Calculating the area of boundary Let $G=\{(x,y,z)|z>x^2+y^2 \ and \ x^2+y^2+z^2<2\}\subset \mathbb{R}^3$.
Calculate the area of the boundary: $\partial G=M$.
I don't even know how to approach this..any help?

The boundary of the area in common have one portion that is a paraboloid, and one portion that is the surface of a sphere.
Lets convert to cylindrical coordinates.
$x = r \cos\theta\\\
y = r \sin\theta\\
z = z $
for one surface:
$x^2 + y^2 = z\\
z = r^2$
and for the other
$x^2 + y^2 + z^2 = 2\\
r^2 + z^2 = 2\\
z = \sqrt {2-r^2}$
$\|dS\| = \|(\frac {\partial x}{\partial \theta},\frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta})\times(\frac {\partial x}{\partial r},\frac {\partial y}{\partial r},\frac {\partial z}{\partial r})\|$
for one surface that will be
$\|(-r\sin\theta, r\cos\theta, 0) \times (\cos\theta, \sin\theta, 2r)\| = r\sqrt {4r^2+1}\ dr\ d\theta$
and the other
$\|(-r\sin\theta, r\cos\theta, 0) \times (\cos\theta, \sin\theta, \frac {r}{\sqrt {2-r^2}})\| = \frac {2}{\sqrt {2-r^2}}$
Check my work on those... If I have that right then.
$\int_0^{2\pi}\int_0^{1} r\sqrt {4r^2+1}+\frac {2}{\sqrt {2-r^2}}\ dr\ d\theta$
