Volumes Simply Add Up

Surface area was tricky because joined faces disappeared. Volume is easier. When two solids are joined, no volume is lost — the volume of the combined solid is exactly the sum of the volumes of its parts. Volume of combination=Volume of part 1+Volume of part 2+…\text{Volume of combination} = \text{Volume of part 1} + \text{Volume of part 2} + \dots

Key Point: For surface area you drop the hidden faces; for volume you just add everything. Different rules — do not mix them up.

Add for a Join, Subtract for a Cavity

If a solid has a hollow or a depression, subtract that empty volume: Actual volume (capacity)=Outer solid−hollowed-out part.\text{Actual volume (capacity)} = \text{Outer solid} - \text{hollowed-out part}.

  • A glass (cylinder) with a hemispherical raised bottom: actual capacity =πr2h−23πr3= \pi r^2 h - \dfrac23\pi r^3.
  • A pen stand (cuboid) with conical depressions: volume of wood =cuboid−(number of cones)×13πr2h= \text{cuboid} - (\text{number of cones})\times\dfrac13\pi r^2 h.

Key Point: "How much it holds" or "how much material is left" ⇒\Rightarrow subtract the empty/removed part.

Familiar Combinations

  • Capsule / gulab jamun (cylinder ++ two hemispheres): V=πr2h+2×23πr3=πr2h+43πr3V = \pi r^2 h + 2\times\dfrac23\pi r^3 = \pi r^2 h + \dfrac43\pi r^3, where h=length−2rh = \text{length} - 2r.
  • Toy (cone ++ hemisphere): V=13πr2h+23πr3V = \dfrac13\pi r^2 h + \dfrac23\pi r^3.
  • Model (cylinder ++ a cone at each end): V=πr2hcyl+2×13πr2hconeV = \pi r^2 h_{\text{cyl}} + 2\times\dfrac13\pi r^2 h_{\text{cone}}.
  • Shed (cuboid ++ half-cylinder roof): V=lbh+12πr2LV = lbh + \dfrac12\pi r^2 L.

[Board Important] Always read off which radius/height belongs to which part — combinations often reuse the letter rr or hh for different pieces.

A capsule made of a cylinder with a hemisphere joined at each end; the total length equals the cylinder length h plus two radii (h plus 2r).

Capacity vs Volume of Material

Be clear on what is being asked:

  • Capacity / volume of air / how much it holds == inner (empty) volume.
  • Volume of material / wood / metal == solid volume −- any hollow.

For a vessel with walls, use the inner radius for capacity. If the question gives outer dimensions and a thickness, capacity uses (outer −- thickness).

Key Point: Decide first: am I measuring the stuff (material) or the space (capacity)? Then add or subtract accordingly.

Solved Examples

Example 1: Glass with a hemispherical bottom

A cylindrical glass of inner diameter 5 cm and height 10 cm has a hemispherical raised bottom. Find its apparent and actual capacity. (π=3.14)\left(\pi=3.14\right)

Solution:

  1. r=2.5r = 2.5 cm. Apparent capacity =πr2h=3.14×6.25×10=196.25= \pi r^2 h = 3.14\times 6.25\times 10 = 196.25 cm3^3.
  2. Hemisphere volume =23πr3=23×3.14×15.625=32.71= \dfrac23\pi r^3 = \dfrac23\times 3.14\times 15.625 = 32.71 cm3^3.
  3. Actual capacity =196.25−32.71=163.54= 196.25 - 32.71 = 163.54 cm3^3.

Final Answer: Apparent 196.25196.25 cm3^3; actual 163.54163.54 cm3^3.

Takeaway: A raised bottom reduces capacity — subtract the hemisphere.

Example 2: Toy volume

A solid toy is a hemisphere surmounted by a cone. The cone's height is 2 cm and the base diameter is 4 cm. Find the toy's volume. (π=3.14)\left(\pi=3.14\right)

Solution:

  1. r=2r = 2 cm. V=23πr3+13πr2hV = \dfrac23\pi r^3 + \dfrac13\pi r^2 h.
  2. =23×3.14×8+13×3.14×4×2=16.75+8.37=25.12= \dfrac23\times 3.14\times 8 + \dfrac13\times 3.14\times 4\times 2 = 16.75 + 8.37 = 25.12 cm3^3.

Final Answer: 25.1225.12 cm3^3.

Takeaway: Volume adds: hemisphere ++ cone.

Example 3: Air in a shed

A shed is a cuboid (15 m×7 m×8 m15\text{ m}\times 7\text{ m}\times 8\text{ m}) topped by a half-cylinder of diameter 7 m and length 15 m. Find the volume of air it can hold. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Cuboid =15×7×8=840= 15\times 7\times 8 = 840 m3^3.
  2. Half-cylinder =12πr2L=12×227×3.52×15=288.75= \dfrac12\pi r^2 L = \dfrac12\times\dfrac{22}{7}\times 3.5^2\times 15 = 288.75 m3^3.
  3. Total =840+288.75=1128.75= 840 + 288.75 = 1128.75 m3^3.

Final Answer: 1128.751128.75 m3^3.

Takeaway: Half-cylinder volume =12πr2L= \tfrac12\pi r^2 L.

Example 4: Volume of wood in a pen stand

A cuboidal pen stand (15×10×3.515\times 10\times 3.5 cm) has 4 conical depressions, each of radius 0.5 cm and depth 1.4 cm. Find the volume of wood. (π=227)\left(\pi=\dfrac{22}{7}\right)

Solution:

  1. Cuboid =15×10×3.5=525= 15\times 10\times 3.5 = 525 cm3^3.
  2. One cone =13πr2h=13×227×0.25×1.4=1130≈0.3667= \dfrac13\pi r^2 h = \dfrac13\times\dfrac{22}{7}\times 0.25\times 1.4 = \dfrac{11}{30}\approx 0.3667 cm3^3; four cones ≈1.4667\approx 1.4667 cm3^3.
  3. Wood =525−1.4667≈523.53= 525 - 1.4667 \approx 523.53 cm3^3.

Final Answer: ≈523.53\approx 523.53 cm3^3.

Takeaway: Volume of material == solid −- (cavities).