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Surface integral z^2=2xy 
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#1
Apr3010, 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] 


#2
Apr3010, 10:09 AM

P: 9

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



#3
May110, 12:04 AM

HW Helper
Thanks
PF Gold
P: 7,575

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. 


#4
May110, 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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