For the vector field $a = [-z,x-y,-2y+2z]^T$ and the area $F$ on the cylinder $x^2 + z^2 = 9$ , which is above the ground plane $z = 0$ , in front of the plane $x = 0$ and between the cross plane $y = 0$ and lies to the their parallel plane $y = 5$ , calculate the following integral:
$\int_{F}^{} \! a\cdot dn \, = ?$
So I use that:
$x=3cos(u)$
$y=v$
$z=3sin(u)$
and than I calculate normal vector.
I get integral $\int_{0}^{5}\int_{0}^{\Pi/2}\begin{pmatrix}-z\\x-y\\-2y+2z\end{pmatrix}\cdot \begin{pmatrix}3sin(u)\\ 0\\ 3sin(u)\end{pmatrix}dudv $
$\int_{0}^{6}\int_{0}^{\Pi/2}\begin{pmatrix}-3sin(u)\\3cos(u)-v\\-2v+6sin(u)\end{pmatrix}\cdot \begin{pmatrix}3sin(u)\\ 0\\ 3sin(u)\end{pmatrix}dudv $
$\int_{0}^{5}\int_{0}^{\Pi/2}-9sin^2(u)-6vsin(u)+18sin^2(u)$
$\int_{0}^{\Pi/2}45sin^2(u)-75sin(u)du$
at the end I get that $45*\Pi/4-75$ which isn't correct result.
Solution is: $-195/2+45/2*\Pi$