1. ## Surface area

Hi there, sorry for the hurry, I had a big problem this weekend and I couldn't study. I have my exam tomorrow, and need help with this. I'm finishing with the topic, and I have almost studied all what I had to study, but this exercise resulted problematic.

This is it.

It asks me to calculate the area of the surface $\displaystyle x^2-y^2-z^2=0$ under the conditions $\displaystyle x\geq{0},y\geq{0},z\geq{0},x\leq{1-z}$

So I've made the parametrization this way:
$\displaystyle \begin{Bmatrix}x=u\\y=u \cos v \\z=u \sin v\end{matrix}$
And I've found: $\displaystyle ||T_u\times{T_v}||=\sqrt[ ]{2}u$

Then I made the parametrization for the plane:
$\displaystyle \begin{Bmatrix}x=u\\y=v \\z=1-u\end{matrix}$

Then I made the integral for the surface, well, I've tried:
$\displaystyle \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^ {1-u}\sqrt[ ]{2}u dudv$

I think that the surface is actually not bounded.

Actually, when I did the intersection I've found a parabola:
$\displaystyle \sqrt[ ]{y^2+z^2}=1-z\longrightarrow{y^2+2z-1=0}$

Which I think makes sense, but the surface would be infinite, and I don't know what to do.

Any help will be thanked.

2. Originally Posted by Ulysses
Then I made the integral for the surface, well, I've tried:
$\displaystyle \displaystyle\int_{0}^{2\pi}\displaystyle\int_{0}^ {1-u}\sqrt[ ]{2}u dudv$

I think that the surface is actually not bounded.

Actually, when I did the intersection I've found a parabola:
$\displaystyle \sqrt[ ]{y^2+z^2}=1-z\longrightarrow{y^2+2z-1=0}$
If you look at your integral, the limits are actually lines on the plane, right? Try using those as bounds and compare to where the parabola comes across them.
So you would have your parabola on the uv-plane bounded by:
$\displaystyle v = 0; v = 1 - u; u = 0; u = 2pi$. Something to note: since you have your integral limit set up as a $\displaystyle u$ to $\displaystyle v$, your integral should reflect this with: $\displaystyle sqrt(2)udvdu$. Order is important. This intersected area should be the region you're finding the surface area for.

Hope this helps, and if anywhere I am mistaken, someone please feel free to correct me and get him pointed in the right direction! =)