Compute the surface integral I always get confused with this...

Let $S$ be the surface $$z=x^2+y^2, z\leq 1,$$ oriented so that the normal vector has positive $z$-coordinate, and let $F$ be the vector field $(yz,-xz+\sin(z), e^{x^2+y^2})$. Compute the surface integral $$\int_SF.n\, dS  \,\,$$

 A: Of course this would be much easier using Gauss divergence theorem, but I wanted to calculate it directly, because it is what the OP asked
Today I feel generous:


*

*Parametrize

*Find the normal vector (according to the given orientation)

*Calculate the integral using the formula


So let's go:


*

*Let's use the following coordinates 
$\Phi:(\theta,z)\in\Omega:=[0,2\pi]\times[0,1]\mapsto\begin{pmatrix}z\cos(\theta) \\ z\sin(\theta) \\ z^2\end{pmatrix}$

*We first compute one of the possible normal vectors  $$\tilde{\vec{n}}=\frac{\Phi_\theta\times \Phi_z}{\|\Phi_\theta\times \Phi_z\|}=\frac{\begin{pmatrix}-z\sin(\theta) \\ z\cos(\theta) \\ 0\end{pmatrix}\times \begin{pmatrix}\cos(\theta) \\ \sin(\theta) \\ 2z\end{pmatrix}}{\|\Phi_\theta\times \Phi_z\|}= \frac{\begin{pmatrix}2z^2\cos(\theta) \\ 2z^2\sin(\theta) \\ -z\end{pmatrix}}{\|\Phi_\theta\times \Phi_z\| }.$$
As the normal vector has to be chosen with positive $z$ coordinate, we thus set $$\vec{n}=-\tilde{\vec{n}}$$

*\begin{align*}\int_SF\cdot dA& =\int_\Omega F\circ\Phi(\theta,z)\cdot \vec{n}\|\Phi_\theta\times \Phi_z\|dzd\theta = \int_\Omega F\circ\Phi(\theta,z)\cdot-(\Phi_\theta\times \Phi_z) \ dzd\theta\\ & =\int_0^{2\pi}\int_0^1 \begin{pmatrix}\sin(\theta) z^3\\ -\cos(\theta)z^3+\sin(z^2) \\ e^{z^2} \end{pmatrix}\cdot \begin{pmatrix}-2z^2\cos(\theta) \\ -2z^2\sin(\theta) \\ z\end{pmatrix} dzd\theta \end{align*}
you can take it from here
A: Let $g(x,y,z)=x^2+y^2-z$  as a result $$\nabla g(x,y,z)=(2x,2y,-1)$$
 and $$n=\frac{\nabla g}{\left\| \nabla g.\overset{\to }{\mathop{k}}\, \right\|}=(2x,2y,-1)$$ 
we have 
\begin{align}
  & \int{F.n\,ds}=\iint\limits_{D}{\left( y({{x}^{2}}+{{y}^{2}})\,,\,-x({{x}^{2}}+{{y}^{2}})+\sin ({{x}^{2}}+{{y}^{2}})\,,\,{{e}^{({{x}^{2}}+{{y}^{2}})}} \right)}.(2x,2y,-1)dA \\ 
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\iint\limits_{D}{[2y\sin ({{x}^{2}}+{{y}^{2}})-{{e}^{({{x}^{2}}+{{y}^{2}})}}]dA} \\ 
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int_{0}^{2\pi }{\int_{0}^{1}{(2{{r}^{2}}\sin \theta \sin ({{r}^{2}})-r{{e}^{{{r}^{2}}}})\,d}}rd\theta  \\ 
\end{align}
