Very nice.
All this so far is very nice, simple and elegant, but I want something that is a complete opposite of that.
I will now demonstrate how to compute the area of a sphere in an overly complicated way just because I want to be a cool kid who uses new fancy libraries for sport.
Here, we are going to use jax and its automatic differentiation capabilities to extract the gradients of vectors normal to the sphere in order to compute the following integral, and thus estimate the surface area:
$$
A = \int_x \int_y \sqrt{\Big(\frac{\partial z}{\partial x}\Big)^2 + \Big(\frac{\partial z}{\partial y}\Big)^2 + 1} dx dy
$$
What is this integral anyway?
Well, we first have to set one of the coordinates, in this case I have chosen $z$, to be a function of other two:
$$
z = z(x, y)
$$
Since we talk about a sphere here, $z$, as a function of $x$ and $y$, is defined as:
$$
z = z(x, y) = \pm \sqrt{r - x^2 - y^2}
$$
where $r$, as always, stands for the radius.
This parametrization yields the definition of the position vector $\vec r$, shown in Fig. 2, written as follows:
$$
\vec r = \vec r(x, y) = x \vec i + y \vec j + z(x, y) \vec k
$$
Figure 2. The position vector connects each point of a sphere to the ball origin. Image source: https://mathworld.wolfram.com/SphericalCoordinates.html
The magnitude of the cross product of the partial derivatives of $\vec r(x, y)$ yields the vector surface element, $\vec{dA}$:
$$
\vec{dA} = \Big|\frac{\partial \vec r(x, y)}{\partial x} \times \frac{\partial \vec r(x, y)}{\partial y}\Big|
$$
By following the defintion of the position vector, the partial derivatives with respect to $x$ and $y$ are respectively written as:
$$
\frac{\partial \vec r}{\partial x} = \vec i + \frac{\partial z(x,y)}{\partial x} \vec k
$$
and
$$
\frac{\partial \vec r}{\partial y} = \vec j + \frac{\partial z(x,y)}{\partial y} \vec k
$$
From here, it follows that
$$
\vec{dA} = \Big|\Big(\vec i + \frac{\partial z(x,y)}{\partial x} \vec k \Big) \times \Big(\vec j + \frac{\partial z(x,y)}{\partial y} \vec k\Big)\Big| = \Big|\Big(-\frac{\partial z(x, y)}{\partial x}, -\frac{\partial z(x, y)}{\partial x}, 1\Big)\Big|
$$
Finally, once we define the magnitude of this expression:
$$
|\vec{dA}| = \sqrt{\Big(\frac{\partial z}{\partial x}\Big)^2 + \Big(\frac{\partial z}{\partial y}\Big)^2 + 1}
$$
we finally have the reasoning behind the above mentioned surface integral.
The expression for $|\vec{dA}|$ can be separated in a following way:
$$
|\vec{dA}| = dA \cdot \vec n
$$
where $\vec n$ is the vector normal to the parameterized surface element $dA = dxdy$.
Okay, so entire outlined math here basically means that if we can somehow are able to compute gradients of the normal vector in each point on our parameterized surface, we can compute the overall surface area. The parametrization of a sphere is a bit tricky for a computer to handle it because it includes square roots, so instead parametrization of an entire sphere, we are going to parametrize only a small positive portion and scale the final area estimate accordingly. Let's begin.
