# Homework Help: Surface Integral involving Paraboloid

1. May 29, 2013

### LunaFly

1. The problem statement, all variables and given/known data

Evaluate the surface integral:

∫∫s y dS

S is the part of the paraboloid y= x2 + z2 that lies inside the cylinder x2 + z2 =4.

2. Relevant equations

∫∫sf(x,y,z)dS = ∫∫Df(r(u,v))*|ru x rv|dA

3. The attempt at a solution

I've drawn the region D in the xz-plane as a circle with radius 2. My limits of integration are r between 0 and 2, and t between 0 and 2∏.

I've parametrized the surface as:
r(r,t)= <rcos(t), r2, rsin(t)>

I've gotten the vectors:
rr=<cos(t), 2r, sin(t)>
rt=<-rsin(t), 0, rcos(t)>

Trying to find dS, I did:
rr X rt = <2r2cos(t), -r, 2r2sin(t)>,
and the magnitude of that is r√(4r2 +1).

Overall, I end up with dS = r2*√(4r2+1)drdt.

The integral becomes:

∫∫r4*√(4r2+1)drdt, with limits as mentioned above.

However this is not the correct answer. I am having a difficult time with these surface integrals and I am not really sure why. I keep getting confused about whether or not I am parametrizing
the surfaces correctly, so that may be where my error is. Any help would be much appreciated! Thanks
-Luna

Last edited: May 29, 2013
2. May 30, 2013

### haruspex

How did the r2 become r4? Not saying it's wrong, ... this is not an area I'm expert on.

3. May 30, 2013

### HallsofIvy

Yes, this is what I get.

What? $\int\int y dS= \int \int (r^2)dS= \int \int (r^2)(r\sqrt{4r^2+ 1})drd\theta$
You should have $r^3$ outside the square root, not $r^4$.

4. May 30, 2013

### LunaFly

I believe it is r4 because:

My paramatrization has y=r2, so f(r(r,t))=r2

dS = |rr X rt|dA = r√(4r2 + 1) * rdrdt

dA = r drdt (because of the relationship between polar and Cartesian coordinates) which is where the r4 comes from.

Altogether, ∫∫sf(r(r,t))dS = ∫∫Dr2*r√(4r2 + 1) * rdrdt= ∫∫Dr4√(4r2+1) drdt

This may be helpful: According to the book, the integral equals (∏/60)(391√(17) +1).

Last edited: May 30, 2013
5. May 30, 2013

### LunaFly

So I tried solving the integral that HallsofIvy thought was correct, and it was! Thanks for the response.

But that brings a new question to light: why does dA become drdt rather than r*drdt?

I had the impression that every time we switched from Cartesian to polar, dA became r*drdt, but it seems that in this problem, that's not the case.

Why?

6. May 30, 2013

### LCKurtz

The formula for surface area element for a surface parameterized as $\vec R = \vec R(u,v)$ is $dS = |\vec R_u \times \vec R_v|dudv$. There is no extra $u$ or $v$, and in your case, no extra $r$.

Suppose your surface had been a circle in the xy plane. You might parameterize it as $\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta,0\rangle$. If you calculate $dS = |\vec R_r\times \vec R_\theta|drd\theta$ you will find that that cross product is where the familiar $r$ in $rdrd\theta$ comes from. You don't add an extra $r$ when you parameterize it from the beginning.