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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!

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  • $\begingroup$ There's a bunch of ways to do this. For instance - sit in polar coordinates, and recognize that the above integral is the flux of the field $\rho^2 \left(\frac{\hat{\rho} + \hat{z}}{\sqrt{2}}\right)$ through $S$. Now use Gauss' theorem. An alternate way is to split the surface into two - let $S_1$ be the bottom disk, and $S_2$ be the cone. Calculate the integrals separately over these two surfaces, and add. $\endgroup$ – stochasticboy321 Apr 21 '16 at 15:42
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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 $$ \iint_{S}x^2+y^2\;dS= \iint_{x^2+y^2\le9}(x^2+y^2)\sqrt{2}\;dA= \int_{0}^3\int_0^{2\pi}r^2\sqrt{2}\;rd\theta dr=\frac{81\pi}{\sqrt{2}} $$

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