Question (refresh if symbols do not display properly):
Let $$\vec F =<y,\ -x,\ z^2>$$ and $$S=\{(x,y,z): z=x^2+y^2, z<1\}$$
. Compute the outward flux of $$\vec F$$.

False Solution:

Treat $$x,y$$ as parameters:

$$\left\{ \begin{array}{ c c c } x & = & x \\ y & = & y \\ z & = & x^2+y^2 \end{array} \right.$$

or write the above as
$$\vec r(x,y)=<x,y,z>=<x,y,x^2+y^2>.$$

Then a normal vector $$\vec N$$ is given by
$$\vec N = \frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y} = <1,0,2x>\times <0,1,2y>=<-2x,-2y,1>.$$

$$\int\int\limits_{S}\vec F\cdot\vec n dS = \int\int\limits_{A}\vec F\cdot\frac{\vec N}{|\vec N|} \left|\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y}\right|dA$$

$$= \int\int\limits_{A}\vec F\cdot\frac{\vec N}{\left|\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y} \right|} \left|\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y}\right|dA$$

$$= \int\int\limits_{A}\vec F\cdot \vec N dA$$

$$= \int\int\limits_{A}(-2xy+2xy+z^2) dA = \int\int\limits_{A} z^2 dA = \int\int\limits_{A}(x^2+y^2)^2 dxdy$$

Now using polar coordinate, the above integral becomes

$$\int_0^{2\pi}d\theta\int_0^1r^4\cdot rdr=\frac{\pi}{3}.$$

The above solution did not give the correct answer. Can you find the mistake(s)?