<s>[INST] Finding flux over a surface If $F = 6z\mathbf i + (2x+y)\mathbf j -x\mathbf k$, evaluate $\int_S \mathbf F \cdot \mathbf n ds$ over the surface bounded by the cylinder $x^2 + z^2 = 9. x=0, y=0, z=0 $ and $ y= 6$
Okay, so I know this is a quarter cylinder in the first octant. Now assuming the surface they're talking about is $z = \sqrt{ 9 - x^2 } $, here's what I've got so far.
$ n = \nabla f $ where $f(x,y,z)= x^2 + z^2 - 9 = 0$ which yields $<2x, 0, 2z>$
now for $\hat n$
$$ \hat n = \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}$$
and do get $ds$, I used $g(x,y) =  \sqrt{ (9 - x^2) } $ where $g_x = \frac {-x}{\sqrt{9-x^2}}$ and $g_y= 0 $ 
therefore $$ds = \sqrt{ \frac {9}{9-x^2}     } = \frac {3}{\sqrt{ {9-x^2} }} dA$$ 
Finally $$\int_S\mathbf F\cdot \hat n ds = <6z, 2x+4, -x> \cdot \frac {<2x, 0, 2z>} {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} }    } $$
$$ = \int\int_R \frac {10x \sqrt{ {9-x^2} }  } {\sqrt {4x^2 + 4z^2 }}\frac {3}{\sqrt{ {9-x^2} }    }dA $$ OR
$$ = \int\int_R \frac {15x } {\sqrt {x^2 + z^2 }   }dxdy = \int_0^6\int_0^3 5x dx dy$$
My question is, is this setup correct or is any of this wrong?
 [/INST] For a direct computation, note that your surface is the union of the five surfaces parameterized by
\begin{align*}
  \mathbf r(x,y)&=\left\langle x,y,\sqrt{9-x^2}\right\rangle & (x,y)&\in[0,3]\times[0,6] \\
  \mathbf s(x,y) &= \langle x,y,0\rangle & (x,y)&\in[0,3]\times[0,6] \\
  \mathbf t(y,z) &=\langle0,y,z\rangle & (y,z)&\in[0,6]\times[0,3] \\
  \mathbf u(r,\theta) &= \langle r\cos\theta,0,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2] \\
  \mathbf v(r,\theta) &= \langle r\cos\theta,6,r\sin\theta\rangle & (r,\theta)&\in[0,3]\times[0,\pi/2]
\end{align*}
Our computation is then
\begin{align*}                                                                                                                                                                                                                                 
  \iint_S(\mathbf F\cdot\mathbf n)\,dS                                                                                                                                                                                                         
  &= \int_0^6\int_0^3\mathbf F\bigl(\mathbf r(x,y)\bigr)\cdot(\mathbf r_x\times\mathbf r_y)\,dx\,dy \\                                                                                                                                         
  &\qquad+\int_0^6\int_0^3\mathbf F\bigl(\mathbf s(x,y)\bigr)\cdot(\mathbf s_y\times\mathbf s_x)\,dx\,dy \\                                                                                                                                    
  &\qquad\qquad+\int_0^3\int_0^6\mathbf F\bigl(\mathbf t(y,z)\bigr)\cdot(\mathbf t_z\times\mathbf t_y)\,dy\,dz \\                                                                                                                              
  &\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf u(r,\theta)\bigr)\cdot(\mathbf u_r\times\mathbf u_\theta)\,dr\,d\theta \\                                                                                                   
  &\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^3\mathbf F\bigl(\mathbf v(r,\theta)\bigr)\cdot(\mathbf v_\theta\times\mathbf u_r)\,dr\,d\theta \\                                                                                             
  &= \int_0^6\int_0^3\left\langle6\sqrt{9-x^2},2\,x+y,-x\right\rangle\cdot\left\langle \frac{x}{\sqrt{9-x^2}},0,1\right\rangle\,dx\,dy \\                                                                                                      
  &\qquad+\int_0^6\int_0^3\langle0,2\,x+y,-x\rangle\cdot\langle0,0,-1\rangle\,dx\,dy \\                                                                                                                                                        
  &\qquad\qquad+\int_0^3\int_0^6\langle6\,z,y,0\rangle\cdot\langle-1,0,0\rangle\,dy\,dz \\                                                                                                                                                     
  &\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta,-r\cos\theta\rangle\cdot\langle0,-r,0\rangle\,dr\,d\theta \\                                                                                                    
  &\qquad\qquad\qquad\qquad+\int_0^{\pi/2}\int_0^{3}\langle6r\sin\theta,2r\cos\theta+6,-r\cos\theta\rangle\cdot\langle0,r,0\rangle\,dr\,d\theta \\                                                                                             
  &= \underbrace{5\int_0^6\int_0^3x\,dx\,dy+\int_0^6\int_0^3x\,dx\,dy-6\int_0^3\int_0^6z\,dy\,dz}_{=0} \\                                                                                                                                      
  &\qquad+\int_0^{\pi/2}\int_0^3-2r^2\cos\theta\,dr\,d\theta+\int_0^{\pi/2}\int_0^3(2 r^2\cos\theta+6r)\,dr\,d\theta \\                                                                                                                                                            
  &= 6\int_0^{\pi/2}\int_0^3r\,dr\,d\theta \\                                                                                                                                                                                               
  &= 6\left(\frac{9}{2}\right)\int_0^{\pi/2}d\theta \\
&= 6\left(\frac{9}{2}\right)\left(\frac{\pi}{2}\right) \\
&=\frac{27}{2}\pi                                                                                                                                                                                                                                      
\end{align*}
The problem is considerably easier if we use the divergence theorem. We are asked to compute
$$
\iint_{S}(\mathbf F\cdot\mathbf n)\,dS
$$
where $\mathbf F(x,y,z)=\langle 6\,z,2\,x+y,-x\rangle$ and $S$ is the quarter cylinder you describe. The divergence theorem states that
$$
\iint_{S}(\mathbf F\cdot\mathbf n)\,dS=\iiint_T(\nabla\cdot\mathbf F)\,dV
$$
where $T$ is the solid enclosed by $S$ and $\displaystyle\nabla=\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle$. Our computation is then
\begin{align*}
  \iint_{S}(\mathbf F\cdot\mathbf n)\,dS
  &=\iiint_T(\nabla\cdot\mathbf F)\,dV \\
  &=\iiint_T\left\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right\rangle\cdot\langle 6\,z,2\,x+y,-x\rangle\,dV \\
  &= \iiint_T(0+1+0)\,dV \\
  &= \iiint_TdV \\
  &= \text{volume of }T \\
  &= \frac{1}{4}\pi\cdot 3^2\cdot 6 \\
  &= \frac{27}{2}\pi
\end{align*}
This verifies that our direct computation was, indeed, correct.
</s>