
surface area problem..
been trying to solve the folowing area calc:
$\displaystyle
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 $\displaystyle z=\sqrt{1x^2}\sqrt{1y^2}$
then to use the formula for area integrals over uneven surfaces
$\displaystyle \int \int \sqrt{ (\frac{df}{dx})^2 + (\frac {df}{dy})^2 + 1} $
i derive as follows:
$\displaystyle
\frac{dz}{dx} = \frac{x}{ \sqrt{1x^2}}
$
$\displaystyle
\frac{dz}{dy} = \frac {y}{\sqrt{1y^2}}
$
and then in to the formula:
$\displaystyle \int \int \sqrt{ (\frac{x}{ \sqrt(1x^2)})^2 + (\frac {y}{\sqrt(1y^2)})^2 + 1}dxdy $
but then i'm stucked, anyone got an idea how to solve this one?

Okay you have,
$\displaystyle
z=\sqrt{1x^2}\sqrt{1y^2}
$
Then,
$\displaystyle \frac{\partial z}{\partial x}=\frac{x}{\sqrt{1x^2}}$
And,
$\displaystyle \frac{\partial z}{\partial y}=\frac{y}{\sqrt{1y^2}}$
Then, square them,
$\displaystyle \frac{x^2}{1x^2}$
And,
$\displaystyle \frac{y^2}{1y^2}$
Add them,
$\displaystyle \frac{x^2}{1x^2}+\frac{y^2}{1y^2}$
Which is,
$\displaystyle \frac{x^2(1y^2)+y^2(1x^2)}{(1x^2)(1y^2)}$
Thus,
$\displaystyle \frac{x^2x^2y^2+y^2x^2y^2}{(1x^2)(1y^2)}$
Combine,
$\displaystyle \frac{x^22x^2y^2+y^2}{(1x^2)(1y^2)}$
Add 1,
$\displaystyle \frac{x^22x^2y^2+y^2}{(1x^2)(1y^2)}+1$
Add fractions,
$\displaystyle \frac{x^22x^2y^2+y^2+1x^2y^2+x^2y^2}{(1x^2)(1y^2)}=\frac{1x^2y^2}{(1x^2)(1y^2)}$
So the surface area is,
$\displaystyle \int_A \int \sqrt{\frac{1x^2y^2}{(1x^2)(1y^2)}} dA$
Now, I did not simplify this integration. I am thinking about it. I tried using a Jacobian but alas not success yet.

yepp, have been going that way, been trying reducing it to
$\displaystyle
\frac{x^2+y^22x^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.

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