surface of the saddle

Question

i need Help.

Determine the surface of the saddle $$S={(x,y,z)∈R^3; x^2+y^2<=2, z=x^2 -y^2}$$

and the flow of $v(x) = x$ , by S plane polar coordinates, dx dy = r dr dφ, are helpful.

Thanx

Answer 1

surface area = $\iint \|dS\| $

$dS = (-\frac {dz}{dx}, -\frac {dz}{dy}, 1)\\ (-2x, 2y, 1)\\ \|dS\| = \sqrt{4(x^2 + y^2) + 1}$

convert to polar:

$\int_0^{2\pi}\int_0^{\sqrt2} r \sqrt{4r^2 + 1} dr d\theta$

and that should be easy enough to integrate.

Flow

$\iint v \|dS\|\\ \iint x \sqrt{4(x^2 + y^2) + 1} \,dy \,dx$

For every $(x,y,z)$ inside the region, $(-x,y,z)$ is also in the region, and that thing is gong to integrate to $0.$