Deriving the formula for the surface area of revolution involves approximating the curved surface with small segments and using calculus to sum these approximations. The core idea is to break down the surface into infinitesimally thin bands, each resembling a frustum of a cone.
The Fundamental Concept
Imagine rotating a curve $y = f(x)$ from $x=a$ to $x=b$ around the x-axis. This generates a 3D surface. To find its area, we:
- Divide the curve into many small segments.
- Rotate each small segment around the x-axis to create a thin band.
- Approximate the area of each band using the formula for the lateral surface area of a frustum of a cone.
- Sum the areas of all these bands.
- Take a limit as the number of segments approaches infinity (and their size approaches zero) to get an integral.
Approximating with Frustums
Consider a tiny segment of the curve with length $dL$. When this segment is rotated around the x-axis, it forms a narrow band. If the segment is small enough, this band is very close to the lateral surface of a frustum of a cone.
The lateral surface area of a frustum is given by $A = \pi (r_1 + r_2) L$, where $r_1$ and $r_2$ are the radii of the two bases and $L$ is the slant height.
For our small band formed by rotating a curve segment, the radii $r_1$ and $r_2$ are the y-values (distances from the x-axis) at the endpoints of the segment. As the segment becomes infinitesimally small, $r_1 \approx r_2 \approx y$, the radius of the curve at that point. The slant height $L$ becomes the arc length element $dL$.
So, the approximate area of a small band, $dS$, is:
$dS \approx \pi (y + y) dL = 2\pi y \, dL$
This represents the circumference of the band ($2\pi y$) multiplied by its width ($dL$).
Incorporating Arc Length
The arc length element $dL$ for a curve $y = f(x)$ is given by:
$dL = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
Substituting this into the approximate area element $dS$:
$dS = 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
As mentioned in the provided reference clip which references the formula including "Times DX", this differential form signifies the element of surface area ready for integration.
The Integral Formula
To find the total surface area of revolution, we sum up these infinitesimal area elements $dS$ along the curve from $x=a$ to $x=b$. This summation is done using a definite integral:
$S = \int{a}^{b} dS = \int{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
Substituting $y = f(x)$:
$S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx$
This is the standard formula for the surface area generated by revolving a curve $y=f(x)$ around the x-axis.
Variations of the Formula
The derivation changes slightly depending on the axis of revolution and how the curve is defined:
-
Revolving $x = g(y)$ around the y-axis:
$S = \int{c}^{d} 2\pi x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy = \int{c}^{d} 2\pi g(y) \sqrt{1 + \left(g'(y)\right)^2} \, dy$
(where the curve is from $y=c$ to $y=d$) -
Revolving $y = f(x)$ around the y-axis:
$S = \int_{a}^{b} 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$
(Here, the radius of rotation is the x-value) -
Revolving $x = g(y)$ around the x-axis:
$S = \int_{c}^{d} 2\pi y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy$
(Here, the radius of rotation is the y-value) -
Parametric Curve $x(t), y(t)$ from $t=\alpha$ to $t=\beta$:
- Around x-axis: $S = \int_{\alpha}^{\beta} 2\pi y(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
- Around y-axis: $S = \int_{\alpha}^{\beta} 2\pi x(t) \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$
In summary, the derivation relies on approximating infinitesimal surface elements as frustums and summing their areas via integration, with the differential element of area involving $2\pi \times (\text{radius of rotation}) \times (\text{arc length element})$. The reference highlighting "Times DX" points to the standard integral form derived from approximating with respect to the x-variable.
