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Calculate the area of the surface $z=x+y$ that is inside the cylinder $x^2+y^4 = 4$.

I was able to find the correct answer by calculating the normal vector (using cross product) at each point on the surface parametrized: $$ \vec{n} = (-r)\vec{i}+(-r)\vec{j}+(r)\vec{k} $$ And then I used polar coordinates to integrate the domain of the parametrized surface: $$ \int_0^{2\pi}\int_0^2 ||\vec{n}|| \text{ } dr \text{ } d\theta = \sqrt{3}\int_0^{2\pi}\int_0^2 r\text{ } dr \text{ } d\theta = 4\pi\sqrt{3} $$ But after that I was asking myself why changing $drd\theta$ to the infinetesimal element of area $r dr d\theta$ is not giving me the correct answer. If I do that, I'll get my integral to be: $$ \sqrt{3}\int_0^{2\pi}\int_0^2 r^2\text{ } dr \text{ } d\theta \neq 4\pi\sqrt{3} $$ When we change our double integral from cartesian coordinates to polar coordinates, the infinetesimal element of area is changed to $dA = rdrd\theta$. But when I do that I get the wrong answer?! What am I doing that is not correct?

Thanks!

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  • $\begingroup$ @DougM I'll correct! sorry! $\endgroup$ – Bruno Reis Jun 22 '17 at 22:11
  • $\begingroup$ I will now ask, do you mean $x^2 + y^2 = 4$? $\endgroup$ – Doug M Jun 22 '17 at 22:12
  • $\begingroup$ @DougM Yeah, I've correct it! $\endgroup$ – Bruno Reis Jun 22 '17 at 22:12
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Asside from definitions of the cylinder

your surface:

$S:z = x+y\\ dS = (-1,-1,1)\ dy\ dx\\ \|dS\| = \sqrt 3\ dy \ dx$

In polar:

$\|dS\| = \sqrt 3 (r\ dr) \ d\theta$

$\int_{-2}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \sqrt 3 \ dy \ dx = \int_0^{2\pi}\int_0^2 \sqrt 3\ r\ dr\ d\theta = 4\pi\sqrt{3}$

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  • $\begingroup$ Wait. In that case the normal is trivial because the surface is a plane. But how did you find it without looking to the equation of the plane? I parametrized the surface of integration in polar coordinates, and then I took $dS = ||\vec{r_r} \times \vec{r_\theta}||$ which gave me $\vec{n} = (-r)\vec{i}+(-r)\vec{j}+(r)\vec{k}$, that when r = 1 is going to give me tha same normal as yours.What's wrong about that? If I kept going, and the place $dA$ I would have $r^2$ in the integration... $\endgroup$ – Bruno Reis Jun 22 '17 at 22:27
  • $\begingroup$ I found the parameterization $dS = (-1,-1,1)$ in cartesian and then flipped to polar, which introduces the $r$ If you find the parmetarization in polar directly $dS = \|r_r \times r_\theta\|$ that introduces the jacobian $r$ and you don't need to do it again. $\endgroup$ – Doug M Jun 22 '17 at 22:45
  • $\begingroup$ That is what I was looking for! the parametrization in polar introduces the jacobian... that's good to know! Thanks mate!!! $\endgroup$ – Bruno Reis Jun 22 '17 at 23:07

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