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I'm trying to figure out how to solve a surface area with surface and line integrals (showing both methods).

The area I'm trying to compute is the area of the shape $$x^2+y^2=9$$ bounded by $z=0$ and $z=y$. (Note: $y \ge 0$).

I've started the problem by making a parametrization: $$ r(u,v) = \langle 3\cos v, 3\sin v, u\rangle $$ from $0 \leq u \leq 3\sin v$ and $-\frac{ \pi}{3} \leq v \leq \pi$. The magnitude of $|r_u \times r_v| = 3$.

Not sure where to go from here.

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  • $\begingroup$ Do you mean the volume? $\endgroup$ – nbubis Apr 16 '14 at 1:34
  • $\begingroup$ No, surface area of the piece of this cylinder. $\endgroup$ – msenevir Apr 16 '14 at 1:35
  • $\begingroup$ Shouldn't it be $3\cos v$ instead? otherwise $x^2+y^2\neq 9$ $\endgroup$ – nbubis Apr 16 '14 at 1:42
  • $\begingroup$ @s3wix A little-known LaTeX tip: You can use \langle and \rangle (in left-and-right pairs) to get $\langle \ \ \ \rangle$ instead of using the inequality symbols < and > which give $<$ and $>$ $\endgroup$ – Fly by Night Apr 16 '14 at 1:42
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The surface integral is given by: $$\iint \left\| {\partial \mathbf{r} \over \partial x}\times {\partial \mathbf{r} \over \partial y} \right\| dx\, dy$$ Now when you transfer to cylindrical coordinates, you must convert your derivatives and add the Jacobian, so you must convert: $${\partial \mathbf{r} \over \partial x}={\partial \mathbf{r} \over \partial u}\cdot{\partial u \over \partial x} + {\partial \mathbf{r} \over \partial v}\cdot{\partial v \over \partial x}$$ And similar for $y$. Also, $$dx\, dy=u\, du\, dv$$ Put everything together and you should get your result.

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