Need Help with Surface Integral

Saketh

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!

HallsofIvy

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.

Saketh

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]