# Surface area of a conical frustum?

I was reading about a conical frustum on Wikipedia, but on Wikipedia does not include the proof of surface area of a conical frustum.

How do we arrive at that formula?

Position the frustrum of the cone so that the bottom center is at the origin. Let the height be $h$, the top radius $r_1$ and the bottom radius $r_2$. Then the frustrum is formed by rotating the line $$y={h\over r_1-r_2}(x-r_2) \iff \underbrace{x={r_1-r_2\over h}\,y+r_2}_{\text{radius of slice at height }y}$$ about the $y$ axis. The surface area of a solid of revolution (when rotated about the $y$ axis) is given by \begin{align} S&=\int_{y_\min}^{y_\max} 2\pi\,r(y)\,\sqrt{1+[r'(y)]^2}\,dy, \quad r(y)=\text{ radius at height } y\\ &=\int_{0}^{h} 2\pi\,\left({r_1-r_2\over h}\,y+r_2\right)\,\sqrt{1+\left[{r_1-r_2\over h}\right]^2}\,dy, \quad r(y)=\text{ radius at height } y\\ &=\pi(r_1+r_2)\sqrt{(r_1-r_2)^2+h^2}. \end{align}
Using the method of disks to compute the volume, we see the volume of a slice at height $y$ of thickness $\Delta y$ is $$V_\text{slice}=\pi (\text{radius of slice})^2\Delta y=\pi \left[{r_1-r_2\over h}y+r_2\right]^2 \Delta y$$ so the total volume is $$V_\text{frustrum}=\int_0^h \pi \left[{r_1-r_2\over h}y+r_2\right]^2 dy={\pi h\over 3}\left[r_1^2+r_1r_2+r_2^2\right].$$