# Surface Integral setup and evaluate

1. Nov 25, 2012

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

Evaluate the surface integral ∫∫f(x,y,z)dS using an explicit representation of the surface.

$f(x,y,z) = x^2 + y^2;\mbox{ S is the paraboloid } z= x^2 + y^2\mbox{ for }0\leq z \leq 4$

2. Relevant equations

$\displaystyle \int \int_{S} f(x,y,z)\ dS = \int \int_{D} f \{x,y,g(x,y)\}\ \sqrt{1 + (\frac {\partial g}{\partial x})^{2} + (\frac {\partial g}{\partial y})^{2}}\ dx\ dy$

3. The attempt at a solution

This is how I have set it up so far. I seem to have gotten stuck, and it makes me think that there is something wrong with my set-up.

$dS=\sqrt{4x^2+4y^2+1}dA$

$\int\int(x^2+y^2)*2*\sqrt{x^2+y^2+\frac{1}{4}}dA$

$2\int_0^{2\pi}\int_0^2(r^2)*(r^2+\frac{1}{4})^{1/2}rdrd\theta$

$2\int_0^{2\pi}\int_0^2(r^3)*(r^2+\frac{1}{4})^{1/2}drd\theta$

Please let me know if you see any errors in my work, and if I am on the right track perhaps a gentle nudge on how to integrate this mess.

Thanks,
Mac

Last edited: Nov 26, 2012
2. Nov 26, 2012

### haruspex

That last step is definitely wrong. try a substitution to get rid of the surd. Something like r = tan(ψ)/2 or sinh(ψ)/2.

3. Nov 26, 2012

Just to clarify; the last correct step was $2\int\int(x^2+y^2)\sqrt{x^2+y^2+\frac{1}{4}}dA$? So converting to polar was incorrect?

Okay, now that I've looked up "surd," (great word by the way), I'll try to proceed. Just wanted to verify what the last correct step was.

*EDIT* Nevermind, I see what you mean in the last step. That was a type error, the r3 was supposed to be outside the square root. I have fixed the original post.

Last edited: Nov 26, 2012
4. Nov 26, 2012

So trying a trig substitution I get,

$2\int\int(r^3)\sqrt{r^2+\frac{1}{4}}drd\theta$

$\mbox{for }a^2+x^2\mbox{ let }x=atan\theta$

$r=\frac{1}{2}tan\theta$

$dr=\frac{1}{2}sec^2\theta d\theta$

$2\int\int(\frac{1}{2}tan\theta)^3\sqrt{\frac{1}{4}tan\theta}\frac{1}{2}sec^2\theta d\theta d\theta$

Problem with this method is that there are two $d\theta$'s, which of course is wrong.

Where did I go wrong? (other then everywhere)

Thanks,
Mac

*This has been edited since the original writing. I figured out why I had an extra "r."

Last edited: Nov 26, 2012
5. Nov 26, 2012

Well, looking above I was obviously tired last night when I wrote that. However, I'm still not positive about this and could use some assistance if possible.

Instead of using $\theta$ again for a substitution, I'll use $\phi$

$2\int\int(r^3)\sqrt{r^2+\frac{1}{4}}drd\theta$

$\mbox{for }a^2+x^2\mbox{ let }x=atan\phi$

$r=\frac{1}{2}tan\phi$

$dr=\frac{1}{2}sec^2\phi d\phi$

$2\int\int(\frac{1}{2}tan\phi)^3\sqrt{\frac{1}{4}tan^2\phi+\frac{1}{4}}(\frac{1}{2}sec^2\phi) d\phi d\theta$

This can be simplified down to,

$8 \phi \int tan^3 \phi sec^3\phi d\phi$

Still not pretty.

Could anyone verify if this is a correct approach?

Thanks again,
Mac

Last edited: Nov 26, 2012
6. Nov 26, 2012

### Dick

It's a correct approach. There's some details wrong I think, you meant 8π outside the integral in the last step and the numbers don't look right. But there is a much easier substitution than the trig one. Just put u=r^2+1/4.

Last edited: Nov 26, 2012
7. Nov 26, 2012

Forgive me for being slow here, but if I let $u=r^2+\frac{1}{4}\mbox{then }du=2rdr\mbox { and I have nothing to account for the } r^3;$

I could do something like this,

$(\frac{1}{2})^3 du=r^3 dr$,

but I've never done that maneuver before, and I honestly don't know if it's valid?

Thanks for the patience and assistance.
Mac

8. Nov 26, 2012

If I work through it with that substitution I end up with the clean answer of $\frac{4\pi}{3}$, so I'm keeping my fingers crossed that I can do that trick.

9. Nov 26, 2012

### Dick

I'm not sure what 'trick' you are talking about but it doesn't look right to me. I don't get that clean an answer. You've got r^3*sqrt(u)dr=r^2*sqrt(u)*rdr. You'll need the rdr to make du. So that leaves you with just a r^2. But since u=r^2+1/4, r^2=u-1/4.

10. Nov 26, 2012

### haruspex

Beauty is in the eye of beholder. Removing the spurious phi in front:
$\int tan^3 \phi sec^3\phi d\phi = \int sin^3 \phi cos^{-6}\phi d\phi = -\int sin^2 \phi cos^{-6}\phi d cos(\phi) = \int (cos^2 \phi - 1)cos^{-6}\phi d cos(\phi)$
$= \int (cos^{-4} \phi - cos^{-6}\phi) d cos(\phi) = [-cos^{-3}(\phi)/3+ cos^{-5}(\phi)/5]$

11. Nov 26, 2012

The trick I was talking about was cubing both sides, like so $(\frac{1}{2})^{3} du = r^3dr$

That accounted for the r3, but it didn't work. I see what your saying in your post, and I can't believe I forgot about simply rearranging my "u" equation. I think I can work through that, but it will not be until later.

Thanks again for the assistance.

Hey, my phi wasn't that spurious! Besides, it did need the eight in front, just with a ∏ instead.

I suppose I'm just not beholden. This question was supposed to be about a surface integral, not some integral that I should be able to simply calculate and have obviously forgotten how.

12. Nov 26, 2012

### Dick

Nah, that would give you $(\frac{1}{2})^{3} {du}^3 = r^3 {dr}^3$ True, but not very useful.