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boneill3
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Homework Statement
Evauate Surface integral
[itex]\int\int_{\sigma}(x^2 + y^2)dS[/itex]
where [itex]\sigma [/itex] is the portion of the sphere [itex]x^2 + y^2 + z^2 = 4 [/itex]
above the plane z = 1.
Homework Equations
[itex]
\int\int_{\sigma} f(x,y) \sqrt{\frac{\partial z}{\partial x}^2+\frac{\partial z}{\partial y}^2+1}
[/itex]
The Attempt at a Solution
The plane and the sphere intercect at z=1 and x2+y2=3.
so
[itex]z=\sqrt{4-x^2-y^2}[/itex]
we have
[itex]
\frac{\partial z}{\partial x}= \frac{-x}{\sqrt{4-x^2-y^2}} [/itex]
and
[itex]\frac{\partial z}{\partial y}= \frac{-y}{\sqrt{4-x^2-y^2}} [/itex]
Z gives a projection of a disk onto the xy plane of x2+y2[itex]\leq 3[/itex]
Which is
[itex] R: 0 \leq \theta \leq 2\pi [/itex] [itex] 0\leq r \leq \sqrt{3} [/itex] in polar co-ordinates
Therefore
[itex]\int\int_{\sigma}(x^2 + y^2)dS[/itex]= [itex]\int\int_{R}(x^2 + y^2)\sqrt{(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2+1} [/itex]
which equals
[itex]\int\int_{R}(x^2 + y^2)dS[/itex]= [itex]\int\int_{R}(x^2 + y^2)\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1} [/itex]
In polar co-ordinates
We have I think...
[itex]\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\left[(r^2)2\sqrt{\frac{-1}{r^2-4}} (r) \right]dr d\theta[/itex]
[itex]\int_{0}^{2\pi}\frac{-10}{3}d\theta[/itex]
[itex] = \frac{-20\pi}{3}[/itex]
Does this look alright ?
I'm not sure about if I've changed the [itex]\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1}[/itex] to polar co-ordinates correctly