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I don't know much about surface integrals and unfortunately I have been asked this problem.

If $$\mathbf{F}(x,y,z)=y^2 \hat i-y \hat j+xyz \hat k$$ evaluate $$\int_S \mathbf{F}\cdot \mathbf{n}\, dS$$ where $S$ is the curved surface of the cylinder $$x^2+y^2=4, \: 0 \leq z \leq 3$$ and $\bf n$ is directed away from the $z$-axis. Any help will be appreciated.

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    Dear Basil, it would be helpful if you tell people what exactly do you know about surface integrals and what is giving you trouble with this problem. For example, do you know what's the usual procedure to find a surface integral?2012-06-18
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    Should $n$ be a vector, namely $\bf n$?2012-06-18

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Let's call $S_c$ the curved surface of the cylindrical solid of radius $2$ about the interval $[0,3]$ along the $z$-axis, $S_t$ the top flat surface, $S_b$ the bottom flat surface, and similarly index the respective outward unit normal vectors. We are trying, then, to find $$\int_{S_c}\mathbf{F}\cdot\mathbf{n}_c\,dS_c.$$

Note that we have $$\mathbf{n}_c=\frac{\nabla(x^2+y^2-4)}{\lVert\nabla(x^2+y^2-4)\rVert}=\frac{2x\hat i+2y\hat j+0\hat k}{\lVert 2x\hat i+2y\hat j+0\hat k\rVert}=\frac{2}{\sqrt{(2x)^2+(2y)^2+0^2}}\left(x\hat i+y\hat j+0\hat k\right)=\frac{1}{\sqrt{x^2+y^2}}\left(x\hat i+y\hat j+0\hat k\right),$$ and recalling that $x^2+y^2=4$ on $S_c$, we have $$\mathbf{n}_c=\frac{1}{2}\left(x\hat i+y\hat j+0\hat k\right).$$ Thus, $$\mathbf{F}\cdot\mathbf{n}_c=\frac{1}{2}(x-1)y^2.$$

Also, $\mathbf{n}_t=\hat k$, and $\mathbf{n}_b=-\hat k$, so $$\mathbf{F}\cdot\mathbf{n}_t=(y^2\hat i-y\hat j+3xy\hat k)\cdot\hat k=3xy,$$ and $$\mathbf{F}\cdot\mathbf{n}_b=(y^2\hat i-y\hat j+0\hat k)\cdot\hat k=0.$$

By divergence theorem, $$\int_V(\nabla\cdot\mathbf{F})\,dV=\int_S(\mathbf{F}\cdot\mathbf{n})\,dS,$$ where $V$ is the cylindrical solid described at the start and $S$ is the full surface of said solid. It is clear that $$\int_S(\mathbf{F}\cdot\mathbf{n})\,dS=\int_{S_c}(\mathbf{F}\cdot\mathbf{n}_c)\,dS_c+\int_{S_t}(\mathbf{F}\cdot\mathbf{n}_t)\,dS_t+\int_{S_b}(\mathbf{F}\cdot\mathbf{n}_b)\,dS_b,$$ so by the work above, $$\int_S(\mathbf{F}\cdot\mathbf{n})\,dS=\int_{S_c}(\mathbf{F}\cdot\mathbf{n}_c)\,dS_c+3\int_{S_t}xy\,dS_t.$$ Thus, since $$\nabla\cdot\mathbf{F}=\left(\frac{\partial}{\partial x}\hat i+\frac{\partial}{\partial y}\hat j+\frac{\partial}{\partial z}\hat k\right)\cdot(y^2\hat i-y\hat j+xyz\hat k)=-x+xy=x(y-1),$$ then we have $$\int_{S_c}(\mathbf{F}\cdot\mathbf{n}_c)\,dS_c=\int_Vx(y-1)\,dV-3\int_{S_t}xy\,dS_t.\qquad (\#)$$

Now, since we're dealing with a cylinder, it makes sense to use cylindrical coordinates to evaluate the volume integral. In particular, $dV=r\,dz\,dr\,d\theta$, $x=r\cos\theta$, $y=r\sin\theta$, $0\leq z\leq 3$, $0\leq r\leq 2$, and $0\leq\theta\leq 2\pi$--hopefully, that should be enough to let you evaluate $\int_Vx(y-1)\,dV$ as a triple integral.

We can readily parameterize $S_t$ by $\mathbf{r}(x,y)=x\hat i+y\hat j+3\hat k$, where $(x,y)$ varies over the region $C=\{(x,y)\in\Bbb R^2:x^2+y^2\leq 4\}$ in the $xy$-plane. Now, $$dS_t=\left\lVert\frac{\partial\mathbf{r}}{\partial x}\times\frac{\partial\mathbf{r}}{\partial y}\right\rVert\,dA,$$ so since $$\frac{\partial\mathbf{r}}{\partial x}\times\frac{\partial\mathbf{r}}{\partial y}=\left|\begin{array}{ccc}\hat i & \hat j & \hat k\\ 1 & 0 & 0\\ 0 & 1 & 0\end{array}\right|=\hat k,$$ so $dS_t=dA$ and $$\int_{S_t}xy\,dS_t=\int_Cxy\,dA.$$ As we are dealing with a circular region $C$, then polar coordinates will be the best way, with $dA=r\,dr\,d\theta$, $x=r\cos\theta$, $y=r\sin\theta$, $0\leq r\leq 2$, and $0\leq\theta\leq 2\pi$--again, hopefully this will be enough for you to calculate $\int_{S_t}xy\,dS_t$ as a double integral.

Once you've calculated the aforementioned triple and double integrals, the desired surface integral will follow from the equation $(\#)$.

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HINT Use gauss divergence theorem and then cylindrical coordinates http://en.wikipedia.org/wiki/Divergence_theorem