# Surface integral z^2=2xy

• Apr 30th 2010, 06:25 AM
onefour
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 $z^2=2xy$ 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
$