Cone: Volume and Surface Area

With V = ⅓·πr²·h: base area πr² times height, one third of it. Example: r = 3 cm and h = 4 cm give V = ⅓·π·9·4 = 12π ≈ 37.7 cm³. The factor ⅓ is the most common stumbling block; without it you compute the cylinder, which holds three times as much. It is also important to use the perpendicular height h and not the slanted line s. If a task only gives s, compute h first via the Pythagorean theorem: h = √(s² − r²). The unit is cubic; for filling amounts again 1000 cm³ = 1 litre.

The height h is perpendicular: it runs from the apex straight down to the centre of the base. The slant height s, in contrast, runs along the outside of the cone, from the apex to the edge of the base circle. Both are connected by the Pythagorean theorem, since r, h and s form a right triangle: s = √(r² + h²). Example: r = 3 cm, h = 4 cm gives s = 5 cm; s is always longer than h. Remember the assignment: the volume needs h (V = ⅓πr²h), the lateral surface needs s (M = πrs). If they are swapped, volume or lateral surface are systematically wrong.

With M = π·r·s, where s is the slant height. If s is not given, first compute s = √(r² + h²). Example: r = 6 cm, h = 8 cm gives s = √(36 + 64) = 10 cm and M = π·6·10 = 60π ≈ 188.5 cm². The formula becomes intuitive if you cut the lateral surface open and roll it out flat: a circular sector with radius s appears whose arc length is exactly the circumference 2πr of the base circle; its share of πs² yields πrs. For the total surface the base circle is added: O = πr² + πrs. For an ice cream cone without a base, M alone suffices.

Because a cone holds exactly one third of the cylinder with equal base and height. You can see it experimentally by pouring: a cone filled with water must be poured three times into the matching cylinder. It was rigorously proven in antiquity by Eudoxus with the method of exhaustion; the cone is the limiting case of a pyramid, and every prism splits into three pyramids of equal volume. With upper-school tools you verify it via the volume of revolution: the line f(x) = (r/h)·x rotates over [0; h], and π·∫(r/h)²x² dx = π·r²/h²·h³/3 = ⅓πr²h. Three routes, one factor.

By rearranging V = ⅓πr²h. For the height: h = 3V/(πr²); the factor 3 compensates the third. Example: V = 100 cm³ and r = 5 cm give h = 300/(π·25) = 300/78.54 ≈ 3.82 cm. For the radius: r = √(3V/(πh)), with a square root, because r appears squared in the formula. Finally check by substituting into the original formula: ⅓·π·25·3.82 ≈ 100 ✓. The typical mistake is forgetting the factor 3 or putting it on the wrong side. In composite tasks (cone on cylinder) rearrange each partial formula separately.