8.3 Surface Area and Volume of Cubes, Cuboids, Cylinders, Cones, Spheres & Melting-Recasting Solids

Cuboids and Cubes

1. Cuboid (Rectangular Parallelepiped)

Defined by length $l$, breadth $b$, and height $h$:

• Volume: $V = l \times b \times h$.
• Total Surface Area (TSA): Sum of all six rectangular faces:

$$\text{TSA} = 2(lb + bh + hl)$$

• Lateral Surface Area (LSA) / Area of 4 Walls:

$$\text{LSA} = 2h(l + b) = \text{Perimeter of Base} \times \text{Height}$$

• Space Diagonal ($d$): The length of the longest straight rod or rail that can fit inside the cuboid:

$$d = \sqrt{l^2 + b^2 + h^2}$$

+─────────────────────────────────────────────────────────────+
|                      CUBOID SPACE METRICS                   |
+─────────────────────────────────────────────────────────────+
| Dimensions: Length = l, Breadth = b, Height = h             |
|                                                             |
| Space Diagonal: d = √(l² + b² + h²)                         |
| (Longest pole/rod that can fit inside the container)        |
|                                                             |
| Important Algebraic Expansion Identity:                     |
| (l + b + h)² = (l² + b² + h²) + 2(lb + bh + hl)             |
| (Sum of Dimensions)² = (Diagonal)² + (Total Surface Area)   |
| TSA = (l + b + h)² - d²                                     |
+─────────────────────────────────────────────────────────────+

2. Cube (Side / Edge $a$)

When all dimensions are equal ($l = b = h = a$):

• Volume: $V = a^3$.
• Total Surface Area (TSA): $\text{TSA} = 6a^2$.
• Lateral Surface Area (LSA): $\text{LSA} = 4a^2$.
• Space Diagonal: $d = a\sqrt{3}$.
• Face Diagonal: $d_{\text{face}} = a\sqrt{2}$.

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Right Circular Cylinders & Hollow Pipes

1. Solid Cylinder (Radius $r$, Height $h$)

• Volume: $V = \pi r^2 h = \text{Base Area} \times \text{Height}$.
• Curved Surface Area (CSA): Area of unrolled rectangular jacket:

$$\text{CSA} = 2\pi r h$$

• Total Surface Area (TSA): CSA plus top and bottom circular bases:

$$\text{TSA} = 2\pi rh + 2\pi r^2 = 2\pi r(r + h)$$

2. Hollow Cylinder (Pipes, Track Drainage Culverts & Tank Shells)

With outer radius $R$, inner radius $r$, height/length $h$, and thickness $t = R - r$:

• Volume of Metal / Material:

$$V = \pi(R^2 - r^2)h = \pi(R + r)(R - r)h$$

• External Curved Surface Area: $\text{CSA}_{\text{ext}} = 2\pi Rh$.
• Internal Curved Surface Area: $\text{CSA}_{\text{int}} = 2\pi rh$.
• Total Surface Area (TSA):

$$\text{TSA} = \text{Ext CSA} + \text{Int CSA} + 2(\text{Annular End Area})$$ $$\text{TSA} = 2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2) = 2\pi(R + r)(h + R - r)$$

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Right Circular Cones & Frustums

+─────────────────────────────────────────────────────────────+
|                      CONE & FRUSTUM GEOMETRY                |
+─────────────────────────────────────────────────────────────+
| RIGHT CIRCULAR CONE:            FRUSTUM OF A CONE:          |
|            ▲ Vertex                        Top Radius = r   |
|           / \                             ┌─────────────┐   |
| Slant    /   \                           /               \  |
| Height  /  h  \                         /        h        \ |
|   (l)  /       \                       /                   \|
|       /    r    \                     └─────────────────────┘
|      └───────────┘                        Bottom Radius = R 
| l = √(r² + h²)                     l = √[h² + (R - r)²]     |
| V = (1/3)πr²h                      V = (1/3)πh(R² + r² + Rr)|
| CSA = πrl                          CSA = π(R + r)l          |
+─────────────────────────────────────────────────────────────+

1. Solid Cone (Base Radius $r$, Vertical Height $h$)

• Slant Height ($l$):

$$l = \sqrt{r^2 + h^2}$$

• Volume: Exactly one-third of the circumscribing cylinder:

$$V = \frac{1}{3}\pi r^2 h$$

• Curved Surface Area (CSA): $\text{CSA} = \pi r l$.
• Total Surface Area (TSA): $\text{TSA} = \pi rl + \pi r^2 = \pi r(l + r)$.

2. Frustum of a Cone (Hopper Chutes, Water Buckets)

When a cone is sliced by a plane parallel to its base, leaving top radius $r$, bottom radius $R$, and vertical height $h$:

• Slant Height ($l$):

$$l = \sqrt{h^2 + (R - r)^2}$$

• Volume:

$$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$$

• Curved Surface Area (CSA):

$$\text{CSA} = \pi (R + r) l$$

• Total Surface Area (TSA):

$$\text{TSA} = \pi (R + r) l + \pi R^2 + \pi r^2$$

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Spheres, Hemispheres & Spherical Shells

1. Solid Sphere (Radius $r$)

• Volume: $V = \frac{4}{3}\pi r^3$.
• Surface Area ($S$): $S = 4\pi r^2$. (Note: A sphere has only one surface area; CSA and TSA are identical).

2. Hemisphere (Radius $r$)

• Volume: Half of the full sphere:

$$V = \frac{2}{3}\pi r^3$$

• Curved Surface Area (CSA): $2\pi r^2$.
• Total Surface Area (TSA): CSA plus flat circular base:

$$\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2$$

3. Spherical Shell (Hollow Sphere)

With outer radius $R$ and inner radius $r$:

• Volume of Metal: $V = \frac{4}{3}\pi(R^3 - r^3)$.
• Outer Surface Area: $4\pi R^2$; Inner Surface Area: $4\pi r^2$.

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Melting, Recasting & Volume Conservation Dynamics

1. The Fundamental Law of Volume Conservation

When one or more solid bodies are melted, crushed, or recast into different solid shapes without loss of material, the total volume remains strictly constant:

$$\text{Initial Volume } (V_{\text{initial}}) = \text{Total Final Volume } (\sum V_{\text{final}})$$

If a single solid of volume $V_{\text{large}}$ is melted and recast into $N$ identical smaller solids each of volume $V_{\text{small}}$:

$$N = \frac{V_{\text{large}}}{V_{\text{small}}}$$

2. Recasting a Sphere into Smaller Spheres

If a solid sphere of radius $R$ is melted into $N$ small spheres of radius $r$:

$$\frac{4}{3}\pi R^3 = N \times \left(\frac{4}{3}\pi r^3\right) \implies R^3 = N \cdot r^3 \implies N = \left(\frac{R}{r}\right)^3$$ $$\frac{\text{New TSA}}{\text{Original TSA}} = \frac{N \times 4\pi r^2}{4\pi R^2} = N \left(\frac{r}{R}\right)^2 = N \left(\frac{1}{N^{1/3}}\right)^2 = N^{1/3} = \frac{R}{r}$$

Melting a large sphere into 64 small spheres ($N = 64$) increases the total surface area by $64^{1/3} = 4$ times!

3. Liquid Flow Rate Through Pipes into Cisterns

When water flows through a cylindrical pipe of cross-sectional radius $r_{\text{pipe}}$ at uniform velocity $v$ for time $t$, the total volume of fluid delivered is:

$$\text{Discharge Volume } (V) = \text{Cross-Sectional Area} \times \text{Velocity} \times \text{Time} = (\pi r_{\text{pipe}}^2) \cdot v \cdot t$$

Equating this to the receiving rectangular reservoir ($l \times b \times H$) gives the time or height reached:

$$(\pi r_{\text{pipe}}^2) \cdot v \cdot t = l \times b \times H$$

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Dimensional Percentage Variations & Scaling Laws

Scaling Factor ($k$)

When all linear dimensions of a solid are scaled by factor $k$:

• Any Linear Dimension (radius, height, diagonal) changes by $k$.
• Any Surface Area (CSA, TSA, Base Area) changes by $k^2$.
• Any Volume changes by $k^3$.

Percentage Changes in Cylinders and Cones

Because Volume $V \propto r^2 h$:

• If radius increases by $x\%$ and height changes by $y\%$:

$$V_{\text{new}} = V_{\text{old}} \times \left(1 + \frac{x}{100}\right)^2 \left(1 + \frac{y}{100}\right)$$

• If radius increases by $+20\%$ and height decreases by $-25\%$:

$$V_{\text{new}} = V_{\text{old}} \times (1.20)^2 \times (0.75) = V_{\text{old}} \times 1.44 \times 0.75 = 1.08 V_{\text{old}}$$

Net volume increases by $+8\%$.

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Worked Examples

Worked Example 1: Parcel Van Space Diagonal & Internal Surface Painting

Problem: A railway divisional parcel brake-van compartment has length $12\text{ m}$, breadth $9\text{ m}$, and height $8\text{ m}$. (a) Calculate the length of the longest rigid steel measuring rail that can be stowed completely inside the compartment. (b) Calculate the total cost of painting the four internal walls and the ceiling (5 surfaces, excluding the floor) at ₹25 per $\text{m}^2$.

• Step 1: Compute Space Diagonal (Longest Rod):

$$d = \sqrt{l^2 + b^2 + h^2} = \sqrt{12^2 + 9^2 + 8^2} = \sqrt{144 + 81 + 64} = \sqrt{289} = 17\text{ m}$$

• Step 2: Compute Surface Area to Paint (4 Walls + Ceiling):

$$\text{Area of 4 Walls} = 2h(l + b) = 2(8)(12 + 9) = 16 \times 21 = 336\text{ m}^2$$ $$\text{Area of Ceiling} = l \times b = 12 \times 9 = 108\text{ m}^2$$ $$\text{Total Surface Area to Paint} = 336 + 108 = 444\text{ m}^2$$

• Step 3: Compute Painting Expenditure:

$$\text{Cost} = 444\text{ m}^2 \times \text{INR } 25/\text{m}^2 = \text{INR } 11,100$$

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Worked Example 2: Recasting a Solid Metal Cylinder into Cones

Problem: A solid cylindrical bronze axle rod of base radius $6\text{ cm}$ and height $14\text{ cm}$ is melted down and completely recast into small solid decorative cones, each having base radius $3\text{ cm}$ and vertical height $7\text{ cm}$. Assuming no wastage of metal during foundry casting, determine how many such cones are formed.

• Step 1: Compute Volume of the Bronze Cylinder:

$$V_{\text{cyl}} = \pi r_1^2 h_1 = \pi \times (6)^2 \times 14 = 36 \times 14 \times \pi = 504\pi\text{ cm}^3$$

• Step 2: Compute Volume of a Single Recast Cone:

$$V_{\text{cone}} = \frac{1}{3}\pi r_2^2 h_2 = \frac{1}{3}\pi \times (3)^2 \times 7 = \frac{1}{3} \times 9 \times 7 \times \pi = 21\pi\text{ cm}^3$$

• Step 3: Calculate Number of Cones ($N$):

$$N = \frac{V_{\text{cyl}}}{V_{\text{cone}}} = \frac{504\pi}{21\pi} = \frac{504}{21} = 24\text{ cones}$$

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Worked Example 3: Cylindrical Pipe Discharge into a Locomotive Reservoir

Problem: Water flows through a cylindrical railway water pipe of internal diameter $14\text{ cm}$ at a velocity of $5\text{ m/s}$ into a large rectangular locomotive reservoir measuring $22\text{ m}$ long and $14\text{ m}$ wide. Using $\pi = 22/7$, determine how many hours it will take for the water level in the reservoir to rise by $3\text{ m}$.

• Step 1: Standardize Units to Meters and Seconds:
  • Pipe radius $r = 14\text{ cm} / 2 = 7\text{ cm} = 0.07\text{ m} = \frac{7}{100}\text{ m}$.
  • Flow velocity $v = 5\text{ m/s}$.
• Step 2: Compute Rate of Water Discharge per Second:

$$\text{Discharge Rate} = \pi r^2 \times v = \frac{22}{7} \times \left(\frac{7}{100}\right)^2 \times 5 = \frac{22}{7} \times \frac{49}{10000} \times 5 = \frac{154 \times 5}{10000} = \frac{770}{10000} = 0.077\text{ m}^3/\text{s}$$

• Step 3: Compute Volume of Water Delivered per Hour ($3600\text{ s}$):

$$\text{Discharge per Hour} = 0.077 \times 3600 = 277.2\text{ m}^3/\text{hour}$$

• Step 4: Compute Required Reservoir Volume:

$$V_{\text{reservoir}} = l \times b \times H = 22 \times 14 \times 3 = 924\text{ m}^3$$

• Step 5: Compute Filling Duration in Hours:

$$\text{Time} = \frac{924\text{ m}^3}{277.2\text{ m}^3/\text{hour}} = \frac{9240}{2772} = \frac{10}{3} \text{ hours} = 3\frac{1}{3}\text{ hours} = 3\text{ hours } 20\text{ minutes}$$

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Calculation Traps & Speed Heuristics

1. The Hemisphere Surface Area Trap (CSA vs. TSA): The curved surface area of a hemisphere is $2\pi r^2$. However, if a problem asks for the total surface area of a solid hemisphere, you must include the flat circular top: $\text{TSA} = 3\pi r^2$. Candidates who blindly use $2\pi r^2$ lose marks instantly.
2. The Cone Slant Height vs. Vertical Height Trap: The volume of a cone uses vertical height $h$ ($V = \frac{1}{3}\pi r^2 h$), but the curved surface area uses slant height $l$ ($\text{CSA} = \pi r l$). Never plug vertical height into the CSA formula without first computing $l = \sqrt{r^2 + h^2}$.
3. Dimensional Scaling Speed Shortcut: If the edge of a cube or radius of a sphere is doubled ($k = 2$):
   • Surface area becomes $2^2 = 4$ times (increases by $300\%$).
   • Volume becomes $2^3 = 8$ times (increases by $700\%$).

Volume conservation in these recasting exercises assumes the same material and density before and after recasting, with no wastage. In physical processing generally, mass conservation alone does not guarantee equal volume if density changes.