1. ## Difficult Surface Integral...

In order to figure this out, I think I have to parametrize the integral correct? If I do, would this parametrization be valid?

x=x
y=x^2

BTW I think the answer will be zero since the flux going in is equal to the flux coming out. This of course is assuming this flux is the same as electrical flux.

2. $\displaystyle \int\int_S ( x^2 e^{-z} + y^2 e^{-z} )dS$

$\displaystyle = \int\int_S (x^2+y^2)e^{-z}~dS$

$\displaystyle x^2 + y^2 = 9 = 3^2 , z \in [0,10]$

if we transform in cylindrinal coordinates ,

$\displaystyle x^2 + y^2$ becomes $\displaystyle r^2$

$\displaystyle dS$ becomes $\displaystyle rd\theta dz$

$\displaystyle \int\int_S (r^2)e^{-z}(rd\theta dz)$

$\displaystyle = \int\int_S r^3 e^{-z} d\theta dz$

but $\displaystyle r = 3$ for every points on the surface

Surface integral $\displaystyle = (3)^3 \int_0^{2\pi} \int_0^{10} e^{-z} dz d\theta$

$\displaystyle = 27(2\pi) (1- e^{-10} )$

do we need to calcuate the integral for the base ?

3. Thanks for the help, one question though... Why did you switch the $\displaystyle d\theta$ for dz?

4. Originally Posted by simplependulum
$\displaystyle \int\int_S ( x^2 e^{-z} + y^2 e^{-z} )dS$

$\displaystyle = \int\int_S (x^2+y^2)e^{-z}~dS$

$\displaystyle x^2 + y^2 = 9 = 3^2 , z \in [0,10]$

if we transform in cylindrinal coordinates ,

$\displaystyle x^2 + y^2$ becomes $\displaystyle r^2$

$\displaystyle dS$ becomes $\displaystyle rd\theta dz$

$\displaystyle \int\int_S (r^2)e^{-z}(rd\theta dz)$

$\displaystyle = \int\int_S r^3 e^{-z} d\theta dz$

but $\displaystyle r = 3$ for every points on the surface

Surface integral $\displaystyle = (3)^3 \int_0^{2\pi} \int_0^{10} e^{-z} dz d\theta$

$\displaystyle = 27(2\pi) (1- e^{-10} )$

do we need to calcuate the integral for the base ?
first of all dS is not necessarily dxdy and so you need to be careful. second the cylinder is closed and so you also need to consider the top and the bottom of the cylinder!