I can't seem to figure out the bounds of a given surface. The problem statement is to approximate the mass of the homogeneous lamina that has the shape of given surface S and density function: $$S = 8-x-4y, z\ge 0, x\ge 0, y\ge 0; \rho = x$$
I found $\vec{r}(u,v)=<u, v, 8-u-4v>$. When I sketched out the surface (which is a plane), I seemed to have found the bounds $0 \le u \le 8$ and $ 0 \le v \le 8-\frac{x}{4}$.
Finding the cross product of the partial derivatives in respect to $u$ and $v$ I get the vector $<1, 4, 1>$ and it's magnitude is $\sqrt{18}$.
So, $\rho(\vec{r}(u,v)) = u$ and $|\vec{r}_u \times \vec{r}_v| = \sqrt{18}$.
Setting up my double integral, $$\int_{0}^{8}\int_{0}^{8-\frac{x}{4}}u\sqrt{18}dvdu$$
After simplifying, I get $\frac{640}{3}\sqrt{18}$ which is obviously wrong.
Did I go setting up the bounds of the surface incorrectly, or am I just applying the integration wrong?
Just for completeness' sake, from the comments we clarified that the surface of integration is the portion of that plane located in the first octant only. All of your work up to this point is correct, the only problem is the final bound where
$$y = \frac{8-x}{4} = 2 - \frac{x}{4} \neq 8 - \frac{x}{4}$$
which gives us the integral
$$\sqrt{18}\int_0^8\int_0^{2-\frac{u}{4}}u\:dv\:du = 3\sqrt{2}\int_0^82u-\frac{u^2}{4}\:du = \boxed{64\sqrt{2}}$$