How much is the curved surface area (in square units) of a cone with a height…

Concept

The curved surface area (CSA) of a right circular cone is CSA = \(\pi r l\), where \(l\) is the slant height, \(l = \sqrt{r^2 + h^2}\). Its volume is \(V = \frac{1}{3}\pi r^2 h\). Both quantities are built only from the radius r and height h of the cone.

Application

1. The height is twice the base radius: \(h = 2r\). Substitute into the volume formula: \(V = \frac{1}{3}\pi r^2 (2r) = \frac{2}{3}\pi r^3\).

2. Solve this equation for r: \(r^3 = \frac{3V}{2\pi}\), so \(r = \left(\frac{3V}{2\pi}\right)^{\frac{1}{3}}\).

3. Find the slant height using \(h = 2r\): \(l = \sqrt{r^2 + h^2} = \sqrt{r^2 + 4r^2} = \sqrt{5r^2} = r\sqrt{5}\).

4. Substitute r and l into the CSA formula: \(CSA = \pi r l = \pi r (r\sqrt{5}) = \pi\sqrt{5}\,r^2\).

5. Write \(r^2\) in terms of V by squaring the result from step 2: \(r^2 = (r^3)^{\frac{2}{3}} = \left(\frac{3V}{2\pi}\right)^{\frac{2}{3}}\). So \(CSA = \pi\sqrt{5}\left(\frac{3V}{2\pi}\right)^{\frac{2}{3}}\).

Cross-check

Test with \(r = 1\) and \(h = 2\) (which satisfies \(h = 2r\)). Then \(V = \frac{1}{3}\pi(1)^2(2) = \frac{2}{3}\pi\), and directly \(l = \sqrt{1+4} = \sqrt{5}\), so \(CSA = \pi(1)(\sqrt{5}) = \pi\sqrt{5}\). Evaluating the derived formula at this V gives \(\left(\frac{3V}{2\pi}\right)^{\frac{2}{3}} = \left(\frac{3\cdot(2\pi/3)}{2\pi}\right)^{\frac{2}{3}} = 1^{\frac{2}{3}} = 1\), so \(CSA = \pi\sqrt{5}\times 1 = \pi\sqrt{5}\) — matching the direct computation.

So the curved surface area is \(CSA = \pi\sqrt{5}\left(\frac{3V}{2\pi}\right)^{\frac{2}{3}}\) square units.