I just came across this exam question:
Let $$\ S = \{(x; y; z) : z = 3 - \sqrt{x^2 + y^2} ; x^2 + y^2 \le 9 \}$$ which is the surface of a cone.
(a) Sketch S.
(b) Evaluate $$\ \iint_S x^2 + y^2 dS $$
For part B I'm just not sure how to do this. If anyone can show me how I'd really appreciate it!
Parametrize $S$: \begin{cases} x=x\\ y=y\quad \quad \quad \quad\quad \quad\mbox{with}\; x^2+y^2\le 9\\ z=3-\sqrt{x^2+y^2} \end{cases} Then compute $$ ||r_x\times r_y||=\sqrt{2} $$ It follows that $$ \iint_{S}x^2+y^2\;dS= \iint_{x^2+y^2\le9}(x^2+y^2)\sqrt{2}\;dA= \int_{0}^3\int_0^{2\pi}r^2\sqrt{2}\;rd\theta dr=\frac{81\pi}{\sqrt{2}} $$