Evauate Surface integral

[boneill3]

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

[Dick]

That actually looks pretty good. Except where did you get that wacky (-1) factor? You are integrating a nonnegative function, x^2+y^2. The answer had better be nonnegative.

[boneill3]

Thanks alot!
I don't think I was concentrating enough.
It is very easy to get lost doing some of these integrals.