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I have been asked to find a parameterization for the surface $9=x^2+z^2,0\leq y\leq4$, and rewrite the surface integral $\iint y dS$ as a double integral.

I believe that the parameterization should look like this.

$y=y,$ $ x=f(y)\cos\theta,$ $z=f(y)\sin\theta,$ $ 0\leq y\leq4 ,$ $0\leq \theta\leq2\pi $

I then should have $\textbf{r}(t)=\langle 3 \cos \theta,y,3\sin\theta\rangle$

However from this point on I'm not sure about how to convert it to arrive at the form $A(S)=\iint| \textbf{r}_u \times \textbf{r}_v|dA$

So building off Mhenni Benghorbal's answer I should arrive at $| \textbf{r}_u \times \textbf{r}_v| = \sqrt{9\cos^2v + 9\sin^2v } = 3$

Thus the surface integral is $\iint y dS = \iint u|\textbf{r}_u \times \textbf{r}_u|dA=\int^{2\pi}_0 \int^4_0 3v$ $du$ $dv = \int^{2\pi}_0 12dv = 24\pi$

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1 Answer

Here is your parametrization of the surface $r(u,v)$ $$ r(u,v)= \langle 3 \cos v, u , 3\sin v \rangle, \quad 0 \leq u \leq 4,\, 0\leq v \leq 2\pi. $$

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