# 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 ${\displaystyle \rho }$ for distance to the axis to avoid the use of ${\displaystyle r}$ in spherical coordinates. Define the vector area differential element ${\displaystyle d{\vec {A}}={\hat {n}}dA}$ using the outward unit normal ${\displaystyle {\hat {n}}}$. From the top down, we see the top of the cylinder in what looks like polar coordinates. On the top circle have a ${\displaystyle dA=\rho d\rho d\varphi }$ Also shown in is the curved side of the cylinder. Here the dimensions of the small rectangle are such that ${\displaystyle dA=\rho d\rho dz}$. The unit vectors ${\displaystyle ({\hat {i}},{\hat {j}},{\hat {k}})}$ are replaced by ${\displaystyle ({\hat {\rho }},{\hat {\varphi }},{\hat {z}})}$. Evaluate the integral over the top surface. On the curved sides, ${\displaystyle F\cdot d{\vec {A}}=F_{\rho }dA}$. Note that the "odd" term vanishes: ${\displaystyle \int _{-a}^{a}z^{n}dz=0{\text{ if n odd}}}$ ${\displaystyle \int _{-a}^{a}z^{n}dz=2\int _{0}^{a}z^{n}dz{\text{ if n even}}}$ 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 ${\displaystyle {\hat {n}}_{top}={\hat {z}}}$ on the top surface but that ${\displaystyle {\hat {n}}_{bot}=-{\hat {z}}}$ on the bottom surface. The result depends on whether ${\displaystyle F_{z}(\rho ,z)}$ is an even or odd function of z.