# A couple of questions regarding to my calculus work.

1. Nov 29, 2011

### JJRKnights

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.

2.How do you show that vector fields are not conservative for certain values?

Ex: F = <-y,x>/|r|^p show it's not conservative for p = 2.

3. How do you find normal vectors to rounded surfaces given a vector field and the function of the rounded surface? (paraboloids, spheres)

2. Dec 1, 2011

### JJRKnights

Can anybody say anything on this?

They should be simple questions to people more skilled than I am.

3. Dec 2, 2011

### HallsofIvy

Yes, that is correct.

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 continous at (0, 0) and so not in any region containing that point.

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.

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 $\phi$.

Last edited by a moderator: Dec 2, 2011