How to Evaluate a Surface Integral on a Portion of a Sphere Above a Plane?

[boneill3]

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

 

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