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**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 ?