# Thread: Surface integral.

1. ## Surface integral.

Calculate $\int {x^2 zds}$ Being S the external surface of $x^2 + y^2 = a^2$ understood between $z = -2$ and $z = 2$

2. Originally Posted by rikelme23
Calculate $\int {x^2 zds}$ Being S the external surface of $x^2 + y^2 = a^2$ understood between $z = -2$ and $z = 2$
$x^2+ y^2$ is a cylinder, with radius a, in three dimensions. It can be written in parametric equations using cylindrical coordinates but taking r= a: $x= a cos(\theta)$, $y= a sin(\theta)$, $z= z$.

The "position vector" of any point on the cylinder can be written as $\vec{p}= a cos(\theta)\vec{i}+ a sin(\theta)\vec{j}+ z\vec{k}$. The two derivatives:
$\vec{p}_\theta= -a sin(\theta)\vec{i}+ a cos(\theta)\vec{j}$ and
$\vec{p}_z= \vec{k}$
are tangent vectors to the cylinder. Their cross product is
$-a sin(\theta)\vec{i}+ a cos(\theta)\vec{j}$ and the "scalar differential of surface area" is given by the length of that: $a d\theta dz$.

(For this simple geometry, you could have just noted that the differential of length around the cylinder is $a d\theta$ so the differential of area is $a d\theta dz$.)

Since $x^2= a^2 cos^2(\theta)$ this integral is
$\int_{z= -2}^2\int_{\theta= 0}^{2\pi} (a^2 cos^2(\theta)) z (a d\theta dz)$.

(In fact, just thinking about the "symmetry" (or, more correctly "asymmetry") makes this problem trivial.)