Surface Area of Cone (And Solid of Revolution in General) 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?

The surface area of a cone is the base surface area (πr^2) and the slanted surface put together. The slanted area flattened looks like an iscosele triangle with a curved base with the identical edges. The straight edges is the slanted height, and the curved edge is the circumference of the base of the cone (2πr). It would be quite hard to work out the slanted area, but that is where the slanted height comes in. Hope this helps! I'm the future, I may include images if you wish.