Surface Area of a surface of revolution

Consider a surface of revolution obtained by rotating the curve y = f(x) about the x-axis, for x from a to b.
Over a very small interval $\Delta(x)$ in x, it seems reasonable to approximate the surface by the frustum of a cone, with radius at one end f(x) and at the other $f(x) + f'(x) \Delta(x)$.
What is the surface area of such a frustum?
If a cone has edge length l (this is the length of a straight line segment on the cone starting at the vertex and ending at the base, perpendicular to the base) and radius at its base r, then if we snip it along a ray coming down from the vertex, perpendicular to the base and unwrap, we find the sector of a circle of radius l and angle $\frac{2\pi r}{l}$, so the area of the cone's curved surface is:

$$\frac{1}{2}l^2 \frac{2\pi r}{l} = \pi r l.$$

Now, let a frustum have inner radius r₁, outer radius r₂, and length l.
Let $R = \frac{r_2 - r_1}{2}$ and $r = \frac{r_1 + r_2}{2}$.
Then r₁ = r - R and r₂ = r + R.
Regard the frustum as part of a cone of total length l₂ with a piece of length l₁ removed.
So we have the relation: l = l₂ - l₁.
So by similar triangles, we have $\frac{r_1}{l_1} = \frac{r_2}{l_2}$.
So l₂(r-R) = l₁(r + R), or R(l₁ + l₂) = rl.
Then the area of the frustum is:

$$\pi r_2l_2 - \pi r_1l_1 = \pi(r + R)l_2 - \pi(r - R)l_1$$

$$= \pi(r(l_2 - l_1) + \pi R(l_1 + l_2) = 2\pi rl.$$

For our very small frustum, the average radius is approximately f(x) and the edge length l is $\Delta s$ the length along the curve.
So the area is approximately $2\pi f(x)\Delta s$.
Summing up such small frustums gives a Riemann sum.
In the limit as $\Delta(x)$ goes to zero, this gives us an integral formula for the area of a surface of revolution about the x-axis:

$$A = \int_{x =a}^{x = b} 2\pi y ds = 2\pi \int_{x =a}^{x = b} 2\pi y \sqrt{(dx)^2 + (dy)^2},$$

$$A = \int_a^b 2\pi f(x)\sqrt{1 + (\frac{df}{dx})^2} dx.$$

If the curve is given parametrically by a formula x = x(t), y = y(t), for t from t₀ to t₁, then the surface area is:

$$A = \int_{t_0}^{t_1} 2\pi y(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } dt.$$

If we rotate about the y-axis instead of the x-axis, then the role of the x and y variables is interchanged:

$$A = 2\pi \int_c^d xds = 2\pi \int_c^d x(y) \sqrt{1 + (\frac{dx}{dy})^2} dy,$$

$$A = \int_{t_0}^{t_1} 2\pi x(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 } dt.$$