Apostol page 429, problem 4(adsbygoogle = window.adsbygoogle || []).push({});

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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# Homework Help: Surface integral z^2=2xy

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