# Surface integral: S half-cylinder

Could anyone help with this problem?

Evaluate $\iint_S \textbf{F}$ $\cdot$ $\textbf{n}$ $d \alpha$.

c) S is the truncated half-cylinder $z=\sqrt{4-y^2}$, $0 \le x \le 1$, $\textbf{n}$ is the outward-pointing normal, and $\textbf{F}(x,y,z)=(x^2+y^2,yz,z^2)$.

Here is what I have done:

Parametrized by $\phi(u,v)=(u,v,\sqrt{4-v^2})$, and found the normal to be $\textbf{n} = (0,-v/\sqrt{4-v^2},1)$. Then integrated $\textbf{F}$ $\cdot$ $\textbf{n} = 4-2v^2$ over $[0,1]\times[-2,2]$, but got the wrong answer. The answer is supposed to be 16.

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Have you tried using the parametrization $(u\quad 2\cos\;v\quad 2\sin\;v)^T$ ? – J. M. Apr 28 '11 at 3:51
That one works.But why didn't the other one work? – user2467 Apr 28 '11 at 4:14

Using your parameterization, I got $\mathbf{n}=\left( 0,v/\sqrt{4-v^2},1\right)$, which gives
$$\mathbf{F}\cdot \mathbf{n}=\left( v\sqrt{4-v^2}\right) \left( v/\sqrt{4-v^2}\right) +(4-v^2)=4,$$
which gives your correct answer of $16$. I'm thinking you just accidentally screwed up the minus sign in your calculation.