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Frustums and Surface Area

A frustum is what remains when you slice the top off a cone or pyramid with a cut parallel to the base — like a truncated cone. Water towers, buckets, and hopper bins are frustum shapes. The formula looks unusual at first but notice that $R^2 + Rr + r^2$ averages the three "sizes" of the cross-section — it reduces to the cone formula ($r=0$) and to a cylinder ($R=r$) as special cases. For the lateral (curved) surface area of a conical frustum, you need the slant height $l=\sqrt{h^2+(R-r)^2}$ — the distance measured along the slanted face. A prismatoid (any solid with parallel top and bottom faces) uses Simpson's Rule: $V=\frac{h}{6}(A_{\text{top}}+4A_{\text{mid}}+A_{\text{bot}})$.

$$V_{\text{cone frustum}}=\frac{\pi h}{3}(R^2+Rr+r^2), \quad l=\sqrt{h^2+(R-r)^2}$$
$$LSA_{\text{cone frustum}}=\pi(R+r)l \qquad V_{\text{prismatoid}}=\frac{h}{6}(A_1+4A_m+A_2)$$

Frustum Volume

Find the volume of a conical frustum with $R=5$ m, $r=2$ m, and $h=6$ m.

$$V=\frac{\pi(6)}{3}(5^2+5(2)+2^2)=78\pi=245.04\text{ m}^3$$

Final answer: $245.04\text{ m}^3$.

★ Lateral Surface Area of a Cone

A traffic cone has a base radius of 15 cm and a slant height of 40 cm. Find the lateral surface area of the cone.

$$LSA=\pi rl=\pi(15)(40)=600\pi=1884.96\text{ cm}^2$$

Final answer: 1884.96 cm².

★ Total Surface Area of a Closed Cylinder

A cylindrical water pipe (both ends closed) has radius 0.5 m and length 3 m. How many square meters of steel are needed to fabricate it?

Total surface area = lateral surface + 2 circular ends.

$$TSA=2\pi rh+2\pi r^2=2\pi(0.5)(3)+2\pi(0.5)^2=3\pi+0.5\pi=3.5\pi=10.996\text{ m}^2$$

Final answer: $10.996\text{ m}^2$ (approximately 11.0 m²).

★★ Frustum of a Square Pyramid

A hopper bin is shaped like a frustum of a square pyramid. The bottom is 1.5 m × 1.5 m, the top is 3 m × 3 m, and the height is 2 m. Find the volume.

For a prismatoid (or using the prismatoid/frustum of pyramid formula): $V=\frac{h}{3}(A_1+\sqrt{A_1 A_2}+A_2)$.

$A_1 = 1.5^2 = 2.25\text{ m}^2$, $A_2 = 3^2 = 9\text{ m}^2$, $h=2$ m.

$$V=\frac{2}{3}\left(2.25+\sqrt{2.25\times9}+9\right)=\frac{2}{3}(2.25+4.5+9)=\frac{2}{3}(15.75)=10.5\text{ m}^3$$

Final answer: $10.5\text{ m}^3$.

★★ Frustum: Finding the Slant Height and LSA

A conical frustum has lower radius $R=6$ m, upper radius $r=3$ m, and vertical height $h=4$ m. Find the slant height and the lateral surface area.

The slant height is the hypotenuse of a right triangle with legs $h$ and $(R-r)$.

$$l=\sqrt{h^2+(R-r)^2}=\sqrt{4^2+(6-3)^2}=\sqrt{16+9}=5\text{ m}$$
$$LSA=\pi(R+r)l=\pi(6+3)(5)=45\pi=141.37\text{ m}^2$$

Final answers: $l=5$ m, LSA = 141.37 m².

★★★ How Much Paint to Coat a Tank?

A closed cylindrical tank with radius 2 m and height 5 m needs two coats of anti-corrosion paint. One liter of paint covers 8 m². How many liters are needed?

Find the total external surface area first.

$$TSA=2\pi rh+2\pi r^2=2\pi(2)(5)+2\pi(2)^2=20\pi+8\pi=28\pi=87.96\text{ m}^2$$

Two coats means double the area equivalent:

$$\text{Liters needed}=\frac{2\times87.96}{8}=\frac{175.93}{8}=22.0\text{ liters}$$

Final answer: approximately 22 liters of paint.

★★★ Water Volume in a Frustum-Shaped Tank

A water reservoir is shaped like a conical frustum with lower radius 4 m, upper radius 7 m, and depth 3 m. The water depth is 2 m from the bottom. Find the volume of water.

At depth $y$ from the bottom, the radius varies linearly: $r(y)=R_{\text{bot}}+\frac{(R_{\text{top}}-R_{\text{bot}})}{H}\cdot y=4+\frac{3}{3}y=4+y$.

At $y=0$: $r=4$ m. At $y=2$ m: $r=6$ m. Use frustum formula with these values and $h=2$.

$$V=\frac{\pi(2)}{3}(4^2+4(6)+6^2)=\frac{2\pi}{3}(16+24+36)=\frac{2\pi}{3}(76)=\frac{152\pi}{3}=159.0\text{ m}^3$$

Final answer: $159.0\text{ m}^3$.

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