Let S be the portion of the cylinder $x^2+y^2=4$ between the planes $z=0$ and $z=x+3$. Compute the following integral:


I know the typical conversion formula is:

$$\displaystyle\iint_{S}{f(x,y,z)dS} = \displaystyle\iint_{D}{f(\Phi(u,v))||T_u x T_v||dudv}$$

My problem for this is I'm not sure how to parametrize this surface. Would I parametrize the base of the cylinder (the disc) so that $\Phi(r,\theta) = (rcos(\theta), rsin(\theta), 0)$ or do $\Phi(u,v) = (u,v,u+3)$ or something completely different? Thanks in advance.

also, although it may be a more complicated method, I could achieve the same result using a standard triple integral correct?


You know that $0\leq z \leq x+3$. Therefore a good parametrization would be

$$\Phi(z,\theta)=(2\cos\theta,2\sin\theta,z),$$ where $$0\leq\theta\leq2\pi \,\,~~~~~~~~~\text{ and }\,\,~~~~~~~ 0\leq z \leq 2\cos\theta+3.$$

With respect to the last question, you could break the cylinder into two surfaces that can be represented as functions. Then you would set a triple integral with cartesian coordinates.

  • $\begingroup$ I was trying to find a way to turn it into cylindrical coordinates for a parametrization but I was never satisfied with what I got. This is what I was looking for. Thanks! $\endgroup$ – infinitylord Dec 5 '14 at 3:24
  • $\begingroup$ Yep, these are cylindrical coordinates :P $\endgroup$ – Vladimir Vargas Dec 5 '14 at 3:26

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