Let’s start with a truncated cone with radii [math]R[/math] (bottom base) and [math]r[/math] (top base), height [math]h[/math], and slant height l. We’ll extend it to a cone with base [math]R[/math] and slant height [math]L[/math].
We then open up this cone into a circular sector. The lateral surface area of truncated cone is equal to the area of larger sector (with radius [math]L[/math]) minus the area of smaller sector (with radius [math]L-l[/math]).
Note that arc length of sector is the circumference of cone’s base.
Bases have area [math]=[/math]
[math]\qquad \pi R^2 + \pi r^2 = \boldsymbol{\pi (R^2+r^2)}[/math]
To find slant height [math]l[/math] in terms of height and radii of bases, drop perpendicular from perimeter of top base to bottom base. This gives us a right triangle with height [math]h[/math], base [math]= R-r[/math], and hypotenuse [math]= l[/math]. Using Pythagorean theorem, we get:
[math]\qquad l^2 = h^2 + (R-r)^2[/math]
[math]\qquad l = \sqrt{h^2 + (R-r)^2}[/math]
For central angle [math]\theta[/math] (measured in radians) we have:
[math]\qquad \text{arc length} = \text{radius} \times \theta[/math]
[math]\qquad \theta = \dfrac{\text{arc length}}{\text{radius}}[/math]
[math]\qquad \theta = \dfrac{2\pi r}{L-l} = \dfrac{2\pi R}{L} \implies L = \dfrac{Rl}{R-r}[/math]
[math]\qquad \dfrac{\theta}{2\pi} = \dfrac{r}{L-l} = \dfrac{r}{\frac{Rl}{R-r}-l} = \dfrac{R-r}{l}[/math]
[math]\begin{align*} \text{Lateral area} & = \text{Area(larger sector)} - \text{Area(smaller sector)} \\ & = \frac{\theta}{2\pi}\cdot\pi L^2 - \frac{\theta}{2\pi}\cdot\pi (L-l)^2 \\ & = \frac{R-r}{l}\cdot\pi \left(L^2-(L-l)^2\right) \\ & = \frac{R-r}{l}\cdot\pi l\left(2L-l\right) \\ & = \pi(R-r)\left(\frac{2Rl}{R-r}-l\right) \\ & = \pi(R+r)l \\ & = \boldsymbol{\pi(R+r)\sqrt{h^2+(R-r)^2}} \end{align*}[/math]
Total surface area :
[math]\qquad S.A = \boxed{\boldsymbol{\pi (R^2+r^2) + \pi(R+r)\sqrt{h^2+(R-r)^2}}}[/math]
We can also calculate surface area using calculus.
By rotating top line (in diagram above) about the [math]x[/math]-axis, we can form a truncated cone.
To find surface of revolution, we use the following formula:
[math]\qquad\displaystyle V = 2\pi \int_{a}^{b} y \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx[/math]
This line rotated about [math]x[/math]-axis to form truncated cone passes through points [math](0,r)[/math] and [math](h,R)[/math] and has equation:
[math]\qquad y = r + \frac{R-r}{h} \; x \qquad 0 \le x \le h[/math]
[math]\qquad \frac{dy}{dx} = \frac{R-r}{h}[/math]
[math]\qquad 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{(R-r)^2}{h^2}[/math]
Lateral surface area:
[math]\begin{align*} L.A. & = 2\pi \int_{0}^{h} \left(r + \frac{R-r}{h} \; x\right) \sqrt{\frac{h^2+(R-r)^2}{h^2}} \, dx \\[1ex] & = \frac{2\pi}{h} \sqrt{h^2+(R-r)^2} \int_{0}^{h} \left(r + \frac{R-r}{h} \; x\right) \, dx \\[1ex] & = \frac{2\pi}{h} \sqrt{h^2+(R-r)^2} \left.\left(rx + \frac{R-r}{2h} \; x^2\right)\right|_{0}^{h} \\[1ex] & = \frac{2\pi}{h} \sqrt{h^2+(R-r)^2} \left(rh + \frac{R-r}{2h} \; h^2\right) \\[1ex] & = \pi (R+r) \sqrt{h^2+(R-r)^2} \end{align*}[/math]