Evaluate $\iint_S (x^2 + y^2 + z^2) dS$ where $S$ is the boundary of the solid enclosed by the cone $z = \sqrt{x^2 + y^2}$ and $z = 4$. I tried to solve this by:
$$\iint_S (x^2 + y^2 + z^2)dS = \iint_{S_1} (x^2 + y^2 + z^2)dS + \iint_{S_2} (x^2 + y^2 + z^2) dS$$
where $S_1$ is the surface around the cone, and $S_2$ is the intersection of the cone and $z = 4$.
So for $S_1$ I have: $$ r(x,y) = \left\langle x,y, \sqrt{x^2 + y^2}\right\rangle $$ $$ r_x = \left\langle 1,0, \frac{x}{\sqrt{x^2+y^2}} \right\rangle $$ $$ r_y = \left\langle 0,1, \frac{y}{\sqrt{x^2+y^2}} \right\rangle $$
So, the cross product of $r_x$ and $r_y$ is:
$$ r_x \times r_y = \left\langle \frac{-x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1 \right\rangle $$ and the magnitude of that is: $\sqrt2 $
So,
$$\iint_{S_1} (x^2 + y^2 + z^2)dS = \iint_{D} (x^2 + y^2 + x^2 + y^2) \sqrt2 \,dS $$ which I've calculated to be $256\pi\sqrt2$
Now I'm trying to calculate $\iint_{S_2} (x^2 + y^2 + z^2) dS$ and I'm extremely confused on how to parametrize it. I know that the surface is $z = 4$ with $x^2 + y^2 \le 16 $ But how would I parameterize it? Would I use the same parameterization as the previous step?
I'd appreciate any help with this, but please explain!