# Evauate Surface integral

## Homework Statement

Evauate Surface integral
$\int\int_{\sigma}(x^2 + y^2)dS$
where $\sigma$ is the portion of the sphere $x^2 + y^2 + z^2 = 4$
above the plane z = 1.

## Homework Equations

$\int\int_{\sigma} f(x,y) \sqrt{\frac{\partial z}{\partial x}^2+\frac{\partial z}{\partial y}^2+1}$

## The Attempt at a Solution

The plane and the sphere intercect at z=1 and x2+y2=3.
so
$z=\sqrt{4-x^2-y^2}$

we have

$\frac{\partial z}{\partial x}= \frac{-x}{\sqrt{4-x^2-y^2}}$
and
$\frac{\partial z}{\partial y}= \frac{-y}{\sqrt{4-x^2-y^2}}$

Z gives a projection of a disk onto the xy plane of x2+y2$\leq 3$

Which is
$R: 0 \leq \theta \leq 2\pi$ $0\leq r \leq \sqrt{3}$ in polar co-ordinates

Therefore
$\int\int_{\sigma}(x^2 + y^2)dS$= $\int\int_{R}(x^2 + y^2)\sqrt{(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2+1}$

which equals

$\int\int_{R}(x^2 + y^2)dS$= $\int\int_{R}(x^2 + y^2)\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1}$

In polar co-ordinates
We have I think.....

$\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\left[(r^2)2\sqrt{\frac{-1}{r^2-4}} (r) \right]dr d\theta$

$\int_{0}^{2\pi}\frac{-10}{3}d\theta$
$= \frac{-20\pi}{3}$

Does this look alright ?
I'm not sure about if I've changed the $\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1}$ to polar co-ordinates correctly

## The Attempt at a Solution

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Dick