Calculating the area of boundary

Question

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?

Answer 1

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.

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