How to Use This Section

This is your practice bank for Surface Areas and Volumes. Keep the base-formula table in view, decide whether a problem is surface area (drop hidden faces), volume (add / subtract cavities), conversion (old volume = new volume), or frustum, and work each step with units. Use π=227\pi=\tfrac{22}{7} unless told 3.143.14.

Example 1: Volume of a sphere of radius 7 cm. Solution: 43×227×343=43123≈1437.33\tfrac43\times\tfrac{22}{7}\times343=\tfrac{4312}{3}\approx1437.33 cm3^3.

Example 2: TSA of a cylinder r=7r=7, h=3h=3 cm. Solution: 2πr(r+h)=2×227×7×10=4402\pi r(r+h)=2\times\tfrac{22}{7}\times7\times10=440 cm2^2.

Example 3: Two cubes of volume 64 cm3^3 each are joined end to end. Find the surface area of the resulting cuboid. Solution: edge =4=4 cm; cuboid 4×4×84\times4\times8; TSA =2(16+32+32)=160=2(16+32+32)=160 cm2^2. Answer: 160 cm2^2.

Example 4: A toy is a cone (radius 3.5 cm) on a hemisphere of the same radius; total height 15.5 cm. Find TSA. (π=227)(\pi=\tfrac{22}{7}) Solution: r=3.5r=3.5, cone h=15.5−3.5=12h=15.5-3.5=12, l=3.52+122=12.25+144=156.25=12.5l=\sqrt{3.5^2+12^2}=\sqrt{12.25+144}=\sqrt{156.25}=12.5; TSA =πrl+2πr2=πr(l+2r)=227×3.5×(12.5+7)=11×19.5=214.5=\pi r l+2\pi r^2=\pi r(l+2r)=\tfrac{22}{7}\times3.5\times(12.5+7)=11\times19.5=214.5 cm2^2. Answer: 214.5 cm2^2.

Example 5: A cubical block of side 7 cm is surmounted by the largest hemisphere. Find the surface area. (π=227)(\pi=\tfrac{22}{7}) Solution: r=3.5r=3.5; TSA =6a2+πr2=6×49+227×12.25=294+38.5=332.5=6a^2+\pi r^2=6\times49+\tfrac{22}{7}\times12.25=294+38.5=332.5 cm2^2. Answer: 332.5 cm2^2.

Example 6: A vessel is a hollow hemisphere (diameter 14 cm) topped by a hollow cylinder; total height 13 cm. Find the inner surface area. (π=227)(\pi=\tfrac{22}{7}) Solution: r=7r=7; cylinder height =13−7=6=13-7=6; inner SA =2πrh+2πr2=2πr(h+r)=2×227×7×13=572=2\pi rh+2\pi r^2=2\pi r(h+r)=2\times\tfrac{22}{7}\times7\times13=572 cm2^2. Answer: 572 cm2^2.

Example 7: From a solid cylinder (h=2.4h=2.4 cm, diameter 1.4 cm) a conical cavity of the same height and diameter is hollowed out. Find the TSA of the remaining solid. (π=227)(\pi=\tfrac{22}{7}) Solution: r=0.7r=0.7; l=0.72+2.42=0.49+5.76=6.25=2.5l=\sqrt{0.7^2+2.4^2}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5; TSA =2πrh+πrl+πr2=πr(2h+l+r)=227×0.7×(4.8+2.5+0.7)=2.2×8=17.6=2\pi rh+\pi rl+\pi r^2=\pi r(2h+l+r)=\tfrac{22}{7}\times0.7\times(4.8+2.5+0.7)=2.2\times8=17.6 cm2^2. Answer: ~17.6 cm2^2.

Example 8: A wooden article is a cylinder (h=10h=10 cm, radius 3.5 cm) with a hemisphere scooped from each end. Find the TSA. (π=227)(\pi=\tfrac{22}{7}) Solution: TSA =2πrh+2(2πr2)=2πr(h+2r)=2×227×3.5×(10+7)=22×17=374=2\pi rh+2(2\pi r^2)=2\pi r(h+2r)=2\times\tfrac{22}{7}\times3.5\times(10+7)=22\times17=374 cm2^2. Answer: 374 cm2^2.

Example 9: A solid is a cone on a hemisphere, both radius 1 cm, cone height 1 cm. Find the volume in terms of π\pi. Solution: 13π(1)2(1)+23π(1)3=π3+2π3=π\tfrac13\pi(1)^2(1)+\tfrac23\pi(1)^3=\tfrac{\pi}{3}+\tfrac{2\pi}{3}=\pi cm3^3. Answer: π\pi cm3^3.

Example 10: A model is a cylinder (length 12 cm total, diameter 3 cm) with a cone (height 2 cm) at each end. Find the volume of air. (π=227)(\pi=\tfrac{22}{7}) Solution: r=1.5r=1.5; cylinder length =12−2(2)=8=12-2(2)=8; V=πr2(8)+2×13πr2(2)=πr2(8+43)=227×2.25×283=66V=\pi r^2(8)+2\times\tfrac13\pi r^2(2)=\pi r^2(8+\tfrac43)=\tfrac{22}{7}\times2.25\times\tfrac{28}{3}=66 cm3^3. Answer: 66 cm3^3.

Example 11: 45 gulab jamuns (cylinder + two hemispheres, length 5 cm, diameter 2.8 cm) contain syrup up to 30% of volume. Find the syrup volume. (π=227)(\pi=\tfrac{22}{7}) Solution: r=1.4r=1.4; cylinder length =5−2.8=2.2=5-2.8=2.2; one V=πr2(2.2)+43πr3=227×1.96×2.2+43×227×2.744=13.55+11.50=25.05V=\pi r^2(2.2)+\tfrac43\pi r^3=\tfrac{22}{7}\times1.96\times2.2+\tfrac43\times\tfrac{22}{7}\times2.744=13.55+11.50=25.05 cm3^3; 45 of them ≈1127.3\approx1127.3; 30% ≈338\approx338 cm3^3. Answer: ~338 cm3^3.

Example 12: Volume of wood in a cuboid pen stand (15×10×3.515\times10\times3.5) with 4 conical depressions (r=0.5r=0.5, depth 1.4 cm). (π=227)(\pi=\tfrac{22}{7}) Solution: 525−4×13×227×0.25×1.4=525−1.47≈523.53525-4\times\tfrac13\times\tfrac{22}{7}\times0.25\times1.4=525-1.47\approx523.53 cm3^3. Answer: ~523.53 cm3^3.

Example 13: A sphere of radius 3 cm is melted into cones (radius 1 cm, height 3 cm). Number of cones? Solution: 36ππ=36\dfrac{36\pi}{\pi}=36. Answer: 36.

Example 14: 27 solid spheres of radius 1 cm are melted into one sphere. Its radius? Solution: R3=27⇒R=3R^3=27\Rightarrow R=3 cm. Answer: 3 cm.

Example 15: A cylinder of radius 6 cm, height 32 cm is recast into a sphere. Radius? Solution: 43πR3=1152π⇒R3=864⇒R≈9.52\tfrac43\pi R^3=1152\pi\Rightarrow R^3=864\Rightarrow R\approx9.52 cm. Answer: ~9.52 cm.

Example 16: A copper rod (diameter 1 cm, length 8 cm) is drawn into a wire of diameter 1 mm. Find the wire length. Solution: Rod V=π(0.5)2(8)=2πV=\pi(0.5)^2(8)=2\pi; wire radius 0.050.05 cm; π(0.05)2L=2π⇒L=20.0025=800\pi(0.05)^2 L=2\pi\Rightarrow L=\dfrac{2}{0.0025}=800 cm =8=8 m. Answer: 8 m.

Example 17: A cone of water (h = 8 cm, top radius 5 cm) loses one-fourth of its water to spherical shots of radius 0.5 cm. Number of shots? Solution: displaced =14×200π3=50π3=\tfrac14\times\tfrac{200\pi}{3}=\tfrac{50\pi}{3}; one shot =π6=\tfrac{\pi}{6}; number =100=100. Answer: 100.

Example 18: A frustum has R=20R=20, r=8r=8, h=16h=16 cm. Find slant height, CSA and volume. (π=3.14)(\pi=3.14) Solution: l=162+122=20l=\sqrt{16^2+12^2}=20; CSA =πl(R+r)=3.14×20×28=1758.4=\pi l(R+r)=3.14\times20\times28=1758.4 cm2^2; V=13×3.14×16×624=10449.92V=\tfrac13\times3.14\times16\times624=10449.92 cm3^3. Answer: 20 cm, 1758.4 cm2^2, 10449.92 cm3^3.

Example 19: A bucket (frustum, R=20R=20, r=8r=8, h=16h=16) capacity in litres. Solution: 10449.92÷1000≈10.4510449.92\div1000\approx10.45 L. Answer: ~10.45 L.

Example 20: A frustum glass: top radius 3 cm, bottom radius 2 cm, height 6 cm. Capacity? (π=3.14)(\pi=3.14) Solution: 13×3.14×6×(9+4+6)=119.32\tfrac13\times3.14\times6\times(9+4+6)=119.32 cm3^3. Answer: ~119.32 cm3^3.

Example 21: A solid iron pole is a cylinder (h=220h=220 cm, diameter 24 cm) with another cylinder (h=60h=60 cm, radius 8 cm) on top. Mass if 1 cm3^3 iron =8=8 g. (π=3.14)(\pi=3.14) Solution: V=3.14×144×220+3.14×64×60=99475.2+12057.6=111532.8V=3.14\times144\times220+3.14\times64\times60=99475.2+12057.6=111532.8 cm3^3; mass =111532.8×8≈892262=111532.8\times8\approx892262 g ≈892.26\approx892.26 kg. Answer: ~892.26 kg.

Example 22: A conical vessel of base radius 5 cm and height 24 cm is full of water, poured into a cylinder of radius 10 cm. Find the water height in the cylinder. (π cancels)(\pi\text{ cancels}) Solution: cone V=13π(25)(24)=200πV=\tfrac13\pi(25)(24)=200\pi; cylinder π(100)h=200π⇒h=2\pi(100)h=200\pi\Rightarrow h=2 cm. Answer: 2 cm.

Example 23: A hemispherical bowl of radius 6 cm is full of liquid poured into cylindrical bottles of radius 1.5 cm, height 4 cm. Number of bottles? Solution: bowl =23π(216)=144π=\tfrac23\pi(216)=144\pi; bottle =π(2.25)(4)=9π=\pi(2.25)(4)=9\pi; number =16=16. Answer: 16.

Example 24: Volume of the largest sphere carved from a cube of edge 7 cm. (π=227)(\pi=\tfrac{22}{7}) Solution: sphere diameter =7⇒r=3.5=7\Rightarrow r=3.5; V=43×227×42.875≈179.67V=\tfrac43\times\tfrac{22}{7}\times42.875\approx179.67 cm3^3. Answer: ~179.67 cm3^3.

Example 25: A cylinder of radius 5 cm, height 4 cm is melted with a cone (radius 5 cm, height 12 cm)… find combined volume. (π=3.14)(\pi=3.14) Solution: π(25)(4)+13π(25)(12)=100π+100π=200π=628\pi(25)(4)+\tfrac13\pi(25)(12)=100\pi+100\pi=200\pi=628 cm3^3. Answer: 628 cm3^3.

Example 26: A solid metallic sphere of radius 10.5 cm is melted into small cones of radius 3.5 cm, height 3 cm. Number? (π cancels)(\pi\text{ cancels}) Solution: sphere =43π(1157.625)=1543.5π=\tfrac43\pi(1157.625)=1543.5\pi; cone =13π(12.25)(3)=12.25π=\tfrac13\pi(12.25)(3)=12.25\pi; number =126=126. Answer: 126.

Example 27: A well of diameter 3 m is dug 14 m deep; the earth is spread as a ring (width 4 m) around it. Find the height of the ring. (π=227)(\pi=\tfrac{22}{7}) Solution: earth =π(1.5)2(14)=31.5π=\pi(1.5)^2(14)=31.5\pi; ring inner r=1.5r=1.5, outer R=5.5R=5.5; π(R2−r2)h=31.5π⇒(30.25−2.25)h=31.5⇒28h=31.5⇒h=1.125\pi(R^2-r^2)h=31.5\pi\Rightarrow(30.25-2.25)h=31.5\Rightarrow28h=31.5\Rightarrow h=1.125 m. Answer: 1.125 m.

Example 28: A sphere of diameter 6 cm is dropped into a cylinder of radius 6 cm containing water. Find the rise in water level. (π cancels)(\pi\text{ cancels}) Solution: sphere =43π(27)=36π=\tfrac43\pi(27)=36\pi; π(36)h=36π⇒h=1\pi(36)h=36\pi\Rightarrow h=1 cm. Answer: 1 cm.

Example 29: The radii of the two circular ends of a frustum bucket are 14 cm and 7 cm and its height is 30 cm. Find its capacity. (π=227)(\pi=\tfrac{22}{7}) Solution: V=13×227×30×(196+49+98)=13×227×30×343=227×10×343=22×34307=10780V=\tfrac13\times\tfrac{22}{7}\times30\times(196+49+98)=\tfrac13\times\tfrac{22}{7}\times30\times343=\tfrac{22}{7}\times10\times343=\tfrac{22\times3430}{7}=10780 cm3^3. Answer: 10780 cm3^3 (~10.78 L).

Example 30: A tent is a cylinder (radius 14 m, height 3 m) surmounted by a cone of the same base and height 4 m. Find the area of canvas. (π=227)(\pi=\tfrac{22}{7}) Solution:

  1. Cone slant l=142+42=196+16=212≈14.56l=\sqrt{14^2+4^2}=\sqrt{196+16}=\sqrt{212}\approx14.56 m.
  2. Canvas =2πrh+πrl=πr(2h+l)=227×14×(6+14.56)=44×20.56≈904.6=2\pi rh+\pi rl=\pi r(2h+l)=\tfrac{22}{7}\times14\times(6+14.56)=44\times20.56\approx904.6 m2^2.

Final Answer: ≈904.6\approx 904.6 m2^2.

Takeaway: Tent canvas == cylinder CSA ++ cone CSA; find the cone's slant height first.