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The reason I believe that the relationship between the first element (Hydrogen) in the periodic table, and the second, (Helium) is 1 to 4 (Hydrogen has one proton, and Helium has 2 protons and 2 neutrons, bringing its mass to a ratio of 1 to 4, is because of Coloomb's law

A sphere can also be defined as the surface formed by rotating a circle about any diameter. If the circle is replaced by an ellipse, and rotated about the major axis, the shape becomes a prolate spheroid, rotated about the minor axis, an oblate spheroid.

Surface area of a sphere

The surface area of a sphere is given by the following formula

$\!A = 4\pi r^2.$

This formula was first derived by Archimedes, based upon the fact that the projection to the lateral surface of a circumscribing cylinder (i.e. the Lambert cylindrical equal-area projection) is area-preserving. It is also the derivative of the formula for the volume with respect to r because the total volume of a sphere of radius r can be thought of as the summation of the surface area of an infinite number of spherical shells of infinitesimal thickness concentrically stacked inside one another from radius 0 to radius r. At infinitesimal thickness the discrepancy between the inner and outer surface area of any given shell is infinitesimal and the elemental volume at radius r is simply the product of the surface area at radius r and the infinitesimal thickness.

At any given radius r, the incremental volume (δV) is given by the product of the surface area at radius r (A(r)) and the thickness of a shell (δr):

$\delta V \approx A(r) \cdot \delta r. \,$

The total volume is the summation of all shell volumes:

$V \approx \sum A(r) \cdot \delta r.$

In the limit as δr approaches zero^[1] this becomes:

$V = \int_0^r A(r) \, dr.$

Since we have already proved what the volume is, we can substitute V:

$\frac{4}{3}\pi r^3 = \int_0^r A(r) \, dr.$

Differentiating both sides of this equation with respect to r yields A as a function of r:

$\!4\pi r^2 = A(r).$

Which is generally abbreviated as:

$\!A = 4\pi r^2.$

Alternatively, the area element on the sphere is given in spherical coordinates by $dA = r^2 \sin\theta\, d\theta\, d\phi.$ . With Cartesian coordinates, the area element $dS=\frac{r}{\sqrt{r^{2}-\sum_{i\ne k}x_{i}^{2}}}\Pi_{i\ne k}dx_{i},\;\forall k$ . More generally, see area element.

The total area can thus be obtained by integration:

$A = \int_0^{2\pi} \int_0^\pi r^2 \sin\theta \, d\theta \, d\phi = 4\pi r^2.$

Equations in R³

In analytic geometry, a sphere with center (x₀, y₀, z₀) and radius r is the locus of all points (x, y, z) such that

$\, (x - x_0 )^2 + (y - y_0 )^2 + ( z - z_0 )^2 = r^2.$

The points on the sphere with radius r can be parametrized via

$\, x = x_0 + r \sin \theta \; \cos \varphi$

$\, y = y_0 + r \sin \theta \; \sin \varphi \qquad (0 \leq \varphi \leq 2\pi \mbox{ and } 0 \leq \theta \leq \pi ) \,$

$\, z = z_0 + r \cos \theta \,$

A sphere of any radius centered at zero is an integral surface of the following differential form:

$\, x \, dx + y \, dy + z \, dz = 0.$

This equation reflects the fact that the position and velocity vectors of a point travelling on the sphere are always orthogonal to each other.

The sphere has the smallest surface area among all surfaces enclosing a given volume and it encloses the largest volume among all closed surfaces with a given surface area. For this reason, the sphere appears in nature: for instance bubbles and small water drops are roughly spherical, because the surface tension locally minimizes surface area. The surface area in relation to the mass of a sphere is called the specific surface area. From the above stated equations it can be expressed as follows:

$SSA = \frac{A}{V\rho} = \frac{3}{r\rho}.$

protons & neutrons

Surface area of a sphere

Equations in R3

Equations in R³