinates with \rho= R, a constant. Then
\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}
The two derivative vectors,
\vec{r}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}
and
\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}
have cross product
\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ R cos(\theta)cos(\phi) & Rsin(\theta)cos(\phi) & -Rsin(\phi) \\ -Rsin(\theta)sin(\phi) & Rcos(\theta)sin(\phi) & 0 \end{array}\right|= R^2cos(\theta)sin^2(\phi)\vec{i}+ R^2sin(\theta)sin^2(\phi)\vec{j}+ R^2sin(\phi)cos(\phi)\vec{k}
(This is called the "fundamental vector product" for this surface.)
Note that the length of that vector is R^2sin(\phi) and that R^2sin(\phi)d\theta d\phi is the differential of surface area for a sphere.
To determing the flux of a vector function, F= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ g(x,y,z)\vec{k} through that surface, convert x, y, and z to \theta and \phi , take the dot product of F with that "fundamental vector product" and integrate with respect to \theta and \phi.