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?
Best Answer
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.