Find the area of the portion of the sphere $ x^2 + y^2 + z^2 = 1$ between the two parallel planes $ z = a$ and $z = b$ where $-1 < a < b < 1$ are parameters.
How to solve this question using surface integral ? i got $0$ when parametrizing the sphere but it can't be right
Best Answer
So between the planes $z=a$ and $z=b$ we have a portion of a sphere. $z=a$ lies below $z=b$ from the fact $-1 < a < b < 1$. Now, we aren't dealing with any sections of the portion of the sphere cut away. Another way of explaining is if you cut through any section of the shape, every cross section would be a circle. So we have $0\le \varphi \le 2\pi$
Now, $\theta$ measures the angle from the positive $z$-axis down to the surface of the sphere at a particular $z$ value.
$$\theta = \arccos\left(\frac{z}{r}\right)$$
For this sphere, $r=1$, so $\theta = \arccos(z)$
For $z=a$ and $z=b$, we have $\theta = \arccos(a)$ and $\theta = \arccos(b)$ respectively
Since the plane $z=b$ lies above the plane $z=a$, the $\theta$ angle to the plane $z=b$ will be smaller than that for the plane $z=a$ so we have $$\arccos(b)\le \theta \le \arccos(a)$$
The Jacobian for spherical coordinates is $r^2 \sin(\theta) = \sin(\theta)$ since $r=1$
The surface integral is therefore $$\int_{\theta = \arccos(b)}^{\theta = \arccos(a)} \int_{\varphi = 0}^{\varphi=2\pi} \sin(\theta) \ \mathrm d \varphi \mathrm d \theta$$