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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 x

^{2}+y

^{2}=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 x

^{2}+y

^{2}[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