Chapter 14: Math of Space: Surface Area and Volume - Vardaan Master Notes

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1. The Geometry of 3D Space: Surface Area vs Volume

While plane geometry deals with 2-dimensional figures containing length and breadth, real-world objects occupy 3-dimensional space having depth/height.

Core Distinction

Surface Area (Boundary): The total 2D measure of all exposed outer surfaces enclosing a solid. Measured in square units ($\text{cm}^2, \text{m}^2$).
Volume / Capacity (Internal Space): The total 3D space occupied by a solid body, or the quantity of fluid a hollow container can hold. Measured in cubic units ($\text{cm}^3, \text{m}^3, \text{litres}$).

$$\text{Metric Conversion Standards:}$$ $$1\text{ m}^3 = 1000\text{ litres} = 1{,}000{,}000\text{ cm}^3$$ $$1\text{ litre} = 1000\text{ cm}^3 = 1000\text{ mL}$$ $$1\text{ cm}^3 = 1\text{ mL}$$

2. Cuboid and Cube

A. Cuboid (Rectangular Prism)

A solid bounded by six rectangular faces having length $l$, breadth $b$, and height $h$:

Total Surface Area (TSA): $2(lb + bh + hl)$
Lateral Surface Area (LSA / Area of 4 Walls): $2(l + b)h$
Volume: $V = l \times b \times h$
Length of Main Diagonal: $d = \sqrt{l^2 + b^2 + h^2}$

B. Cube (Regular Hexahedron)

A special cuboid where all edges are equal ($l = b = h = a$):

Total Surface Area (TSA): $6a^2$
Lateral Surface Area (LSA): $4a^2$
Volume: $V = a^3$
Main Diagonal: $d = a\sqrt{3}$

3. Right Circular Cylinder

Generated by revolving a rectangle around one of its sides. Base radius is $r$ and height is $h$:

$$\text{Curved Surface Area (CSA): } 2\pi rh$$ $$\text{Total Surface Area (TSA): } 2\pi rh + 2\pi r^2 = 2\pi r(r + h)$$ $$\text{Volume: } V = \pi r^2 h$$

UNROLLING THE CYLINDER (PHYSICAL INTUITION)

If you cut the curved surface of a hollow cylinder along its height and unroll it flat on a table, it becomes a single flat rectangle with length equal to the base circumference ($2\pi r$) and breadth equal to the cylinder height ($h$).
Hence: $\text{Curved Surface Area} = \text{Area of Rectangle} = 2\pi r \times h = 2\pi rh$!

Hollow Cylinder (Pipe / Tube)

External radius $R$, internal radius $r$, height $h$:

Volume of Material: $V = \pi(R^2 - r^2)h$
Total Surface Area: $2\pi Rh + 2\pi rh + 2\pi(R^2 - r^2) = 2\pi(R + r)h + 2\pi(R^2 - r^2)$

4. Right Circular Cone

Formed by revolving a right-angled triangle about one of its perpendicular legs. Radius $r$, vertical height $h$, slant height $l$:

$$\text{Slant Height: } l = \sqrt{r^2 + h^2}$$ $$\text{Curved Surface Area (CSA): } \pi rl$$ $$\text{Total Surface Area (TSA): } \pi rl + \pi r^2 = \pi r(l + r)$$ $$\text{Volume: } V = \frac{1}{3}\pi r^2 h$$

THE 1/3 RATIO RELATIONSHIP

A cone with base radius $r$ and height $h$ holds exactly one-third the volume of a cylinder with the identical radius and height: $V_{\text{cone}} = \frac{1}{3} V_{\text{cylinder}}$. Filling a cone with water and pouring into the cylinder takes exactly 3 full cones to fill the cylinder!

5. Regular Pyramids

Pyramid Geometry

A polyhedron whose base is a polygon and whose lateral faces are triangles meeting at a single common vertex (apex).

$$\text{Volume of Any Regular Pyramid: } V = \frac{1}{3} \times \text{Base Area} \times h$$ $$\text{Total Surface Area: } \text{Base Area} + \text{Sum of Lateral Triangular Faces}$$

6. Sphere and Hemisphere

A. Solid Sphere

A perfectly round 3D solid where every surface point is equidistant ($r$) from the centre:

$$\text{Surface Area of Sphere: } A = 4\pi r^2$$ $$\text{Volume of Sphere: } V = \frac{4}{3}\pi r^3$$

B. Solid Hemisphere (Half-Sphere)

Obtained by slicing a solid sphere through its center plane:

$$\text{Curved Surface Area (CSA): } 2\pi r^2$$ $$\text{Flat Circular Top Area: } \pi r^2$$ $$\text{Total Surface Area (TSA): } 2\pi r^2 + \pi r^2 = 3\pi r^2$$ $$\text{Volume: } V = \frac{2}{3}\pi r^3$$

7. Master Mensuration Formula Matrix

Solid Shape	Curved / Lateral Surface Area	Total Surface Area (TSA)	Volume ($V$)
Cuboid ($l, b, h$)	$2(l + b)h$	$2(lb + bh + hl)$	$lbh$
Cube ($a$)	$4a^2$	$6a^2$	$a^3$
Right Cylinder ($r, h$)	$2\pi rh$	$2\pi r(r + h)$	$\pi r^2 h$
Right Cone ($r, h, l$)	$\pi rl$	$\pi r(l + r)$	$\frac{1}{3}\pi r^2 h$
Sphere ($r$)	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$
Hemisphere ($r$)	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$

8. Principles of Reshaping and Fluid Flow

Conservation of Volume (Recasting): When a solid of one shape is melted and recast into another shape, the total volume remains strictly conserved: $$V_{\text{initial}} = V_{\text{final}}$$
Fluid Flow Through Pipes: If liquid flows through a cylindrical pipe of internal cross-section area $A = \pi r^2$ at speed $v$ for time $t$: $$\text{Volume of Water Discharged} = \pi r^2 \times v \times t$$

9. Solved Master Examples (NCERT Ganita Manjari Pattern)

Example 1: Conical Circus Tent Canvas Calculation NCERT

Problem: A conical tent is $10\text{ m}$ high and the radius of its base is $24\text{ m}$. Find:

The slant height of the tent.
The cost of the canvas required to make the tent, if the cost of $1\text{ m}^2$ canvas is ₹70 (Take $\pi = \frac{22}{7}$).

Solution:

Given: Height $h = 10\text{ m}$, Base radius $r = 24\text{ m}$.

(1) Slant height $l$: $$l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\text{ m}$$

(2) Canvas required (Curved Surface Area only, floor is not covered): $$\text{CSA} = \pi rl = \frac{22}{7} \times 24 \times 26 = \frac{13728}{7}\text{ m}^2$$

Total Cost: $$\text{Cost} = \text{Area} \times \text{Rate} = \frac{13728}{7} \times 70 = 13728 \times 10 = \text{\rupee } 137{,}280$$ Answer: Slant height is $26\text{ m}$, and total cost is ₹1,37,280.

Example 2: Recasting / Melting Sphere into Cones HOTS

Problem: A solid metallic sphere of radius $10.5\text{ cm}$ is melted and recast into several smaller solid cones, each of base radius $3.5\text{ cm}$ and height $3\text{ cm}$. Find the total number of cones formed.

Solution:

By Principle of Conservation of Volume: $$\text{Total Volume of Sphere} = n \times \text{Volume of one cone}$$ $$\frac{4}{3}\pi R^3 = n \times \left(\frac{1}{3}\pi r^2 h\right)$$ Cancelling $\frac{1}{3}\pi$ on both sides: $$4R^3 = n \times r^2 h$$ Substitute values: $R = 10.5 = \frac{21}{2}\text{ cm}$, $r = 3.5 = \frac{7}{2}\text{ cm}$, $h = 3\text{ cm}$: $$4 \times \left(\frac{21}{2}\right)^3 = n \times \left(\frac{7}{2}\right)^2 \times 3$$ $$4 \times \frac{9261}{8} = n \times \frac{49}{4} \times 3$$ $$\frac{9261}{2} = n \times \frac{147}{4}$$ $$n = \frac{9261 \times 4}{2 \times 147} = \frac{9261 \times 2}{147} = 63 \times 2 = 126$$ Answer: Exactly 126 cones are formed.

10. Chapter Summary Checklist

Core Quick Revision

Cuboid TSA $= 2(lb + bh + hl)$, Volume $= lbh$.
Cylinder CSA $= 2\pi rh$, TSA $= 2\pi r(r+h)$, Volume $= \pi r^2 h$.
Cone slant height $l = \sqrt{r^2 + h^2}$, CSA $= \pi rl$, Volume $= \frac{1}{3}\pi r^2 h$.
Pyramid Volume $= \frac{1}{3} \times \text{Base Area} \times h$.
Sphere Surface Area $= 4\pi r^2$, Volume $= \frac{4}{3}\pi r^3$.
Hemisphere CSA $= 2\pi r^2$, TSA $= 3\pi r^2$, Volume $= \frac{2}{3}\pi r^3$.
Recasting always preserves volume ($V_1 = V_2$).