How do I evaluate the surface integral $∫_s$ $F\cdot ds $
where $S$ is given by $\psi(r,\theta)=r\theta$i + $sin\theta$j$+r$k
for $0\le r \le 1,$ $ 0\le \theta \le2\pi$ and F $=2xz$ i+$2y$j
Do I start with $∫_s$ $F\cdot ds = ∫_0(x\dot{\,x}+y\dot{\,y})ds $ ?
Please correct me if Im wrong, Im not confident in calculus 3.
For parametric surfaces, there is this formula of $\iint_{S} F \cdot N dS = \iint_{D} F \cdot (r_u \times r_v) dA$. First calculate the partial derivatives of $\phi (r, \theta)$ with respect to r and $\theta$, which is $\phi_r = \theta i + k$ and $\phi_{\theta} = ri + cos \theta j$. Then compute the cross product of the two vectors $\phi_r$ and $\phi_{\theta}$, resulting in the vector $\phi_r \times \phi_{\theta} = -cos (\theta) i + rj+\theta cos(\theta) k$. Then just plug in the information to the formula. The right hand side will turn out to be $\int_{0}^{2\pi} \int_{0}^{1} -2r^{2} \theta cos (\theta) + 2rsin(\theta) dr d\theta$. Applying integration by parts on the left term for the second integral and just integrating the right term will yield an answer of zero.
