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?

  • 1
    $\begingroup$ Can you give a context? $\endgroup$ Dec 2 '18 at 20:39
  • $\begingroup$ I am not really sure actually..apparently there is a difference in the ways to calculate such integral..that's what I am trying to get behind ^^ $\endgroup$ Dec 2 '18 at 21:03
  • $\begingroup$ "$do$" is not a standard notation. You will have to look it up in your book to find out what it means. We don't know anything about where it came from (that "context" thing Tito asked for), so we can't help you, $\endgroup$ Dec 3 '18 at 3:47
  • $\begingroup$ I added it to the text :) $\endgroup$ Dec 3 '18 at 17:39

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 ($\rho, u_1$) $$x=\rho\cos u_1,\quad x=\rho \sin u_1,\quad z=1,\quad \mathrm ds = \rho\,\mathrm d\rho\,\mathrm du_1,$$ $$S_{ceiling} = \int\limits_0^{2\pi}\int\limits_0^r(\rho^2\cos^2u_1)\rho\,\mathrm d\rho\,\mathrm du_1.$$

  • $\begingroup$ thanks for the answere !! what is about both of the cover surfaces? $\endgroup$ Dec 5 '18 at 18:18
  • 1
    $\begingroup$ @wondering1123 If you mean the internal and the external surfaces, you can use factor 2. If the ceiling and the floor, this can be easy calculated too. $\endgroup$ Dec 5 '18 at 18:57
  • $\begingroup$ yes, I do mean the ceiling and the floor! $\endgroup$ Dec 5 '18 at 19:50

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