# MyOpenMath/Solutions/c19SurfaceIntegralCALCULUS Review of dot product and a reminder that unit vectors are orthogonal and have unit magnitude. Since we are in cylindrical coordinates, we use $\rho$ for distance to the axis to avoid the use of $r$ in spherical coordinates. Define the vector area differential element $d{\vec {A}}={\hat {n}}dA$ using the outward unit normal ${\hat {n}}$ . From the top down, we see the top of the cylinder in what looks like polar coordinates. On the top circle have a $dA=\rho d\rho d\varphi$ Also shown in is the curved side of the cylinder. Here the dimensions of the small rectangle are such that $dA=\rho d\rho dz$ . The unit vectors $({\hat {i}},{\hat {j}},{\hat {k}})$ are replaced by $({\hat {\rho }},{\hat {\varphi }},{\hat {z}})$ . Evaluate the integral over the top surface. On the curved sides, $F\cdot d{\vec {A}}=F_{\rho }dA$ . Note that the "odd" term vanishes: $\int _{-a}^{a}z^{n}dz=0{\text{ if n odd}}$ $\int _{-a}^{a}z^{n}dz=2\int _{0}^{a}z^{n}dz{\text{ if n even}}$  When doing the integral over the entire surface, it is customary to put a little circle at the center of the integral sign. When adding the top and bottom integrals we have to remember that ${\hat {n}}_{top}={\hat {z}}$ on the top surface but that ${\hat {n}}_{bot}=-{\hat {z}}$ on the bottom surface. The result depends on whether $F_{z}(\rho ,z)$ is an even or odd function of z.