How Do I Evaluate This Surface Integral on a Paraboloid?

[Saketh]

The Problem
Evaluate the surface integral of

<br /> G(x, y, z) = \frac{1}{1 + 4(x^2+y^2)}<br />

where z is the paraboloid defined by

<br /> z = x^2 + y^2<br />,

from z = 0 to z = 1.
My Work

I rewrote G(x, y, z) as

\frac{1}{1+4z}.

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

<br /> \int \!\!\! \int_R \frac{1}{1+4z} \sqrt{1+4z} \,dA = \int \!\!\! \int_R \frac{1}{\sqrt{1+4z}}<br />.

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 (dx, dy?). 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 z= x^2+ y^2, x^2+ y^2- z= 0 and we can think of that as a "level surface" of F(x,y,z)= x^2+ y^2- z. 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 \sqrt{4x^2+ 4y^2+ 1}dxdy is the diffrential of surface area. Of course it doesn't help to write integrand in terms of z now. When z= 1, x^2+ y^2= 1 and that projects down to the xy-plane as the unit circle. The integral is
\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
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): \frac{x}{y}i+ j- \frax{1}{2y}k dxdz is the vector differential and its length is \sqrt{\frac{x^2}{y^2}+ 1+ \frac{1}{4y^2}}dxdz= \frac{\sqrt{x^2+ y^2+ 1}}{2y} dxdz[/tex]<br /> Hmm, that looks more complicated so let&#039;s drop that!<br /> <br /> Because of the circular symmetry of the parabola, it would make sense to use polar coordinates to get parametric equations for it: let x= r cos(\theta), y= r sin(\theta), and z= x^2+ y^2= r^2. The &quot;position vector&quot; of any point on the parabola is<br /> v= r cos(\theta)i+ r sin(\theta)j+ r^2 k. Differentiate with respect to the two parameters: v_r= cos(\theta)i+ sin(\theta)j+ 2rk and v_\theta= -rsin(\theta)i+ rcos(\theta)j.<br /> The &quot;fundamental vector product&quot; for this surface is the cross product of the two: -2r^2 cos(\theta)i- 2r^2 sin(\theta)j+ r k. The length of that is r\sqrt{4r^2+ 1} and so the differential of surface area in terms of r and \theta is r\sqrt{4r^2+1}drd\theta. The integral is <br /> \int_{r= 0}^1\int_{\theta= 0}^{2\pi}\frac{r}{\sqrt{4r^2+1}}d\theta dr<br /> which is easy to integrate.

 

[Saketh]

Thank you for the explanation!

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