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Vector integrals

  1. Feb 3, 2006 #1
    First question: Im trying to show the fundamental thm for gradients here with this function: T = x^2 + 4xy + 2yz^3 . a = (0,0,0) , b = (1,1,1) . For one path it works out as expected to 7. However when i use the path (0,0,0)->(0,0,1)->(0,1,1)->(1,1,1) , I end up with a value of 1 (0 for first step, 0 for second step and 1 for third step). There's also a path I have to check where z=x^2, y=x^2, dont know how to do this; Do i let dz = dxdx ? How do I deal with this?


    I havent covered these in calculus yet, just in my E-M text. Would like to know so I can complete my assignment, Thanks.
     
  2. jcsd
  3. Feb 3, 2006 #2

    siddharth

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    Check your second and third steps (Hint: for example, don't put x=0 for the third step. Can you see why that is wrong?).

    In fact, why don't you post your work? It will be easier to help that way.
     
    Last edited: Feb 3, 2006
  4. Feb 3, 2006 #3
    Well first step isnt it integral from 0 to 1 of 6yz^2 dz , but since y= constant zero, the integral equals zero. Similar for second step (because x=0). Third step isnt it integral from 0 to 1 of 2x dx which is one?
     
    Last edited: Feb 3, 2006
  5. Feb 3, 2006 #4

    siddharth

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    The fundamental theorem is
    [tex] \int_C \nabla f.d\vec{r} = f(b) - f(a) [/tex]
    For the given f, did you calculate the gradient? Then what will dr be in each step?

    Your first step is right, but what will [tex] \nabla f.d\vec{r} [/tex] for the second step?
     
    Last edited: Feb 3, 2006
  6. Feb 3, 2006 #5
    well for the second step doesnt dx and dz = 0 and you only have to deal with df/dy * dy which is....o ok my bad forgot about the third term with the z :)
     
  7. Feb 3, 2006 #6
    So as my second question, how do i deal with the z = x^2 ? ; what do i put in for dz?

    Here's what i got so far.

    Dell(x,y,z) = (2x+4y,4x+2z^3,6yz^2)

    when you dot it with (dx,dy,dz) and substitute all y for x all z for x^2 i get 6xdx + (4x+2x^6)dx + (6x^5)dx^2 . The last term confuses me.
     
    Last edited: Feb 3, 2006
  8. Feb 8, 2006 #7

    siddharth

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    Hope you don't mind the late reply.

    You have,
    [tex] z=x^2 , y=x^2[/tex]

    If you use 't' as your parameter, you will get

    [tex] x=t, y=t^2, z=t^2 [/tex],

    Now, you can write the position vector of a point P(x,y,z) on the curve as a function of t. This will let you calculate 'dr' in terms of 't' and 'dt'. Now, if you write the Gradient (2x+4y,4x+2z^3,6yz^2) in terms of 't' you will easily be able to get your answer.
     
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