Calculating surface area of $x^2 + y^2 + z^2 = 4$, $z \geq 1$ I'm asked to calculate the surface area of 
$$D : x^2 + y^2 + z^2 = 4 , \quad z \geq 1.$$
My attempt
Let 
$$x=2\sin{\theta}\cos{\phi}$$
$$y=2\sin{\theta}\sin{\phi}$$
$$z=2\cos{\theta}$$
where 
$$ 0 \leq \theta \leq \frac{\pi}{3}$$ 
$$0\leq\phi\leq2\pi.$$
I realize the normal vector to the surface is $(x,y,z)$ which has length $\sqrt{x^2 + y^2 + z^2} = 2$.
Surface area is calculated by integrating over the area D, with the length of the normal vector as the integrand. But since I changed the surface D to the surface given by the spherical coordinate system (call this surface E), I need to add a factor to compensate (Jacobi Determinant).  
But since I've got a variable substitution with 3 functions and 2 variables, this won't be a square matrix so I won't be able to take the determinant of it. 
What am I doing wrong? 
 A: The simpler solution is to note that the searched area is a spherical cap with height $h=1$, and as bsis a circle of radius  $a=\sqrt{3}$ in a sphere of radius $r=2$. So its surface is given by the formula:
$$
S=2 \pi r h=\pi(a^2+h^2)= 4\pi
$$
Note that this formula can be proved in an ''elementary'' way, as you  can see here:Archimedes' derivation of the spherical cap area formula.
If you  want to use the integral calculus, than the surface element in spherical coordinates is $ dS=r^2\sin \theta d\theta d\varphi$, so the integral is:
$$
\int_0^{2\pi}\int_0^{\pi/3}4 \sin \theta d\theta d \varphi
$$
where $\pi/3$ is the angle between the $z$ axis and the radius of the sphere at the basis of the cap. And you can verify that  this gives the same result.
A: The short answer is that $z$ is a dependent variable and $x$ and $y$ are independent variables.  Therefore, the Jacobian must be between $x$ and $y$ and $\phi$ and $\theta$, without $z$.
This is similar to  what the OP suggested, just set up in a different way.  Consider the function:
$$
z=f(x,y)=\sqrt{4-x^2-y^2}.
$$
We can see, directly, that $z\geq 1$ means that $4-x^2-y^2\geq 1$ or that $3\geq x^2+y^2$.  This describes the disk centered at the origin of radius $\sqrt{3}$.  Let $D$ be the disk of radius $\sqrt{3}$.
The surface area formula is
$$
\iint_D\sqrt{f_x(x,y)^2+f_y(x,y)^2+1}dA.
$$
In this case,
\begin{align}
f_x&=\frac{-x}{\sqrt{4-x^2-y^2}}&f_y&=\frac{-y}{\sqrt{4-x^2-y^2}}.
\end{align}
Therefore, the surface area integral becomes
$$
\iint_D\sqrt{\frac{x^2}{4-x^2-y^2}+\frac{y^2}{4-x^2-y^2}+1}dA
=
\iint_D\sqrt{\frac{4}{4-x^2-y^2}}dA=\iint_D\frac{2}{\sqrt{4-x^2-y^2}}dA.
$$
Here's the $2$ that you found in your computation.
Now, let's switch to polar, the disk $D$ is defined by $0\leq\theta\leq 2\pi$ and $0\leq r\leq \sqrt{3}$, so, by switching to polar, we have
$$
\int_0^{2\pi}\int_0^{\sqrt{3}}\frac{2r}{\sqrt{4-r^2}}drd\theta=\int_0^{\sqrt{3}}\frac{4\pi r}{\sqrt{4-r^2}}dr
$$
Now, use the $u$-substitution $u=4-r^2$ to get your final answer.
If you were to compute the determinant that you need to convert $x$ and $y$ to $\theta$ and $r$ (remember that $z$ is a dependent variable in your setup), you should get the same formula.  
With spherical coordinates, as you set up, you'll get a somewhat more complicated formula by substituting in your formulae for $x$ and $y$ in terms of $\phi$ and $\theta$ and taking the determinant of the change of variables from $x$ and $y$ to $\phi$ and $\theta$.  The integral that you should get is
$$
\int_0^{\pi/3}\int_0^{2\pi}\frac{8|\cos(\theta)\sin(\theta)|}{\sqrt{4-4\sin^2(\theta)}}d\phi d\theta=
\int_0^{\pi/3}\frac{16\pi|\cos(\theta)\sin(\theta)|}{\sqrt{4-4\sin^2(\theta)}}d\phi
$$
where the absolute value of the determinant of the Jacobian is $4|\cos(\theta)\sin(\theta)|$.
A: Consider the riemannian metric $$g = dx\otimes dx+dy\otimes dy+dz\otimes dz$$ on $\mathbb{R}^3$. Spherical coordinates are given by the $$F(r,\theta,\phi) = (r\sin(\phi)\cos(\theta),r\sin(\phi)\sin(\theta),r\cos(\phi))$$ Applying the pullback to the tensor and noting $r=2$, we obtain the riemannian metric for the sphere $$F^*g = \begin{bmatrix} 4 & 0 \\ 0 & 4\sin^2(\phi) \end{bmatrix}$$ The riemannian volume form (well its area since we have 2-dimensions) is $$dS = \sqrt{\det(F^*g)}d\phi \wedge d\theta$$ Integrating we see that the surface area of the portion of the sphere is $$SA = \int_0^{2\pi}\int_0^{\pi/3}4\sin(\phi)d\phi \wedge d\theta $$
