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What do i do with z?

  1. Jul 26, 2004 #1
    what do i do with z!?

    ok no idea what to do with this z in the this problem, cause im teaching this to myself i don't know what to do and im asking for some help thanks anyone...

    ok well im trying to compute this problems with Stroke's therom

    the surface is [tex] z = \sqrt{4-x^2-y^2} [/tex] above the xy-plane

    and [tex] \vec{F} = <2x-y,yz^2,y^2z>[/tex]

    ok using stoke's i do [tex]\nabla X \vec{F} = \vec{k}[/tex]
    then instead of finding the normal vector with the surface z, i did, [tex]f(x,y,z) = 4-x^2-y^2-z^2[/tex] which [tex]\vec{n} = \nabla f = <-2x,-2y,-2z>[/tex] but we want a normal vector pointing out from the surface, so..(also note i didn't but the normalizing sqrt cause it comes out in the dS -> dA change) [tex] -\vec{n} = <2x,2y,2z>[/tex] and then fallowing stoke's...[tex]\vec{k} \bullet\vec{n} = 2z -> \int\int_{A}2zdydx[/tex] but my understanding of a surface intergral is that z is 0 so that the surface is actually an area projected on to the xy-plane, what am to do in these situations? or did i just do it wrong?
    Last edited: Jul 26, 2004
  2. jcsd
  3. Jul 26, 2004 #2
    Ok, after a bit of contemplation about doing surface integrals in different ways, I've see you've made a mistake in your calculations, which I will come back to. This does not affect your main question though. The answer is that you are just projecting the surface onto the xy plane, and it is on the surface that the integrating is actually happening. Here (on the surface), z is not zero in general. Remember that here z = sqrt(4 - x^2 - y^2).

    Now to my first point. It has to do with you mixing two methods of calculating the unit normal vector. First you put the equation for the surface into the form: f(x,y,z) = C, and asserted that:

    [tex] $\mathbf{ \hat{n}}$ = \frac{ \nabla f }{| \nabla f|} [/tex]

    which is fine, but then the the square root factors do not quite cancel anymore, because:

    [tex] \cos \theta =
    $\mathbf{ \hat{n} \cdot k}$ = \frac{ \frac{ \partial f }{ \partial x } $\mathbf{i}$ + \frac{ \partial f }{ \partial y } $\mathbf{j}$ + \frac{ \partial f }{ \partial z } $\mathbf{k}$ }{ \sqrt{ \left( \frac{ \partial f }{ \partial x } \right)^2 + \left( \frac{ \partial f }{ \partial y} \right)^2 + \left( \frac{ \partial f }{ \partial y} \right)^2 }} $\mathbf{ \cdot k }$ [/tex]

    \Rightarrow \frac{1}{ $\mathbf{ \hat{n} \cdot k}$ } = \frac{ \sqrt{ \left( \frac{ \partial f }{ \partial x } \right)^2 + \left( \frac{ \partial f }{ \partial y} \right)^2 + \left( \frac{ \partial f }{ \partial y} \right)^2 }}{ \frac{ \partial f }{ \partial z }} [/tex]

    The square roots only cancel when we keep the equation for the surface in the form z = f(x,y). So we have to divide the integrand by 2z, making it 1, and your surface integral is just the area of the circle below. Your line integral should show that.

    P.S. Apologies for the latex which won't show ATM. It's late now, but I'll fix it ASAP. It just shows why the roots don't cancel.
    Last edited: Jul 28, 2004
  4. Jul 26, 2004 #3

    oh oh, i see like [tex]dS = \sqrt{f_{x}^2+f_{y}^2+1}[/tex] is not...the normalizer of the f function, thanks much, thou i don't know if the 2z will drop because of that to 1, but thats good ill take a look and post if that worked
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