Sphere

In Euclidean space, a sphere (গোলক) is a perfectly symmetrical three-dimensional solid bounded by a single continuous curved surface such that every point on the surface is at an equal distance from a fixed point inside it called the center ($O$).

• Generation: A sphere is generated by rotating a planar semicircle through $360^\circ$ (one full revolution) about its fixed diameter as the axis of rotation.
• Radius ($r$): The distance from the center $O$ to any point on the spherical surface.
• Diameter ($d$): Any straight line segment passing through the center $O$ with both endpoints lying on the spherical surface: $d = 2r$.
• Great Circle (মহাবৃত্ত): The circular intersection formed when a plane passes directly through the center of the sphere. The radius of a great circle equals the radius of the sphere ($r$), and its area is $\pi r^2$. Every other plane slicing the sphere produces a small circle of radius $< r$.

Archimedes' Cylinder Proof: Archimedes of Syracuse proved that if a sphere of radius $r$ is enclosed tightly inside a circumscribed right circular cylinder of radius $r$ and height $h = 2r$, the curved surface area of the cylinder is: $$\text{CSA of Cylinder} = 2\pi r h = 2\pi r (2r) = 4\pi r^2$$ Remarkably, the surface area of the sphere is identically equal to the curved surface area of this circumscribed cylinder!

The cubic volume of space enclosed by a sphere of radius $r$ is given by: $$\mathbf{\text{Volume } V = \frac{4}{3}\pi r^3}$$ In terms of diameter $d = 2r$: $$V = \frac{4}{3}\pi \left(\frac{d}{2}\right)^3 = \frac{4}{3}\pi \frac{d^3}{8} = \frac{1}{6}\pi d^3$$ Archimedes proved that the volume of the sphere is exactly $\frac{2}{3}$ of the volume of the circumscribed cylinder: $$\text{Volume of Cylinder} = \pi r^2 (2r) = 2\pi r^3 \implies \frac{2}{3}(2\pi r^3) = \frac{4}{3}\pi r^3$$

A hollow sphere (e.g., hollow iron cannonball, metal float ball) is bounded by two concentric spherical surfaces with the same center $O$ but different radii:

• Internal radius ($r$): Radius of the inner spherical hollow cavity.
• External radius ($R$): Radius of the outer spherical boundary ($R > r$).
• Thickness ($t$): Thickness of the spherical wall: $t = R - r$.
• Volume of Material / Metal: $$V_{\text{metal}} = \text{External Volume} - \text{Internal Volume} = \mathbf{\frac{4}{3}\pi (R^3 - r^3)}$$
• Mass of Hollow Sphere: $\text{Mass} = V_{\text{metal}} \times \text{Density} = \left[\frac{4}{3}\pi (R^3 - r^3)\right] \times \rho$.
• External Surface Area: $4\pi R^2$; Internal Surface Area: $4\pi r^2$.

When a solid sphere of radius $r$ is cut into two equal halves by a plane passing through its center $O$, each half is called a solid hemisphere (নিরেট অর্ধগোলক):

• 1. Curved Surface Area (বক্রতলের ক্ষেত্রফল): Exactly half the surface area of the original sphere: $$\text{CSA} = \frac{1}{2}(4\pi r^2) = \mathbf{2\pi r^2}$$
• 2. Flat Circular Base Area (সমতলের ক্ষেত্রফল): The cut exposes a planar circular base of radius $r$: $$\text{Base Area} = \mathbf{\pi r^2}$$
• 3. Total Surface Area of Solid Hemisphere (সমগ্রতলের ক্ষেত্রফল): $$\text{TSA} = \text{Curved Surface Area} + \text{Flat Base Area} = 2\pi r^2 + \pi r^2 = \mathbf{3\pi r^2}$$
• 4. Volume of Solid Hemisphere: Exactly half the volume of the original sphere: $$\text{Volume} = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) = \mathbf{\frac{2}{3}\pi r^3}$$

Consider an open hollow hemispherical bowl with inner radius $r$ and outer radius $R$:

• Outer Curved Surface: $2\pi R^2$
• Inner Curved Surface: $2\pi r^2$
• Circular Rim (Annular Ring at Top): $\pi R^2 - \pi r^2 = \pi(R^2 - r^2)$
• Total Surface Area of Open Bowl: $$\text{TSA} = 2\pi R^2 + 2\pi r^2 + \pi(R^2 - r^2) = \mathbf{\pi(3R^2 + r^2)}$$
• Volume of Metal in Bowl: $$V = \mathbf{\frac{2}{3}\pi (R^3 - r^3)}$$

When solid metal objects are melted and recast into a different solid shape, the total volume remains strictly constant (assuming zero loss of metal):

• Case 1: Three small spheres melted into one large sphere: $$\frac{4}{3}\pi r_1^3 + \frac{4}{3}\pi r_2^3 + \frac{4}{3}\pi r_3^3 = \frac{4}{3}\pi R^3 \implies \mathbf{R^3 = r_1^3 + r_2^3 + r_3^3}$$
• Case 2: One large sphere melted into $N$ identical small spheres: $$\frac{4}{3}\pi R^3 = N \times \left(\frac{4}{3}\pi r^3\right) \implies \mathbf{N = \left(\frac{R}{r}\right)^3}$$
• Case 3: Sphere recast into a right circular cone or cylinder: $$\frac{4}{3}\pi r_{\text{sph}}^3 = \pi r_{\text{cyl}}^2 h_{\text{cyl}}$$

When a solid sphere of radius $r$ is completely submerged in water inside a right circular cylindrical vessel of base radius $R_{\text{cyl}}$:

• By Archimedes' Principle, the volume of water displaced equals the volume of the submerged sphere.
• The displaced water forms a cylindrical volume of height $h$ and radius $R_{\text{cyl}}$: $$\pi R_{\text{cyl}}^2 \times h = \frac{4}{3}\pi r^3 \implies \mathbf{h = \frac{4 r^3}{3 R_{\text{cyl}}^2}}$$