Calculating the mass of a surface? I have equation for a surface, $ z = 2/3(x^{3/2} + y^{3/2}) $ with $0 \leq x \leq 1$, $0 \leq y \leq 1$ which has the density of $g(x,y,z) = 120$.
I need to set up a double integral for the mass of this surface. Basically I'm not sure where the $120$ comes into play and what to do with it, or even how to begin the problem.
 A: If you have a small section of your surface of area $\Delta S$ where the density doesn't change much, then its mass is going to be roughly the density times the area, i.e. $g \Delta S$. Summing over all such small pieces gives us an approximate total mass of $\sum g\Delta S$, and taking the limit as $\Delta S \rightarrow 0$ leaves us with:
Mass = $\displaystyle \int_S g \, dS$
A: I'm going to assume that $g(x,y,z)$ represents the mass per unit area of the surface.  In that case, to find the total mass of the surface, you need to integrate
$$
\int_{x=0}^{1}\int_{y=0}^{1}g(x,y,z(x,y))dS,
$$
where $z(x,y)=\frac{2}{3}\left(x^{3/2}+y^{3/2}\right)$, and the area element $dS$ is
$$
dS=dx\;dy\;\sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}.
$$
In your case
$$
\frac{\partial z}{\partial x}=\sqrt{x}, \qquad \frac{\partial z}{\partial y}=\sqrt{y},
$$
so
$$
dS=dx \;dy\;\sqrt{1+x+y}.
$$
The total mass is therefore
$$
m=120\int_{0}^{1}dx\int_{0}^{1}dy\sqrt{1+x+y}.
$$
You should be able to take it from here.
