I want to determine
$ \int_F f do $
with $f(x,y,z) = x^2z $
$F$ is cylinder surface (lateral surface) with
$ F= \{ (x,y,z) \in \mathbb{R}^3 : x^2+y^2= 4 , 0 \leq z \leq 1 \} $
My Idea to solve this surface integral is so far:
Using cylindric coordinates
$\phi(u_1, u_2) = \begin{pmatrix} r cos (u_1) \\ rsin(u_1) \\u_2 \end{pmatrix} $
$\phi: (0,2 \pi) \times (o,z) $
$0 \leq u_1 \leq 2\pi , 0\leq u_2 \leq z $
the derivates:
$ \phi_{u_1}= \begin{pmatrix} -r sin (u_1) \\ rcos(u_1) \\ 0 \end{pmatrix}$
and
$ \phi_{u_2}=\begin{pmatrix} 0 \\ 0 \\1 \end{pmatrix}$
the cross product comes to :
$ \phi_{u_1} \times \phi_{u_2} = \begin{pmatrix} r cos (u_1) \\ rsin(u_1) \\0 \end{pmatrix}$
so $ ||\phi_{u_1} \times \phi_{u_2} || = r $
so, it comes to calculate following integral:
$$ \int_0^{2 \pi} \int_0^z (r cos^2 u_1 u_2) r du_2 du_1 $$
dont I miss both of the cover surfaces, or is that not needed?
Best Answer
Notation is strange, the reasoning is correct. Really, $$x=r\cos u_1,\quad z=u_2,\quad \mathrm ds = r\,\mathrm du_1\,\mathrm du_2.$$ But the reasoning leads to the expression for the side surface $$S_{side}=\int\limits_0^{2\pi}\int\limits_0^{\not\hspace{1pt} z\color{red}1}(r^{\not\hspace{1pt} 1\color{red}2}\cos^2u_1\,u_2)r\,du_1\,du_2.$$
On the "floor" surface ($z=0$), the integrand is zero.
On the "ceiling" surface ($z=1$), there are the plain circle with the polar coordinates ($\rho, u_1$) $$x=\rho\cos u_1,\quad x=\rho \sin u_1,\quad z=1,\quad \mathrm ds = \rho\,\mathrm d\rho\,\mathrm du_1,$$ $$S_{ceiling} = \int\limits_0^{2\pi}\int\limits_0^r(\rho^2\cos^2u_1)\rho\,\mathrm d\rho\,\mathrm du_1.$$