Solid Geometry – Problem 24: Find: (1) the slant height;

A right circular frustum has bottom radius $R=2$, top radius $r=1$, and height $h=3$. Find:

(1) the slant height;

(2) the volume;

(3) the total surface area.

Use a coordinate/section model with the frustum axis as the $z$-axis. A meridian cross-section is an isosceles trapezoid with height $h=3$ and half-widths $R=2$ and $r=1$.

(1) The slant height $s$ is the length of the trapezoid’s leg: $$s=\sqrt{h^2+(R-r)^2}=\sqrt{3^2+(2-1)^2}=\sqrt{10}.$$

(2) The frustum volume is $$V=\frac13\pi h\,(R^2+Rr+r^2)=\frac13\pi\cdot 3\,(4+2+1)=7\pi.$$

(3) The lateral area is $$S_{\text{lat}}=\pi(R+r)s=\pi(2+1)\sqrt{10}=3\pi\sqrt{10}.$$ Add the two base areas: $$S_{\text{bases}}=\pi R^2+\pi r^2=\pi(4+1)=5\pi.$$ So the total surface area is $$S_{\text{tot}}=S_{\text{lat}}+S_{\text{bases}}=3\pi\sqrt{10}+5\pi=\pi(5+3\sqrt{10}).$$

Slant height $s=\sqrt{10}$, volume $V=7\pi$, total surface area $S_{\text{tot}}=\pi(5+3\sqrt{10})$.