- #1
Froskoy
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Homework Statement
Let [itex]E[/itex] be the ellipsoid
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}+z^2=1[/tex]
where [itex]a>\sqrt{2}[/itex] and [itex]b>\sqrt{2}[/itex]. Let S be the part of the surface of [itex]E[/itex] defined by [itex]0\le x\le1, 0\le y\le1, z>0[/itex] and let [itex]\mathbf{F}[/itex] be the vector field defined by [itex]\mathbf{F}=(-y,x,0)[/itex]. Given that the surface area element of [itex]S[/itex] is given by
[tex]d \mathbf{S} = \left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 } \right) dxdy [/tex]
find [itex]\int_S\mathbf{F}.d\mathbf{S}[/itex] in the case [itex]a/ne b[/itex]
Homework Equations
Scalar product
The Attempt at a Solution
[tex]
\int_S\mathbf{F}.d\mathbf{S}=\int_S(-y,x,0).\left({ \frac{x}{a^2z}, \frac{y}{b^2z}, 1 }\right) dxdy
=\int_{y=0}^1\int_{x=0}^1\frac{-xy}{a^2z}+\frac{xy}{b^2z}dxdy
=\left({\frac{1}{b^2}-\frac{1}{a^2}}\right)\int_{y=0}^1\int_{x=0}^1\frac{xy}{z}dxdy
[/tex]
It's at this point I'm not sure what to do with the parameter [itex]z[/itex]. I tried continuing, treating [itex]z[/itex] as constant to get
[tex]
\int_S \mathbf{F}.d\mathbf{S}= \frac{1}{4z} \left({ \frac{1}{b^2}- \frac{1}{a^2}}\right)
[/tex]
but don't like the fact there is a [itex]z[/itex] there? Would converting to spherical coordinates help? If so, how would you do it?
With very many thanks,
Froskoy.
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