1. ## surface area problem..

been trying to solve the folowing area calc:
$
S=\{(x,y,z) | x^2 + z^2 = 1, y^2+z^2\leq 1. z\geq 0\}
$

so it should be the area of the surface $z=\sqrt{1-x^2}-\sqrt{1-y^2}$

then to use the formula for area integrals over uneven surfaces
$\int \int \sqrt{ (\frac{df}{dx})^2 + (\frac {df}{dy})^2 + 1}$

i derive as follows:

$
\frac{dz}{dx} = \frac{-x}{ \sqrt{1-x^2}}
$

$
\frac{dz}{dy} = \frac {y}{\sqrt{1-y^2}}
$

and then in to the formula:

$\int \int \sqrt{ (\frac{-x}{ \sqrt(1-x^2)})^2 + (\frac {y}{\sqrt(1-y^2)})^2 + 1}dxdy$

but then i'm stucked, anyone got an idea how to solve this one?

2. Okay you have,
$
z=\sqrt{1-x^2}-\sqrt{1-y^2}
$

Then,
$\frac{\partial z}{\partial x}=-\frac{x}{\sqrt{1-x^2}}$
And,
$\frac{\partial z}{\partial y}=\frac{y}{\sqrt{1-y^2}}$
Then, square them,
$\frac{x^2}{1-x^2}$
And,
$\frac{y^2}{1-y^2}$
$\frac{x^2}{1-x^2}+\frac{y^2}{1-y^2}$
Which is,
$\frac{x^2(1-y^2)+y^2(1-x^2)}{(1-x^2)(1-y^2)}$
Thus,
$\frac{x^2-x^2y^2+y^2-x^2y^2}{(1-x^2)(1-y^2)}$
Combine,
$\frac{x^2-2x^2y^2+y^2}{(1-x^2)(1-y^2)}$
$\frac{x^2-2x^2y^2+y^2}{(1-x^2)(1-y^2)}+1$
$\frac{x^2-2x^2y^2+y^2+1-x^2-y^2+x^2y^2}{(1-x^2)(1-y^2)}=\frac{1-x^2y^2}{(1-x^2)(1-y^2)}$
So the surface area is,
$\int_A \int \sqrt{\frac{1-x^2y^2}{(1-x^2)(1-y^2)}} dA$

Now, I did not simplify this integration. I am thinking about it. I tried using a Jacobian but alas not success yet.

3. yepp, have been going that way, been trying reducing it to
$
\frac{x^2+y^2-2x^2y^2}{(1-(y^2+x^2)+x^2y^2)} + 1
$

and then switch to polar coordinates but it get's even mesier that way, but it doesn't seem to be a very suitable area for polar coords anyway.

4. just a thought, since $y^2+z^2 \leq 1$ doesn't that mean that the half tube in the zy plane is filled? witch makes
$
z=\sqrt{1-x^2}-\sqrt{1-y^2}
$

incorrect?
since $z=\sqrt{1-x^2}-\sqrt{1-y^2}$ makes something that looks like a hyperbolic paraboloid..