Surface Areas and Volumes - Surface Area and Volume of Combinations of Solids

In Class 9, you learned how to calculate the surface areas and volumes of isolated three-dimensional geometrical shapes—cubes, cuboids, right circular cylinders, right circular cones, and spheres. However, the physical objects we encounter daily rarely exist as isolated geometric forms. A circus tent is a combination of a cylinder and a cone; a medicine capsule is a cylinder bounded by two hemispheres; an ice-cream cone consists of a cone surmounted by a hemisphere; and an oil tanker is a cylinder with hemispherical ends.

In Class 10 Mathematics, the focus shifts to calculating the total surface area and volume of such combined 3D solids. Mastering this concept requires visualizing how individual shapes join together, identifying which surfaces remain exposed, and applying precise algebraic formulas. This concept forms a significant portion of the Class 10 CBSE Board Examination and builds foundational skills for engineering, design, architecture, and physics.

---

1. Mathematical Foundation: Reference Table of Basic 3D Solids

Before analyzing combined solids, let us review the fundamental measurement formulas for individual shapes.

---

2. In-Depth Conceptual Breakdown

2.1 Total Surface Area of Combined Solids

The single most critical concept to master when dealing with combined solids is:

$\text{Surface Area of Combined Solid} \neq \text{TSA of Solid 1} + \text{TSA of Solid 2}$

When two or more solids are joined together to form a new solid, the surfaces along which they are joined become internal boundaries and are no longer exposed to the outside. Because "surface area" refers exclusively to the region accessible from the exterior, you must sum only the exposed (visible) outer surfaces.

Fundamental Rule for Surface Area of Joined Solids:

To find the Total Surface Area (TSA) of a combined solid, identify all exposed individual surfaces and add their areas:

$\text{TSA}_{\text{combined}} = \sum \text{Exposed Curved Surface Areas (CSA)} + \sum \text{Exposed Base Areas}$

Common Combinations & Surface Area Formulas:

1. Cone Mounted on a Hemisphere (e.g., Toy, Top, Ice-cream Cone)

   • Exposed surfaces: Curved surface of the cone + Curved surface of the hemisphere.
   • Interface: Circular base of the cone and top face of the hemisphere are hidden. $\text{TSA}_{\text{toy}} = \text{CSA}_{\text{cone}} + \text{CSA}_{\text{hemisphere}} = \pi r l + 2\pi r^2 = \pi r (l + 2r)$

2. Cylinder with Hemispherical Ends (e.g., Capsule, Storage Tank)

   • Exposed surfaces: Curved surface of the cylinder + Curved surface of two identical end hemispheres. $\text{TSA}_{\text{capsule}} = \text{CSA}_{\text{cylinder}} + 2 \times \text{CSA}_{\text{hemisphere}} = 2\pi r h + 2(2\pi r^2) = 2\pi r (h + 2r)$

3. Cylinder Surmounted by a Cone (e.g., Tent, Silo)

   • Exposed surfaces: Curved surface of the cylindrical base + Curved surface of the top cone.
   • Interface: Common circular boundary between cylinder and cone is inside the tent and not exposed. $\text{TSA}_{\text{tent}} = \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}} = 2\pi r h_{\text{cyl}} + \pi r l$

4. Solid Carved Out or Scooped (e.g., Wooden Block with Conical or Hemispherical Cavity)

   • When a shape is carved out of a parent solid, the total surface area increases because a new interior boundary is created and exposed to the outside. $\text{TSA}_{\text{remaining}} = \text{TSA}_{\text{parent solid}} - \text{Area of opening removed} + \text{CSA}_{\text{carved shape}}$

---

2.2 Volume of Combined Solids

Unlike surface area, volume is a measure of space occupied by matter. Therefore, volume is strictly additive and subtractive.

Fundamental Rules for Volume:

1. For Joined/Attached Solids: $\text{Volume}_{\text{total}} = \text{Volume}_{\text{Solid 1}} + \text{Volume}_{\text{Solid 2}} + \dots + \text{Volume}_{\text{Solid } n}$

2. For Carved/Hollowed Solids: $\text{Volume}_{\text{remaining}} = \text{Volume}_{\text{parent solid}} - \text{Volume}_{\text{removed cavity}}$

Common Combinations & Volume Formulas:

1. Cone Surmounting a Hemisphere: $V_{\text{combined}} = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 (h + 2r)$

2. Cylinder with Hemispherical Ends: $V_{\text{combined}} = V_{\text{cylinder}} + 2 \times V_{\text{hemisphere}} = \pi r^2 h + 2\left(\frac{2}{3}\pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)$

3. Cylindrical Block with Conical Cavity Carved Out: $V_{\text{remaining}} = V_{\text{cylinder}} - V_{\text{cone}} = \pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h$

---

3. Real-World Applications

1. Pharmaceutical Capsule Design

Pharmaceutical manufacturers need to optimize the mass-to-volume ratio of drug delivery capsules. A capsule consists of a central cylindrical barrel capped by two hemispherical shells. Calculating exact surface area dictates the coating material needed (e.g., gelatin or enteric coating), while volume calculations determine the internal liquid/powder capacity.

+----+-------------------+----+
| (  |     Cylinder      |  ) |  <- Hemisphere + Cylinder + Hemisphere
+----+-------------------+----+

2. Grain Storage Silos and Industrial Tanks

Civil and agricultural engineers design storage silos using a cylindrical base topped by a conical roof. Calculating volume tells the storage capacity (in cubic meters or metric tonnes), whereas calculating total surface area gives the cost of sheet metal or anti-rust paint needed for weather protection.

3. Architecture and Tent Construction

Circus tents or temporary disaster relief shelters combine a cylindrical base with a conical roof. Calculating the total canvas material required relies strictly on adding the curved surface area of the cylinder and the curved surface area of the cone, while excluding the ground floor area and the internal horizontal circular join.

---

4. Step-by-Step Solved Textbook Examples

Example 1: Total Surface Area of a Combined Toy (Hemisphere + Cone)

Problem Statement: A wooden toy is in the form of a cone mounted on a hemisphere with the same base radius. The radius of the hemispherical base is $3.5\text{ cm}$, and the total height of the toy is $15.5\text{ cm}$. Find the total surface area of the toy. (Take $\pi = \frac{22}{7}$)

         /\
        /  \     <- Cone
       /    \
      /______\
     (________)  <- Hemisphere

Solution:

Step 1: Identify given dimensions.

• Radius of hemisphere ($r$) = $3.5\text{ cm} = \frac{7}{2}\text{ cm}$
• Radius of cone base ($r$) = $3.5\text{ cm} = \frac{7}{2}\text{ cm}$
• Total height of toy ($H$) = $15.5\text{ cm}$

Step 2: Calculate the vertical height of the conical part ($h$). Since the height of the hemispherical part equals its radius ($r = 3.5\text{ cm}$): $h = H - r = 15.5 - 3.5 = 12\text{ cm}$

Step 3: Calculate the slant height ($l$) of the cone. $l = \sqrt{r^2 + h^2}$ $l = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\text{ cm}$

Step 4: Formulate Total Surface Area of the toy. $\text{TSA}_{\text{toy}} = \text{CSA}_{\text{cone}} + \text{CSA}_{\text{hemisphere}}$ $\text{TSA}_{\text{toy}} = \pi r l + 2\pi r^2 = \pi r (l + 2r)$

Step 5: Substitute values and calculate. $\text{TSA}_{\text{toy}} = \frac{22}{7} \times \frac{7}{2} \times (12.5 + 2 \times 3.5)$ $\text{TSA}_{\text{toy}} = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5\text{ cm}^2$

Final Answer:

The total surface area of the toy is $214.5\text{ cm}^2$.

---

Example 2: Hemisphere Surmounting a Cube

Problem Statement: A decorative block is made of two solids—a cube and a hemisphere. The base of the block is a cube with edge $7\text{ cm}$, and the hemisphere fixed on top has a diameter of $7\text{ cm}$. Find the total surface area of the block. (Take $\pi = \frac{22}{7}$)

Solution:

Step 1: Identify given dimensions.

• Edge of the cube ($a$) = $7\text{ cm}$
• Diameter of hemisphere ($d$) = $7\text{ cm} \implies \text{Radius } r = \frac{7}{2}\text{ cm} = 3.5\text{ cm}$

Step 2: Understand the surfaces involved.

• The surface area of the 5 faces of the cube completely exposed = $5a^2$
• The top face of the cube has area $a^2$, but the circular base of the hemisphere covers a portion of it with area $\pi r^2$.
• Exposed area of top face = $a^2 - \pi r^2$
• Curved surface area of hemisphere added on top = $2\pi r^2$

Step 3: Combine the expression. $\text{TSA}_{\text{block}} = \text{TSA}_{\text{cube}} - \text{Base area of hemisphere} + \text{CSA}_{\text{hemisphere}}$ $\text{TSA}_{\text{block}} = 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$

Step 4: Substitute and compute values. $6a^2 = 6 \times (7)^2 = 6 \times 49 = 294\text{ cm}^2$ $\pi r^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} = 38.5\text{ cm}^2$

$\text{TSA}_{\text{block}} = 294 + 38.5 = 332.5\text{ cm}^2$

Final Answer:

The total surface area of the decorative block is $332.5\text{ cm}^2$.

---

Example 3: Volume of a Medicine Capsule

Problem Statement: A medicine capsule is in the form of a cylinder with two hemispheres stuck to each of its ends. The entire length of the capsule is $14\text{ mm}$, and the diameter of the capsule is $5\text{ mm}$. Find its volume. (Take $\pi = \frac{22}{7}$)

Solution:

Step 1: Extract dimensional values.

• Diameter ($d$) = $5\text{ mm} \implies \text{Radius } r = \frac{5}{2}\text{ mm} = 2.5\text{ mm}$
• Total length of capsule ($L$) = $14\text{ mm}$

Step 2: Calculate height ($h$) of the cylindrical section. Since each hemispherical end extends outward by its radius $r$: $h = L - 2r = 14 - 2(2.5) = 14 - 5 = 9\text{ mm}$

Step 3: Formulate total volume. $\text{Volume}_{\text{capsule}} = \text{Volume}_{\text{cylinder}} + 2 \times \text{Volume}_{\text{hemisphere}}$ $V = \pi r^2 h + 2 \times \left(\frac{2}{3}\pi r^3\right) = \pi r^2 \left(h + \frac{4}{3}r\right)$

Step 4: Compute value. $V = \frac{22}{7} \times \left(\frac{5}{2}\right)^2 \times \left(9 + \frac{4}{3} \times \frac{5}{2}\right)$ $V = \frac{22}{7} \times \frac{25}{4} \times \left(9 + \frac{10}{3}\right)$ $V = \frac{11 \times 25}{14} \times \left(\frac{37}{3}\right) = \frac{275}{14} \times \frac{37}{3} = \frac{10175}{42} \approx 242.26\text{ mm}^3$

Final Answer:

The volume of the medicine capsule is approximately $242.26\text{ mm}^3$.

---

Example 4: Carved-Out Conical Cavity in a Cylinder

Problem Statement: From a solid cylinder whose height is $2.4\text{ cm}$ and diameter is $1.4\text{ cm}$, a conical cavity of the same height and same diameter is carved out. Find the total surface area and the volume of the remaining solid.

Solution:

Step 1: Identify given parameters.

• Height of cylinder ($h$) = Height of cone ($h$) = $2.4\text{ cm}$
• Diameter ($d$) = $1.4\text{ cm} \implies \text{Radius } r = 0.7\text{ cm} = \frac{7}{10}\text{ cm}$

Step 2: Slant height of the carved cone ($l$). $l = \sqrt{r^2 + h^2} = \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76} = \sqrt{6.25} = 2.5\text{ cm}$

Step 3: Calculate Total Surface Area of remaining solid. The remaining exposed surfaces are:

1. Outer curved surface area of cylinder = $2\pi r h$
2. Area of one flat circular base (bottom base) = $\pi r^2$
3. Inner curved surface area of conical cavity = $\pi r l$

$\text{TSA}_{\text{remaining}} = 2\pi r h + \pi r^2 + \pi r l = \pi r (2h + r + l)$ $\text{TSA}_{\text{remaining}} = \frac{22}{7} \times 0.7 \times [2(2.4) + 0.7 + 2.5]$ $\text{TSA}_{\text{remaining}} = 2.2 \times [4.8 + 0.7 + 2.5] = 2.2 \times 8.0 = 17.6\text{ cm}^2$

Step 4: Calculate Volume of remaining solid. $\text{Volume}_{\text{remaining}} = \text{Volume}_{\text{cylinder}} - \text{Volume}_{\text{cone}}$ $V_{\text{remaining}} = \pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi r^2 h$ $V_{\text{remaining}} = \frac{2}{3} \times \frac{22}{7} \times (0.7)^2 \times 2.4$ $V_{\text{remaining}} = \frac{2}{3} \times \frac{22}{7} \times 0.49 \times 2.4 = \frac{2}{3} \times 22 \times 0.07 \times 2.4$ $V_{\text{remaining}} = 2 \times 22 \times 0.07 \times 0.8 = 2.464\text{ cm}^3 \approx 2.46\text{ cm}^3$

Final Answer:

The total surface area of the remaining solid is $17.6\text{ cm}^2$, and its volume is $2.464\text{ cm}^3$.

---

5. Common Student Mistakes to Avoid

Pitfall 1: Incorrectly Adding Total Surface Areas

• Mistake: Calculating $\text{TSA}_{\text{combined}} = \text{TSA}_{\text{solid 1}} + \text{TSA}_{\text{solid 2}}$.
• Correction: When solids are attached, their joining face is hidden. Calculate only the exposed outer surfaces. For example, for a cone on a cylinder, use $(\text{CSA}_{\text{cone}} + \text{CSA}_{\text{cylinder}} + \text{Base Area}_{\text{cylinder}})$, not their TSAs.

Pitfall 2: Confusing Radius with Diameter

• Mistake: Directly substituting diameter ($d$) into formulas requiring radius ($r$).
• Correction: Always double-check whether the question gives diameter or radius. Convert immediately using $r = \frac{d}{2}$ in your initial step.

Pitfall 3: Subtraction in Surface Area for Carved-Out Solids

• Mistake: Subtracting the surface area of a carved-out portion when calculating Total Surface Area.
• Correction: Hollowing out an object exposes new internal surfaces. Surface area increases when a cavity is hollowed out, whereas volume decreases.

Pitfall 4: Unit Mismatch and Liquid Conversions

• Mistake: Mixing $\text{cm}$ and $\text{m}$, or forgetting capacity conversion factors.
• Correction: Always convert all dimensions to a common unit before calculating. Remember key capacity conversions:
  • $1\text{ cm}^3 = 1\text{ mL}$
  • $1000\text{ cm}^3 = 1\text{ Litre}$
  • $1\text{ m}^3 = 1000\text{ Litres} = 1\text{ kilolitre}$

---

6. Practice Questions for Self-Assessment

Question 1

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the conical top is $2.8\text{ m}$, find the area of canvas used for making the tent. Also, find the cost of canvas of the tent at the rate of ₹$500$ per $\text{m}^2$. (Note that the base of the tent will not be covered with canvas).

<details> <summary>Click to view Solution</summary>

Step 1: Identify parameters.

• Diameter of cylinder = $4\text{ m} \implies r = 2\text{ m}$
• Height of cylinder ($h$) = $2.1\text{ m}$
• Slant height of cone ($l$) = $2.8\text{ m}$

Step 2: Formula for area of canvas. $\text{Area of canvas} = \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}} = 2\pi r h + \pi r l = \pi r (2h + l)$

Step 3: Calculation. $\text{Area} = \frac{22}{7} \times 2 \times [2(2.1) + 2.8] = \frac{44}{7} \times [4.2 + 2.8] = \frac{44}{7} \times 7 = 44\text{ m}^2$

Step 4: Calculate Cost. $\text{Total Cost} = 44\text{ m}^2 \times ₹500/\text{m}^2 = ₹22,000$

Answer: Area of canvas required = $44\text{ m}^2$; Total Cost = ₹$22,000$.

</details>

---

Question 2

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2\text{ cm}$ and the diameter of the base is $4\text{ cm}$. Determine the volume of the toy. (Take $\pi = 3.14$)

<details> <summary>Click to view Solution</summary>

Step 1: Identify parameters.

• Radius ($r$) = $\frac{4}{2} = 2\text{ cm}$
• Height of cone ($h$) = $2\text{ cm}$

Step 2: Total Volume Formula. $V_{\text{toy}} = V_{\text{cone}} + V_{\text{hemisphere}} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2 (h + 2r)$

Step 3: Calculation. $V_{\text{toy}} = \frac{1}{3} \times 3.14 \times (2)^2 \times [2 + 2(2)]$ $V_{\text{toy}} = \frac{1}{3} \times 3.14 \times 4 \times 6 = 3.14 \times 8 = 25.12\text{ cm}^3$

Answer: Volume of the toy = $25.12\text{ cm}^3$.

</details>

---

Question 3

A hemispherical depression is cut out from one face of a cubical wooden block of edge $a$ such that the diameter $d$ of the hemisphere is equal to the edge $a$. Determine the surface area of the remaining solid in terms of $a$.

<details> <summary>Click to view Solution</summary>

Step 1: Parameters.

• Cube edge = $a$
• Hemisphere diameter $d = a \implies r = \frac{a}{2}$

Step 2: Area Formulation. $\text{TSA} = \text{TSA}_{\text{cube}} - \text{Base area of hemisphere} + \text{CSA}_{\text{hemisphere}}$ $\text{TSA} = 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2$

Step 3: Substitute $r = \frac{a}{2}$. $\text{TSA} = 6a^2 + \pi \left(\frac{a}{2}\right)^2 = 6a^2 + \frac{\pi a^2}{4} = \frac{a^2}{4}(24 + \pi)\text{ sq. units}$

Answer: The surface area of the remaining solid is $\frac{a^2}{4}(24 + \pi)\text{ sq. units}$.

</details>

---

Question 4

A Gulab Jamun contains sugar syrup up to about $30\%$ of its total volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two hemispherical ends with length $5\text{ cm}$ and diameter $2.8\text{ cm}$. (Take $\pi = \frac{22}{7}$)

<details> <summary>Click to view Solution</summary>

Step 1: Parameters.

• Diameter $d = 2.8\text{ cm} \implies r = 1.4\text{ cm}$
• Total length $L = 5\text{ cm}$
• Height of cylinder $h = 5 - 2(1.4) = 5 - 2.8 = 2.2\text{ cm}$

Step 2: Volume of 1 Gulab Jamun. $V_1 = \pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 \left(h + \frac{4}{3}r\right)$ $V_1 = \frac{22}{7} \times (1.4)^2 \times \left(2.2 + \frac{4}{3} \times 1.4\right)$ $V_1 = \frac{22}{7} \times 1.96 \times \left(2.2 + \frac{5.6}{3}\right) = 6.16 \times \left(\frac{6.6 + 5.6}{3}\right) = 6.16 \times \frac{12.2}{3} \approx 25.05\text{ cm}^3$

Step 3: Total volume of 45 Gulab Jamuns. $V_{total} = 45 \times 25.051 = 1127.3\text{ cm}^3$

Step 4: Calculate syrup volume ($30\%$ of total). $\text{Volume of syrup} = 0.30 \times 1127.3 \approx 338.19\text{ cm}^3 \approx 338\text{ cm}^3$

Answer: Syrup contained in 45 gulab jamuns $\approx$ $338\text{ cm}^3$.

</details>

---

7. Board Exam Strategy & Frequently Asked Questions (FAQs)

FAQ 1: How do I know when to use Curved Surface Area (CSA) versus Total Surface Area (TSA)?

Answer: Ask yourself: "If I dip this combined object in paint, which parts get wet?"

• If a face is glued to another shape or hollowed out internally, that original flat surface is not exposed on the outside.
• Generally, for joined solids (e.g., cone on top of cylinder), calculate the CSA of individual solids and sum them up along with any exposed end bases.

FAQ 2: Should I calculate intermediate decimal answers step-by-step or keep everything in algebraic fractions?

Answer: Keep expressions in fractional form with $\pi$ factored out as long as possible! Factor common terms like $\pi r$ or $\pi r^2$ first. Calculate final numbers only at the last step. This saves time, reduces rounding errors, and simplifies multi-step calculations.

FAQ 3: What step-by-step structure guarantees full marks in CBSE Board Exams?

To achieve top marks on 4-mark or 5-mark subjective questions:

1. Given Data: List all dimensions explicitly ($r, h, l, d$) with their units.
2. Formula Statement: Write down the general formula for the surface area or volume of the combined solid before plugging in numbers.
3. Algebraic Simplification: Factor out common variables (e.g., $\pi r$).
4. Calculations with Units: Show clear substitution, and state the final result with appropriate units ($\text{cm}^2, \text{m}^3, \text{Litres}$, etc.). Highlight the final answer clearly.