Surface integral z^2=2xy


by twotwelve
Tags: area, integral, parametrization, projection, surface integral
twotwelve
twotwelve is offline
#1
Apr30-10, 08:15 AM
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 [tex]z^2=2xy[/tex] lying above the [tex]xy[/tex] plane and bounded by [tex]x=2[/tex] and [tex]y=1[/tex].

2. Relevant equations
[tex]
S=r(T)
=\bigg(
X(x,y),Y(x,y),Z(x,y)
\bigg)
=\bigg(
x,y,\sqrt{2xy}
\bigg)
[/tex]
[tex]
\frac{\partial r}{\partial x}=(1,0,\frac{\sqrt{2y}}{2\sqrt{x}})
[/tex]
[tex]
\frac{\partial r}{\partial y}=(0,1,\frac{\sqrt{2x}}{2\sqrt{y}})
[/tex]
[tex]
\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)
[/tex]
[tex]
\left|\left|\frac{\partial r}{\partial x}\times\frac{\partial r}{\partial t}\right|\right|
=\sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}
[/tex]
[tex]
a(S)=\int_0^1 \int_0^2 \sqrt{1+\frac{2y}{4x}+\frac{2x}{4y}}\,dx\,dy
[/tex]
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twotwelve
twotwelve is offline
#2
Apr30-10, 10:09 AM
P: 9
Did I perhaps set something up wrong? Could this be parameterized somehow as an Elliptic Paraboloid?
LCKurtz
LCKurtz is offline
#3
May1-10, 12:04 AM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,177
You are on the right track. It's actually easy to continue. Add the terms under the radical and rewrite is as

[tex]\frac 1 {\sqrt 2}\sqrt{\frac {(x+y)^2}{xy}}[/tex]

and take the root in the numerator.

twotwelve
twotwelve is offline
#4
May1-10, 06:47 AM
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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