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.