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Homework Help: Vector Analysis

  1. Aug 28, 2004 #1
    Hi All,

    I got a couple questions that I need some help getting started on. Any tips would be appreciated.

    1. Derive Gauss and Stokes theorems for the field B = Ap(r), where A is a constant vecotr and p (rho)is a scalar field. r is the unit vector.

    2. Compute the flux of the field A(r)=(y^2, 2xy, 3z^2-x^2) through the surface of a rectangle defined by the four points (b,a,0) (0,a,0) (0,0,a) (b,0,a)

    Last edited: Aug 28, 2004
  2. jcsd
  3. Aug 28, 2004 #2


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    Gauss' theorem, also called the divergence theorem,(in the form in which I am looking at it now) says that, if P and Q are scalar functions on R2, then [tex]\int\int_\Omega \(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\)dxdy= \integral Pdx+ Qdy [/tex] where the second integral is over the boundary of Ω
    Stokes theorem says essentially the same thing except that instead of being in R2, Ω is now some surface in R3.
    I'm not sure why you have reference to "r" and "rho". In problem two, you appear to be using "r" to represent the general (x,y,z) vector but surely you are not asking for a general proof of these thwo theorems?

    Number 2 is not too hard. You are given that A= (y2,2xy, 3z2-x2). Its curl is <0, 2x, 0>.
    The equation of the plane described is y+ z= a. Projecting into the xy-plane, we have ndS= <0, 1, 1> dxdy so curl A.n dS= 4x dx dy. The double integral has limits x=0 to x= b, y= 0 to y= a. The integral is simply (2b)(a)= 2ab.
    Last edited by a moderator: Aug 29, 2004
  4. Aug 28, 2004 #3
    I'm baffled by question 1 also.....as for question 2 I overlooked determining the equation of the plane.

    Thanks for your help.
  5. Aug 28, 2004 #4
    Could you explain the part about projecting onto the xy plane for me. I'm not sure why or how you did it.
  6. Aug 29, 2004 #5
    By the way, I calculated the curl to be <0,2x,0>

    How does that sound?
  7. Aug 29, 2004 #6


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    Since you are integrating over a surface, in order to reduce it to terms of 2 parameters so you need to project to one of the coordinate surfaces (you could write the plane in terms of two parametric equations but then calculating the differential would be harder).
    Think of a surface as given by f(x,y,z)= constant. Then it is a "level surface" for f and grad f is normal to the surface (at each point). It is easy to show that the length of grad f is the differential of area of the surface and I prefer to think of grad f as being the "vector" differential of area. In order to write the integral in x,y,z, "project" down to the xy-plane by dividing the vector by the z component (so that the z component becomes 1). Alternatively, you can project to the yz-plane or xz-plane. Then integrate over the figure in the plane that the surface projects to.

    In this problem, we can write the surface as y+z= a with f(x,y,z)= y+z. Then grad f= <0, 1, 1> . Since the z coordinate is already 1, projecting to the xy plane, dS= <0, 1, 1> dxdy. (Since the y coordinate is already 1, we could project onto the xz plane as dS= <0, 1, 1>dxdz. Since the x coordinate is 0, we could not project to the yz plane- the projection in that direction reduces to a single line.)

    Yes!! I don't why I gave <-2x,2x,-2x>! I went back and calculated it again and it is <0, 2x, 0>.

    The integral [tex]\int_{x=0}^b\int_{y=0}^a 2x dxdy= ab2.
  8. Aug 29, 2004 #7
    Thanks. It all makes sense now.

  9. Oct 4, 2004 #8
    I'm trying to do the same Question as Galipop,

    and while I have managed to do question 2 he posted I havent gotton Question One yet.

    Any idea how to derive Gauss and stokes for that field?

    It seems to want us to derive the theorems rather than proove them
  10. Oct 5, 2004 #9
    Anyone???? Please???
  11. Oct 6, 2004 #10
    Okay our lecturer said that we should try to simplify both theorems when A is a constant vector, not a function of space coordinate r=(x,y,z)).

    Does that help someone to work out how do do this question, if someone could show me how to do it I would be very happy
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