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!
Best Answer
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:
$\|dS\| = \sqrt 3 (r\ dr) \ d\theta$
$\int_{-2}^2\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \sqrt 3 \ dy \ dx = \int_0^{2\pi}\int_0^2 \sqrt 3\ r\ dr\ d\theta = 4\pi\sqrt{3}$