Right Circular Cylinder

In Euclidean solid geometry, a right circular cylinder (লম্ব বৃত্তাকার চোঙ) is a three-dimensional surface bounded by two parallel planar circular bases and a curved lateral surface. It is formally generated by taking a planar rectangle $ABCD$ and rotating it through $360^\\circ$ (one full revolution) about one of its sides (say side $AB$) which is kept fixed as the central axis of rotation.

• The fixed side $AB$ is called the axis of the cylinder. Its length represents the height ($h$) or longitudinal length of the cylinder.
• The adjacent revolving side $AD$ (or $BC$) sweeps out two congruent parallel circular discs, called the bases of the cylinder. The length of this side is the radius ($r$) of the circular base.
• The side $CD$ parallel to the axis sweeps through space to generate the smooth curved surface (lateral surface).
• Why 'Right'?: The cylinder is termed a right cylinder because its central axis is strictly perpendicular ($90^\\circ$) to the plane of the circular bases. If the axis meets the base at an oblique angle, the solid is termed an oblique cylinder (not in the WBBSE Class 10 syllabus).

To understand the curved surface area of a right circular cylinder, imagine making a straight longitudinal cut along the height $h$ of a hollow cylindrical paper tube and unrolling it completely flat onto a planar table:

• The unrolled curved surface forms an exact rectangle.
• The height of this rectangle corresponds directly to the height of the cylinder: $\\text{Breadth} = h$.
• The horizontal length of this rectangle was wrapped completely around the circular base, meaning it is exactly equal to the circumference of the base: $\\text{Length} = 2\\pi r$.

Since the area of a rectangle is $\\text{Length} \\times \\text{Breadth}$: $$\\text{Curved Surface Area (CSA)} = (2\\pi r) \\times h = 2\\pi r h$$ Units of curved surface area are square units ($\text{cm}^2, \text{m}^2, \text{sq. units}$).

Depending on whether the cylinder is closed at both ends, open at one end, or open at both ends, the total surface area is computed as follows:

• 1. Completely Closed Solid Cylinder (both ends closed): $$\\text{TSA} = \\text{CSA} + 2 \\times \\text{Area of Circular Base} = 2\\pi r h + 2\\pi r^2 = \\mathbf{2\\pi r(h + r)}$$
• 2. Open-Topped Cylinder (e.g., bucket, beaker, water cistern open at top): $$\\text{Surface Area} = \\text{CSA} + 1 \\times \\text{Area of Bottom Base} = 2\\pi r h + \\pi r^2 = \\mathbf{\\pi r(2h + r)}$$
• 3. Open Pipe (open at both ends, hollow tube of negligible thickness): $$\\text{Surface Area} = \\text{CSA only} = \\mathbf{2\\pi r h}$$

By Cavalieri's Principle and the fundamental definition of a prismatoid volume, the volume of any uniform solid having identical parallel cross-sections throughout its height is given by: $$\\text{Volume} = \\text{Area of Base} \\times \\text{Height}$$ For a right circular cylinder with circular base of radius $r$ and height $h$: $$\\text{Volume} = (\\pi r^2) \\times h = \\mathbf{\\pi r^2 h}$$ Units of volume are cubic units ($\text{cm}^3, \text{dm}^3, \text{m}^3$).

In Madhyamik board word problems involving water tanks, milk containers, and diesel tanks, volume must be converted to liquid capacity. The standard metric conversion factors are:

A hollow right circular cylinder (e.g., metal pipe, gun barrel, hollow pillar) is bounded by two coaxial cylindrical surfaces of the same height $h$, having different radii:

• Internal radius ($r$): Radius of the inner hollow cylindrical bore.
• External radius ($R$): Radius of the outer outer cylindrical boundary ($R > r$).
• Thickness ($t$): Radial thickness of the solid material: $t = R - r$.
• Base of hollow cylinder: Each base is an annular ring (washer shape) between two concentric circles. Area of each annular base $= \\pi R^2 - \\pi r^2 = \\pi(R^2 - r^2)$.

• 1. Curved Surface Area (both inner and outer lateral surfaces): $$\\text{Total CSA} = \\text{Outer CSA} + \\text{Inner CSA} = 2\\pi R h + 2\\pi r h = \\mathbf{2\\pi(R + r)h}$$
• 2. Total Surface Area of Closed Hollow Cylinder: $$\\text{TSA} = \\text{Total CSA} + 2 \\times \\text{Area of Base Rings} = 2\\pi(R + r)h + 2\\pi(R^2 - r^2)$$ Factoring $R^2 - r^2 = (R + r)(R - r)$: $$\\text{TSA} = 2\\pi(R + r)[h + (R - r)] = \\mathbf{2\\pi(R + r)(h + t)}$$
• 3. Volume of Material in Hollow Cylinder: $$\\text{Volume of metal} = \\text{External Volume} - \\text{Internal Volume} = \\pi R^2 h - \\pi r^2 h = \\mathbf{\\pi(R^2 - r^2)h}$$
• 4. Mass / Weight of Pipe: $$\\text{Mass} = \\text{Volume of Material} \\times \\text{Density} = [\\pi(R^2 - r^2)h] \\times \\rho$$

A heavy cylindrical road roller levels ground by rolling forward without slipping. Consider a roller of radius $r$ (diameter $d = 2r$) and length $h$:

• In one complete revolution ($360^\\circ$ rotation), the roller travels forward by a linear distance equal to its circular base circumference ($2\\pi r$).
• The ground area compacted in 1 revolution is equal to its curved surface area: $$\\text{Area leveled in 1 revolution} = 2\\pi r h$$
• In $n$ complete revolutions, the total ground area leveled is: $$\\text{Total Area Leveled} = n \\times 2\\pi r h$$

A classic Madhyamik examination problem involves digging a cylindrical well of radius $r_1$ and depth $h_1$:

1. Volume of earth excavated: $$V_{\\text{earth}} = \\pi r_1^2 h_1$$
2. This earth is spread evenly all around the mouth of the well to a width $w$ to form an embankment of height $H$.
3. The embankment forms a hollow cylinder with inner radius $r = r_1$ and outer radius $R = r_1 + w$.
4. Area of the annular embankment base $= \\pi(R^2 - r^2) = \\pi[(r_1 + w)^2 - r_1^2]$.
5. By conservation of excavated volume: $$\\pi(R^2 - r^2) \\times H = \\pi r_1^2 h_1 \\implies H = \\frac{r_1^2 h_1}{R^2 - r^2}$$