# Surface Area of Cone (And Solid of Revolution in General) [duplicate]

From my understanding, a cone/frustum is a stack of circles with radius $r(h)$, $r$ being a linear function, and h the height.

Cone:

The surface area of a circle with radius $r$ (slice), by my understanding, should be $2πr$ $dh$.

Therefore, adding up this surface area should give us $2π\int_{H_1}^{H_2}r(h)$ $dh$, where $H_1$ and $H_2$ are the lower and upper heights (for a cone the lower bound is 0, for a frustum, >0 because the cone is truncated by a similar cone).

Apparently in the formula for the cone and general solids of revolution, the slant height is involved. How does this work?