# Surface integral z^2=2xy

by twotwelve
Tags: area, integral, parametrization, projection, surface integral
 P: 9 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 $$z^2=2xy$$ lying above the $$xy$$ plane and bounded by $$x=2$$ and $$y=1$$. 2. Relevant equations $$S=r(T) =\bigg( X(x,y),Y(x,y),Z(x,y) \bigg) =\bigg( x,y,\sqrt{2xy} \bigg)$$ $$\frac{\partial r}{\partial x}=(1,0,\frac{\sqrt{2y}}{2\sqrt{x}})$$ $$\frac{\partial r}{\partial y}=(0,1,\frac{\sqrt{2x}}{2\sqrt{y}})$$ $$\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial y} =\bigg( -\frac{\sqrt{2y}}{2\sqrt{x}},-\frac{\sqrt{2x}}{2\sqrt{y}},1 \bigg)$$ $$\left|\left|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial t}\right|\right| =\sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}$$ $$a(S)=\int_0^1 \int_0^2 \sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}\,dx\,dy$$
 P: 9 Did I perhaps set something up wrong? Could this be parameterized somehow as an Elliptic Paraboloid?
 HW Helper Thanks PF Gold P: 6,992 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.
P: 9

## Surface integral z^2=2xy

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

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