A surface intergal question with detailed (wrong) solution

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)?

Last modified: Friday, 27 July 2012, 11:43 PM