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Example Evaluate $\iint_{S}f(x,y,z)dS$, where $S$ is the surface of the paraboloid $x^2 + y^2 + z = 4$ above the $xy$ plane and

$\displaystyle f(x,y,z) = \frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}$

Answer Since $S : x^2 + y^2 + z = 4$, we find the position vector ${\bf r}$ . Then

$\displaystyle {\bf r} = x\:{\bf i} + y\:{\bf j} + (4 - x^2 - y^2)\:{\bf k} $

We next find the normal vector of $S$. Find ${\bf r}_{x} \times {\bf r}_{y}$. Then
$\displaystyle {\bf r}_{x} \times {\bf r}_{y}$ $\displaystyle =$ $\displaystyle ({\bf i} -2x\:{\bf k}) \times ({\bf j} -2y\:{\bf k})$  
  $\displaystyle =$ $\displaystyle \left\vert\begin{array}{ccc} {\bf i} & {\bf j} & {\bf k}\\ 1 & 0 & -2x\\ 0 & 1& -2y \end{array}\right\vert = 2x\:{\bf i} + 2y\:{\bf j} + {\bf k}$  

Thus,

$\displaystyle \vert{\bf r}_{x} \times {\bf r}_{y}\vert = \sqrt{4x^2 + 4y^2 + 1}$

It has a sum of squares. So, we use the polar coordinate to transform $\displaystyle{\Omega : 4 - (x^2 + y^2) \geq 0}$
$\displaystyle \iint_{S} f(x,y,z) dS$ $\displaystyle =$ $\displaystyle \iint_{\Omega}\frac{2y^2 + z}{(4x^2 + 4y^2 + 1)^{1/2}}\vert{\bf r}_{x} \times {\bf r}_{y}\vert dx dy$  
  $\displaystyle =$ $\displaystyle \iint_{\Omega}(2y^2 + z)dx dy = \int_{0}^{2\pi}\int_{0}^{2}(4 - r^2 + 2r^2 \sin^{2}{\theta}) r dr d\theta$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2 \pi}\left[2r^2 - \frac{r^2}{4} + \frac{r^4}{2}\sin^{2... ...i} d\theta = \int_{0}^{2\pi}(4 + 8\sin^{2}\theta)d\theta = 8 \pi + 8\pi = 16\pi$