Surface area of a sphere above the plane $z=1$ The question is: Find the surface area of the part of the sphere $x^2+y^2+z^2=4$ that lies above the plane $z=1$. I got $4\pi(\sqrt3-1)$ but the answer key says $4\pi(\sqrt2-1)$. Am I doing something wrong or is the answer key wrong?
Thanks in advance.
Edit: This is my work thus far:
We can write the surface in terms of z as $z = \sqrt{4-x^2-y^2}$ (where we take the positive root since $z\geq1 \rightarrow z\geq0$), which can then be parameterized to $\mathbf r(x,y) = <x,y,\sqrt{4-x^2-y^2}>$ with the condition that $x^2+y^2\leq3$ (because $z$ ranges from 1 to 2). 
$\mathbf r_x = <1,0,\frac{-x}{\sqrt{4-x^2-y^2}}>$  
$\mathbf r_y = <0,1,\frac{-y}{\sqrt{4-x^2-y^2}}>$
$\mathbf r_x \times \mathbf r_y = <\frac{x}{\sqrt{4-x^2-y^2}},\frac{y}{\sqrt{4-x^2-y^2}},1>$
and $\| \mathbf{\mathbf r_x \times \mathbf r_y} \| = \frac{2}{\sqrt{4-x^2-y^2}}$
So the surface area $A$ equals:
$A = \iint_S1dS = \iint_R\frac{2}{\sqrt{4-x^2-y^2}}dA = \int_0^{2\pi} \int_0^{\sqrt3}\frac{2r}{\sqrt{4-r^2}}drd\theta$ 
$= \int_0^{2\pi}[-2\sqrt{4-r^2}]_1^\sqrt3d\theta$
$= \int_0^{2\pi}2(\sqrt3-1)d\theta$
= $4\pi(\sqrt3-1)$

For those who want to know why the answer is $4\pi$, let's do it!
First, plot the figure:

Since the goal is to find that area, the integral is just:
$$\iint_S dS,$$
with the appropriate limits. It will be easier if we use spherical coordinates:
$$x = 2 \sin \phi \cos \theta, \quad y = 2 \sin \phi \sin \theta, \quad z = 2 \cos \phi.$$
We know that $0\le \theta \le 2 \pi$, so we have to find the limits for $\phi$. How? That surface area is for $z \ge 1$, so let's find the angle $\phi$ for $z = 1$:
$$
\begin{align}
x^2 + y^2 &= 3, \\
4 \sin^2 \phi \, \cos^2 \theta + 4 \sin^2 \phi \, \sin^2 \theta  &= 3, \\
4 \sin^2 \phi &= 3, \\
\sin \phi &= \dfrac{\sqrt 3}{2}, \\
\phi &= \dfrac{\pi}{3}.
\end{align}
$$
Therefore, the limits for $\phi$ are $0$ and $\pi/3$. The integral is:
$$4 \int_0^{2\pi} \! \int_0^{\pi/3} \sin \phi \, d\phi \, d\theta = 4\pi.$$
