How Do You Calculate the Surface Area of x^2-y^2-z^2=0 with Given Constraints?

[Telemachus]

[Urgent] Surface area

Homework Statement

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 [tex]x^2-y^2-z^2=0[/tex] under the conditions [tex]x\geq{0},y\geq{0},z\geq{0},x\leq{1-z}[/tex]

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

Then I made the parametrization for the plane:
[tex]\begin{Bmatrix}x=u\\y=v \\z=1-u\end{matrix}[/tex]

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

I think that the surface is actually not bounded.

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

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

Any help will be thanked.

 

[tiny-tim]

Hi Telemachus!

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

Yes, the original surface is a cone with generators at 45°, and the intersecting plane is also at 45°, so in one direction they meet "at infinity".

But the conditions x,y,z ≥ 0 eliminate that direction, and the other direction is ok.

 

[Telemachus]

Thank you tim.