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A couple of questions regarding to my calculus work.

  1. Nov 29, 2011 #1
    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. jcsd
  3. Dec 1, 2011 #2
    Can anybody say anything on this?

    They should be simple questions to people more skilled than I am.
  4. Dec 2, 2011 #3


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    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, [itex]\phi[/itex] such that [itex]\nabla \phi= F[/itex]. In that case, if F= <f(x,y), g(x,y)>, we must have [itex]\partial \phi/\partial x= f(x, y)[/itex] and [itex]\partial \phi/\partial y= g(x,y)[/itex]. And now the "mixed derivatives", [itex]\partial^2\phi/\partial x\partial y= f_y(x,y)[/itex] and [itex]\partial^2\phi/\partial x\partial y= g_x[/itex], must be equal so we must have [itex]f_y= g_x[/itex]. 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: [itex]x= cos(\theta)[/itex] and [itex]y= sin(\theta)[/itex] so that [itex]d\vec{r}= (-sin(\theta)\vec{i}+ cos(\theta)\vec{j})d\theta[/itex] and the function is [itex]-y\vec{i}+ x\vec{j}= -sin(\theta)\vec{i}+ cos(\theta)\vec{j}[/itex] and the integral becomes
    [tex]\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[/tex]
    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: [itex]\vec{r}= f(u, v)\vec{i}+ g(u,v)\vec{j}+ h(u,v)\vec{k}[/itex]. The two derivative vectors, [itex]\vec{r}_u= f_u\vec{i}+ g_u\vec{j}+ h_u\vec{k}[/itex] and [itex]\vec{r}_v= f_v\vec{i}+ g_v\vec{j}+ h_v\vec{k}[/itex] 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 [itex]\rho= R[/itex], a constant. Then
    [tex]\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}[/tex]

    The two derivative vectors,
    [tex]\vec{r}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}[/tex]
    [tex]\vec{r}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{k}[/tex]
    have cross product
    [tex]\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}[/tex]
    (This is called the "fundamental vector product" for this surface.)

    Note that the length of that vector is [itex]R^2sin(\phi)[/itex] and that [itex]R^2sin(\phi)d\theta d\phi[/itex] is the differential of surface area for a sphere.

    To determing the flux of a vector function, [itex]F= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ g(x,y,z)\vec{k}[/itex] through that surface, convert x, y, and z to [itex]\theta[/itex] and [itex]\phi[/itex] , take the dot product of F with that "fundamental vector product" and integrate with respect to [itex]\theta[/itex] and [itex]\phi[/itex].
    Last edited by a moderator: Dec 2, 2011
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