Finding mass of surface with non-uniform density

Consider the surface given by $z = \sqrt{2xy}$ with $1 \leq x \leq 5$ and $1 \leq y \leq 4$.

What is the mass of the surface given that the density is $\sigma(x,y,z) = 6z$?

Tried solving this with the double integral: $$\int_1^5 \int_1^4 6\sqrt{2xy}dydx = \frac{56}{3}\sqrt{2}\big(5\sqrt{5}-1\big)$$ Which seems to be the wrong answer.

• You integrand is the mass of a infinitesimal piece of the correct density, but with area $da=dx\ dy$. The surface, however, has a differential area that depends on $x$ and $y$. – Kajelad Mar 20 '18 at 21:17

If $D$ is a compact domain on which $z=f(x,y)$ is integrable and $\sigma(x,y,z)$ is a continuous density function, the total mass of the surface defined by $f$ is
$$M=\int_D\sigma(x,y,z)\,dA.$$ The symbol $dA$ represents the area of the infinitesimal piece of the surface and is determined by
$$dA=\sqrt{f_x^2+f_y^2+1}\,dx\,dy.$$ This makes intuitive sense as the area of an infinitesimal rectangle in the domain $D$ is $dx\,dy$, and the factor $\sqrt{f_x^2+f_y^2+1}$ describes exactly how much $f$ stretches that rectangle to get $dA$. In your particular problem we have $$D=[1,5]\times[1,4],\quad z=f(x,y)=\sqrt{2xy},\quad\text{and}\quad\sigma(x,y,z)=6z=6\sqrt{2xy},$$ which will yield $$M=\int_1^4\int_1^56(x+y)\,dx\,dy.$$ Verify that this integral is correct, then evaluate it for your answer.