## Surface integral z^2=2xy

Apostol page 429, problem 4

Is there a better way to set up this problem or have I made a mistake along the way?
(ie easier to integrate by different parameterization)

1. The problem statement, all variables and given/known data
Find the surface area of the surface $\displaystyle z^2=2xy$ lying above the $\displaystyle xy$ plane and bounded by $\displaystyle x=2$ and $\displaystyle y=1$.

2. Relevant equations
$\displaystyle S=r(T) =\bigg( X(x,y),Y(x,y),Z(x,y) \bigg) =\bigg( x,y,\sqrt{2xy} \bigg)$
$\displaystyle \frac{\partial r}{\partial x}=(1,0,\frac{\sqrt{2y}}{2\sqrt{x}})$
$\displaystyle \frac{\partial r}{\partial y}=(0,1,\frac{\sqrt{2x}}{2\sqrt{y}})$
$\displaystyle \frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y} =\bigg( -\frac{\sqrt{2y}}{2\sqrt{x}},-\frac{\sqrt{2x}}{2\sqrt{y}},1 \bigg)$
$\displaystyle \left|\left|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial t}\right|\right| =\sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}$
$\displaystyle a(S)=\int_0^1 \int_0^2 \sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}\,dx\,dy$