Let´s try to solve the following exercise :
Calculate the flux of the vector field: f$(x,y,z) = (x^2,y^2,z^2)$ through the surface $S = \{x^2+y^2=z^2,0\leq\ z \leq\ 1\} $ (orientation fixed).
My attempt
I tried to solve it using Gauss' Theorem. The closed surface of the volume would be $ S $ $ \cup$ $S_1 $ where $S_1 $ is the surface that describes the circle $ x^2 +y^2 =1 $ in the plane $ z=1$. We are interested in calculating the integral:
$ \iint_S $ f $\cdot\ $n $dS = \iiint_V div $ f $dxdydz - \iint_{S_1} $ f $\cdot\ $n $dS_1 $
In my case:
$div$ f $ =2x+2y+2z $
To calculate the first integral, I decided to use cylindrical coordinates:
$x=\rho \cos\varphi$
$y = \rho \sin \varphi$
$z = z$
Using that $x^2+y^2=z^2 $ on the cone, we have that $\rho = z$. The limits of integration could be:
$ 0\leq\ \rho \leq\ 1\ $
$ 0\leq\ \varphi \leq\ 2\pi $
$ \rho\leq\ z \leq\ 1 $
Putting it all together:
$ \iiint_V div$ f $dxdydz = 2 \cdot \int _{0}^{1} \int _{0}^{2\pi} \int _{\rho}^{1} (\rho (\cos \varphi + \sin \varphi) +z) \cdot \rho \cdot d \rho d\varphi dz $
For the second integral, we parametrize the surface: $r_1(x,y) = (x,y,1)$
$ \iint_{S_1} $ f $\cdot\ $n $dS_1 = \iint_Q (x^2,y^2,1) \cdot (0,0,1) dxdy = \mu (Q) = \pi $
where $Q = \{(x,y) \in R^2|x^2+y^2 \leq\ 1\} $
My questions are:
Is it correct? What can we say about the orientation? Is there a simpler way?
Any suggestions are welcome.