- #1
twotwelve
- 9
- 0
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)
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].
[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]
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)
Homework Statement
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].
Homework 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]
Last edited: