Originally Posted by
andreas I did the following(for the first problem):
I imagined cylinder$\displaystyle x^2+y^2\leq1$ , $\displaystyle 0\leq Z\leq1$
First I consider the pipe without top and bottom ( $\displaystyle x^2+y^2=1$ and $\displaystyle 0<Z<1$ )
Since $\displaystyle F$ for $\displaystyle i$,$\displaystyle j$ has components 1 and 1. The integral for pipe will be simply $\displaystyle 2\pi$ Mr F says: I don't think so. See below.
Now I consider top and bottom discs: integral for bottom disc will be $\displaystyle 0$ because it's at $\displaystyle z=0$.The most interesting is top disc ( $\displaystyle x^2+y^2\leq 1$ and $\displaystyle Z=1$). For area of top disc I will have the following integral $\displaystyle \int z(x^2+y^2)^2 dA$ but the top z is 1 so integral is $\displaystyle \int (x^2+y^2)^2 dA$. dA is a little area at the top disc, and I think it is $\displaystyle 2\pi rdr$ where $\displaystyle r=\sqrt{x^2+y^2}$ So finally my integral is $\displaystyle \int (r^2)^2 2\pi rdr=2 \pi\int r^5 dr$ Taking it from 0 to 1 I get $\displaystyle \frac{\pi}{3}$.
So the finall result is:$\displaystyle \frac{\pi}{3}+2\pi=\frac{7\pi}{3}$
I have to submit both problems very soon. Can you check if I did first correctly? How should I prove second one?