1. ## Surface area

Hi there, I have to compute the surface area for
$\displaystyle V:\{ -2(x+y)\leq{}z\leq{}4-x^2-y^2 \}$

I have a problem on finding the surface area for the paraboloid limited by the plane. I've parametrized the plane in polar coordinates, I thought it would be easier this way, but also tried in cartesian coordinates with the same result. The problem I have is to set the limits of integration on such a way that determines the area of the paraboloid limited by the plane.

So the paraboloid is parametrized by:

$\displaystyle \begin{Bmatrix}{ x=r\sin \theta}\\y=r \cos \theta \\z=4-r^2\end{matrix}$

Then $\displaystyle T_r=( \sin \theata, \cos \theta,-2r);T_{\theta}=(r\cos\theta,-r\sin\theta,0)$
$\displaystyle T_r \times T_{\theta}=(-2r^2\sin^2 \theta,-2r^2\cos^2\theta,-r)$
$\displaystyle ||T_r \times T_{\theta}||=\sqrt[ ]{4r^4+r^2}$

Now the surface area is determined by: $\displaystyle \displaystyle\int_{D} ||T_r \times T_{\theta}||drd\theta$
Now D is the region inside the disc determined by the intersection of the surfaces. So D:
$\displaystyle -2x-2y=4-x^2-y^2\rightarrow (x-1)^2+(y-1)^2=6$

And this is my region of integration for my paraboloid. Now it doesn't seem so easy to express the region for the parametrization I choose.

So I'm trying to solve this in cartesian coordinates, this is the integral for the surface in cartesian coord:

$\displaystyle \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx$

But this integral isn't so easy to solve. I tried to go from here to cylindrical coordinates, using the substitution $\displaystyle x=1+r\cos\theta,y=1+r\sin\theta$ that helps a bit with the limits of integration.

Then I get: $\displaystyle \displaystyle\int_{1-\sqrt[ ]{6}}^{1+\sqrt[ ]{6}}\int_{-\sqrt[ ]{6-(x-1)^2}+1}^{\sqrt[ ]{6-(x-1)^2}+1} \sqrt[ ]{4x^2+4y^2+1} dydx=\displaystyle\int_{0}^{\sqrt[ ]{6}}\int_{0}^{2\pi}r\sqrt[ ]{16+8r\cos\theta+8r\sin\theta+4r^2}d\theta dr$

Thats the best expression I get, but still too complicated to integrate by hand.

2. ## Re: Surface area

If it were me, I'd want a rough estimate before doing too much. This might suggest whether one had wandered off.

The intersection of the two surfaces suggests $\displaystyle 12\pi$ as a lower bound (don't forget the bottom of your container) and not much more than that as an upper bound. Certainly less than $\displaystyle 24\pi$

A quick check of your Rectangular Coordinates results in about $\displaystyle 25\pi$ and your last expression comes in at a whopping $\displaystyle 30\pi$.

Seems like we don't quite have the right things going on just yet.

On the other hand, maybe I'm coming off the wrong medication and making far less sense than I think.

3. ## Re: Surface area

You think I've made a mess with the limits of integration or something like that? it's a possibility, I didn't have too much care when doing it, so it's a possibility. But I don't see any mistake. I'll try to look harder. How did you make your estimations?

4. ## Re: Surface area

Limits: Doesn't strike me, right away either.

In this case, the estimate seems pretty easy. The intersection is a circle. Beneath the circle is somewhat greater, due to its oblique orientation when compared to the circle. The parabolic piece should be more than that. Since this paraboloid is pretty flat near the top, I would expect it to be only a little larger.

5. ## Re: Surface area

Various other estimates could be made. A sphere with radius $\displaystyle \sqrt{6}$ comes in at $\displaystyle 24\pi$, for example. There's a better upper bound.

6. ## Re: Surface area

I did it again in a different fashion and I get to:
$\displaystyle \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^ {\sqrt[ ]{6}}r\sqrt[ ]{9+4r^2+8r(\sin\theta+\cos\theta)}drd\theta$

What do you think? I don't know how to solve it anyway :P

7. ## Re: Surface area

Still $\displaystyle 25\pi$-ish. I guess the next thing would be to challenge my estimates.

8. ## Re: Surface area

Where you think I could be committing the mistake? I've proceeded similarly as I did at first, just changed the parametrization from the beggining, I set $\displaystyle x=r\cos\theta +1,y=r\sin\theta+1,z=4-(r\cos\theta +1)^2-(r\sin\theta+1)^2$ and proceeded as well.

9. ## Re: Surface area

Okay, one last estimate. The volume of the object is easily calculate to be 56.549 unit^3. A sphere of this same volume has a radius of 2.381 and a surface area of 71.247 or about $\displaystyle 22.7\pi$. I keep whittling down the upper bound, but still not quite getting anywhere. Any revelations, yet?