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Need Help with Surface Integral

  1. Oct 14, 2006 #1
    The Problem
    Evaluate the surface integral of

    [tex]
    G(x, y, z) = \frac{1}{1 + 4(x^2+y^2)}
    [/tex]

    where [itex]z[/itex] is the paraboloid defined by

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

    from [itex]z = 0[/itex] to [itex]z = 1[/itex].
    My Work

    I rewrote [itex]G(x, y, z)[/itex] as

    [tex]\frac{1}{1+4z}[/tex].

    Then, I evaluated the surface integral (I'm skipping a few steps in the evaluation here):

    [tex]
    \int \!\!\! \int_R \frac{1}{1+4z} \sqrt{1+4z} \,dA = \int \!\!\! \int_R \frac{1}{\sqrt{1+4z}}
    [/tex].

    My Confusion
    I do not understand how to evaluate this integral properly. I am not experienced in multiple integration, but I have not found an issue with it until now.

    Basically, what are my differential elements supposed to be ([itex]dx, dy[/itex]?). Am I supposed to use polar coordinates here?

    If someone could put me on the correct track, I would appreciate it. Thanks!
     
  2. jcsd
  3. Oct 14, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You dropped "dA" from the last integral. One of the things you need to decide is how to write the differential of surface area- thre are several different ways to do that. In particular, you need to decide whether you will project onto the xy-plane, the yz-plane, the xz-plane, or use parametric equations for the surface. I can't possibly decide whether you are "supposed" to use polar coordinates- but it might be a good idea to use them!

    Projecting to xy-plane: The way I like to do it is use the gradient of the equation of the surface. Since [itex]z= x^2+ y^2[/itex], [itex]x^2+ y^2- z= 0[/itex] and we can think of that as a "level surface" of [itex]F(x,y,z)= x^2+ y^2- z[/itex]. The divergence of that, 2xi+ 2yj- k, is perpendicular to the surface at each point. We can "normalize" to the xy-plane by multiplying by -1 (so that the k component is 1): (-2xi- 2yi+ k)dxdy is the vector 'differential of surface area' and its length [itex]\sqrt{4x^2+ 4y^2+ 1}dxdy[/itex] is the diffrential of surface area. Of course it doesn't help to write integrand in terms of z now. When z= 1, [itex]x^2+ y^2= 1[/itex] and that projects down to the xy-plane as the unit circle. The integral is
    [tex]\int_{x= -1}^1\int_{y= -\sqrt{1-x^2}}^\sqrt{1- x^2} \frac{\sqrt{4x^2+ 4y^2+1}}{1+ 4x^2+ 4y^2}dxdy[/tex]
    which is easy. You can, of course, convert that to polar coordinates if you want to.

    Since, as you point out, the integrand can be written as a function of z only, it might make sense to project to the xz-plane. Again we can use the grad F= 2xi+ 2yj- k. We "normalize" to the xz-plane by dividing by 2y (to make the j component 1): [itex]\frac{x}{y}i+ j- \frax{1}{2y}k dxdz[/itex] is the vector differential and its length is [itex]\sqrt{\frac{x^2}{y^2}+ 1+ \frac{1}{4y^2}}dxdz= \frac{\sqrt{x^2+ y^2+ 1}}{2y} dxdz[/tex]
    Hmm, that looks more complicated so let's drop that!

    Because of the circular symmetry of the parabola, it would make sense to use polar coordinates to get parametric equations for it: let [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex], and [itex]z= x^2+ y^2= r^2[/itex]. The "position vector" of any point on the parabola is
    [itex]v= r cos(\theta)i+ r sin(\theta)j+ r^2 k[/itex]. Differentiate with respect to the two parameters: [itex]v_r= cos(\theta)i+ sin(\theta)j+ 2rk[/itex] and [itex]v_\theta= -rsin(\theta)i+ rcos(\theta)j[/itex].
    The "fundamental vector product" for this surface is the cross product of the two: [itex]-2r^2 cos(\theta)i- 2r^2 sin(\theta)j+ r k[/itex]. The length of that is [itex]r\sqrt{4r^2+ 1}[/itex] and so the differential of surface area in terms of r and [itex]\theta[/itex] is [itex]r\sqrt{4r^2+1}drd\theta[/itex]. The integral is
    [tex]\int_{r= 0}^1\int_{\theta= 0}^{2\pi}\frac{r}{\sqrt{4r^2+1}}d\theta dr[/tex]
    which is easy to integrate.
     
    Last edited: Oct 14, 2006
  4. Oct 14, 2006 #3
    Thank you for the explanation!

    I am posting this for my own convenience - the LaTeX source was invalid:
    [tex]\int_{r= 0}^1\int_{\theta= 0}^{2\pi}\frac{r}{\sqrt{4r^2+1}} \,d\theta \,dr[/tex]
     
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