## Mass of a Parametrized Surface

The mass of a surface parametrized with parameters
$u,v$
and surface density
$\rho (x,y,z)=\rho(u,v)$
varying from point to point is
$M= \int_u \int_v \rho(u,v) | \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| dv du$

Example: A surface is defined by
$\mathbf{r} = u \mathbf{i} + v \mathbf{j} + (u^2 +v^2) \mathbf{k}, 1 \leq u^2 +v^2 \leq 9$
and has surface density
$\rho (u,v) = \alpha \sqrt{u^2 +v^2}$

$\frac{\partial \mathbf{r}}{\partial u} = \mathbf{i}+2 u \mathbf{k}$

$\frac{\partial \mathbf{r}}{\partial v} = \mathbf{j}+2 v \mathbf{k}$

$| \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}| = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 2u \\ 0 & 1 & 2v \end{array} \right| = |-2u \mathbf{i} -2v \mathbf{j} + \mathbf{k}| = \sqrt{1+ 4u^2 + 4v^2}$
The mass is
$= \int_{1 \leq u^2 +v^2 \leq 9} \alpha (u^2 +v^2) \sqrt{1+4 u^2 +4v^2} dv du$

Transform into polar coordinates.
\begin{aligned} M &= \alpha \int^{2 \pi}_0 \int^3_1 r \sqrt{1+4r^2} r dr d \theta \\ &= 4 \alpha \int^{2 \pi}_0 \int^3_1 r^2 \sqrt{1/4 +r^2} dr d \theta \\ &= 4 \alpha \int^{2 \pi}_0 [\frac{r}{4} \sqrt{(r^2+1/4)^3} - \frac{1}{32} \sqrt{r^2+1/4} - \frac{1}{128} ln (r+ \sqrt{r^2+1/4}) ]^3_1 d \theta \\ &= \frac{ \alpha}{4} \int^{2 \pi}_0 (\frac{81}{8} \sqrt{37} - \frac{9}{8} \sqrt{5} + \frac{1}{4} ln (\frac{ 2+ \sqrt{5}}{6+ \sqrt{37}}) d \theta \\ &= \frac{\alpha \pi}{2} (\frac{81}{8} \sqrt{37} - \frac{9}{8} \sqrt{5} + \frac{1}{4} ln (\frac{ 2+ \sqrt{5}}{6+ \sqrt{37}}) \end{aligned}