To find area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ How to find the area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$ ? I am completely stuck ; please help . Thanks in advance 

First, lets see where both surfaces intersect. In polar coordinates, we have
$$
z=3-r = 1+ r^2\quad \Rightarrow r=1.
$$
So the intersection is the circle $r=1$ at height $z=2$. 
The area of your surface is the area $A_1$ of $z=3-\sqrt{x^2+y^2}$ above this circle (let $S_1$ be this surface), plus the area $A_2$ of $z=1+x^2+y^2$ beneath this circle (let $S_2$ be this surface).
You can parametrize $S_1$ with the vectorial function $f$:
$$
x=x, \quad y=y, \quad z=3-\sqrt{x^2+y^2}, 
$$
and $S_2$ with $g$:
$$
x=x, \quad y=y, \quad z=3+x^2+y^2,
$$
with $ (x,y)\in D=\{(r,\theta)\;|\;0\le r\le 1, 0\le \theta \le 2\pi\}$. Therefore, the total area equals
$$
A_1+A_2=\iint_D ||f_x\times f_y|| dA + \iint_D ||g_x\times g_y|| dA\\
=\sqrt{2}\int_0^{2\pi}\int_{0}^1rdrd\theta+\int_0^{2\pi}\int_{0}^1r\sqrt{4r^2+1}drd\theta\\
=\sqrt{2}\pi+\frac{5\sqrt{5}-1}{6}\pi
$$
