Could Surface Integration of z^2=2xy Be Simplified?

[twotwelve]

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)

Homework Statement

Find the surface area of the surface $$z^2=2xy$$ lying above the $$xy$$ plane and bounded by $$x=2$$ and $$y=1$$.

Homework Equations

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

 

[twotwelve]

Did I perhaps set something up wrong? Could this be parameterized somehow as an Elliptic Paraboloid?

 

[LCKurtz]

You are on the right track. It's actually easy to continue. Add the terms under the radical and rewrite is as

$$\frac 1 {\sqrt 2}\sqrt{\frac {(x+y)^2}{xy}}$$

and take the root in the numerator.

 

[twotwelve]

Yes, thank you. I was actually inquiring if anyone could find a better way to represent the surface area.