Area of surface using polar coordinates

Question

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!

Answer 1

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:

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

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