# Supplementary mathematics/sphere

A sphere is a geometric object that is similar to a two-dimensional circle in three dimensions. The sphere is a set of points in space that are at a certain distance from a point to the center of the sphere, which is called the radius of the circle. The radius of the sphere is equal to bar. A sphere is a geometric volume and is considered a type of spherical-polyhedral object.

## BASIC MODIFICATIONS

The radius of the sphere is denoted by R. The radius of the sphere extends from the center of the circle to the points of the sphere. If we extend the radius from the center of the circle of the sphere to a point of the sphere that bisects the base, the diameter is created. The diameter is twice the radius, which is denoted by the symbol 2r. Planets, stars, the earth and the sun are spherical and are similar to each other. These heavenly bodies do not have the same radius, and because they are spheres, they are similar to each other. Mathematicians consider the sphere to be a two-dimensional closed surface embedded in Euclidean three-dimensional space. They distinguish a sphere and a ball, which is a three-dimensional manifold with a boundary containing the volume contained in the sphere. An "open ball" excludes the sphere itself, while a "closed ball" includes the sphere: a closed ball is a combination of the open ball and the sphere, and a sphere is the boundary of a ball (closed or open). The distinction between "ball" and "sphere" has not always been preserved, and especially older mathematical sources speak of the sphere as a solid. The distinction between "circle" and "disc" is similar in the plane.

Small spheres are sometimes called spheroids, for example Mars.

### Tips about sphere

1. The volume of the sphere is two thirds of the volume of the cylinder
2. Its area is four times the area of the circle.
3. Korea is a multifaceted type.
4. The sphere is of the geometric type.
5. If we multiply the volume of two spheres with the same radius, the result is equal to four times the volume of the sphere by the square of the radius of the two spheres.
6. If we multiply the area of two spheres with the same radius, the result is equal to four times the area of the sphere by the square radius of the two spheres.
7. The sphere is used in 3D coordinates.

## Equations

In analytical geometry, let us have a sphere with center (x0,y0,z0) and the coordinates of the sphere (x,y,z) be equal to the location of the entire radius of the sphere. It can be expressed as a quadratic polynomial, a sphere is a quadrilateral surface, a kind of algebraic surface. (x - x_0)^2 + (y - y_0)^2 + (z - z_0) ^2 = r^2.[/itex]

Let's assume that a, b, c, d, e are real numbers with "a" ≠ 0 and put them in the equation of the first degree: x_0 = \frac{-b}{a}, \quad y_0 = \frac {-c}{a}, \quad z_0 = \frac{-d}{a}, \quad \rho = \frac{b^2 +c^2+d^2 - ae}{a^2}. [/itex]Then the equation is equal to${\displaystyle f(x,y,z)=a(x^{2}+y^{2}+z^{2})+2(bx+cy+dz)+e=0}$

### Parametric equation

If we obtain the equation of the sphere as a parameter so that the radius of the sphere is equal to ${\displaystyle r>0}$ and the center of our sphere is equal to ${\displaystyle (x_{0},y_{0},z_{0})}$ as follows is written

{\displaystyle {\begin{aligned}x&=x_{0}+r\sin \theta \;\cos \varphi \\y&=y_{0}+r\sin \theta \;\sin \varphi \\z&=z_{0}+r\cos \theta \,\end{aligned}}}

The symbols used here are the same symbols used in spherical coordinates. r is constant, while θ varies from 0 to π and ${\displaystyle \varphi }$ also varies from 0 to ${\displaystyle 2\pi }$.r

## ballpoint pencil(Spherical)

A sphere is uniquely defined by four points that are not coplanar. In general, a sphere is uniquely determined by four conditions such as passing through a point, being tangent to a plane, etc. This property is similar to the property that three non-linear points form a unique circle in a plane. Set. Consequently, a sphere is determined exclusively by a circle and a point that is not in the plane of that circle (i.e., passes through it). By examining the common solutions of the equations of two spheres, it can be seen that two spheres intersect in a circle and the plane containing it is called a radical circle. Page ' Crossed Spheres. Although the radical plane is a real plane, the circle may be imaginary (the spheres have no real common point) or consist of a point (the spheres are tangent at that point).

f(x,y,z) = 0 and g(x,y,z ) = 0

Then the equations of the two spheres are separate, so: ${\displaystyle \lambda f(x,y,z)+\mu g(x,y,z)=0}$

It is also the equation of a sphere for arbitrary values of parameters λ and μ. The set of all spheres that satisfy this equation is called a "pencil of spheres", defined by the two principal spheres. In this definition, the sphere is allowed to be a plane (radius at infinity, center at infinity) and all pencil spheres are planes if both principal spheres are planes, otherwise only one plane (radical plane) exists. has it. Pencil.

## properties

A sphere can be constructed as a surface formed by revolving a circle around each of its diameters. This is basically the traditional definition of sphere as given in Euclid's Elements. Since a circle is a special type of ellipse, a sphere is also a special type of ellipse of revolution. By replacing the circle with an ellipse that is rotated around its main axis, the shape becomes a prolate sphere. Rotation about the minor axis, an oblique sphere.

A sphere is uniquely defined by four points that are not coplanar. More generally, a sphere is uniquely defined by four conditions such as passing through a point, being tangent to a plane, etc.

Consequently, a sphere is determined exclusively by a circle and a point that is not in the plane of that circle (i.e., passes through it).

By examining the common solutions of the equations of two spheres, it can be seen that two spheres intersect in a circle and the plane containing that circle is called the "radical plane" of intersecting spheres. Although the radical plane is a real plane, the circle may be imaginary (the spheres have no real common point) or consist of a point (the spheres are tangent at that point).

The angle between two spheres at a real point of intersection is the dihedral angle determined by the planes tangent to the sphere at that point. Two spheres intersect at an angle at all points of their circle of intersection. They intersect at right angles (are orthogonal) if and only if the square of the distance between their centers equals the sum of the squares of their radii.

### Eleven Properties of Butter

These properties have been described by two European mathematicians named David Hilbert and Stephen Cohen-Wossen in their geometry book, and these properties have been proven.

1. Points on the sphere are all distant from a fixed point. Also, the ratio of the distance of its points from two fixed points is constant.
The first part is the usual definition of Korea and defines it uniquely. The second part is easily deduced and follows the same result of Apollonius Perga for the circle. This second part is also true for airplanes.
2. "The lines and sections of the plane of the sphere are circles."
This attribute uniquely defines the sphere.
3. "The sphere has constant width and constant circumference."
The width of a surface is the distance between pairs of parallel tangent planes. Many other closed convex surfaces have constant width, for example the Meissner field. The circumference of a surface is the perimeter of the boundary of its orthogonal projection on a plane. Each of these features implies the other.
4. "All the points of a sphere are navel."
at any point of the surface a direction normal to the surface is at right angles because in the sphere these are the lines that go out from the center of the sphere. The intersection of a plane that is normal to the surface forms a curve, which is called a "normal" section, and the curvature of this curve is "normal curvature". For most points on most surfaces, different segments have different curvatures. The maximum and minimum values are called the main curve. Each closed surface has at least four points called "navel points". In a navel, all cross-sectional curves are equal. In particular, the principal curves are equal. The navel points can be considered as the points whose surface is approximately close to a sphere.
For the sphere, the curvature of all normal segments is equal, so every point is a navel. Sphere and plane are the only surfaces that have this property.
5. "The sphere does not have a surface of centers."
For a normal section, there is a circle of curvature equal to the curvature of the section, tangent to the surface, and whose center lines lie along the normal line. For example, the two centers corresponding to the maximum and minimum curvature of the section are called "focal points" and the collection of all these centers constitutes the focal surface.
For most planes, the focal plane forms two planes, each one plane, meeting at the navel points. A few special ones are:
* For channel surfaces, one sheet is a curve and the other sheet is a surface
* For cones, cylinders, nets and cichlids, both form curved sheets.
* For a sphere, the center of each reflecting circle is at the center of the sphere and the focal plane forms a single point. This feature is unique to Korea.
6. "All geodesics on the sphere are closed curves."
Geodesics are curves on a surface that show the shortest distance between two points. They are generalizations of the concept of a straight line on a page. For Korea, geodesics are great circles. Many other levels share this feature.

Among all the solids that have a certain volume, the sphere has the smallest surface area. Among all the solids that have a specific surface, the sphere has the largest volume.

1. It follows from the isoperimetric inequality. These properties uniquely define a sphere and can be seen in soap bubbles: a soap bubble occupies a fixed volume and minimizes the surface tension of its surface for that volume. So a free-floating soap bubble approaches a sphere (although external forces such as gravity change the shape of the bubble slightly). It is also seen in planets and stars that minimize the surface gravity of large celestial bodies.
2. "The sphere has the smallest average total curvature among all convex solids with a given surface."
The average curvature is the average of the two principal curvatures, which is constant because the two principal curvatures are constant at all points on the sphere.
3. The sphere has a constant average curvature.
The sphere is the only latent surface that has no boundary or singularity with constant positive mean curvature. Other immersed surfaces have a constant average curvature, such as minimum surfaces.
4. "The sphere has a constant positive Gaussian curvature."
Gaussian curve is the product of two original curves. This is an intrinsic property that can be determined by measuring lengths and angles and is independent of how the surface is positioned in space. Hence, bending a surface does not change the Gaussian curvature, and other surfaces with constant positive Gaussian curvature can be obtained by cutting a small slot in the sphere and bending it. All of these other surfaces will have a boundary, and the sphere is the only surface that does not have a boundary with constant positive Gaussian curvature. A pseudosphere is an example of a surface with constant negative Gaussian curvature.
5. "The sphere is transformed by a three-parameter family of rigid motions."
Rotation about any axis of a single sphere at the origin maps the sphere onto itself. Any rotation around a line from the origin can be expressed as a combination of rotations around three coordinate axes (See Euler's angles). Thus, there exists a three-parameter family of rotations such that each rotation transforms the sphere into itself. This family is the rotation group SO(3). The plane is the only other surface that has a three-parameter transformation family (translation along the x - and y -axes and rotation around the origin). Circular cylinders are the only surfaces with two-parameter families of rigid motions, and rotary and helicoidal surfaces are the only surfaces with one-parameter families.

## curve on a sphere

### circular curve (ring)

Circles on a sphere are like circles on a plane consisting of all points at a certain distance from a fixed point on the sphere. The intersection of a sphere and a plane is a circle, a point or a void. Great circles are the intersection of a sphere with a plane passing through the center of a sphere: the rest are called small circles.

More complex surfaces may also intersect a sphere circularly: the intersection of a sphere with a surface of revolution whose axis includes the center of the sphere (they are also axis) consists of circles and/or points if they are not empty. Is. For example, the diagram on the right shows the intersection of a sphere and a cylinder, which consists of two circles. If the radius of the cylinder is the radius of the sphere, the intersection will be a unit circle. If the radius of the cylinder was greater than the radius of the sphere, the intersection was empty.

### Loxodrome

In navigation, a rhombus or loxodrome is an arc that crosses all meridians of longitude at the same angle. The loxodrome is the same as the straight lines in Mercator's plan. A rhombus is not a spherical spiral. Except for some simple ones, the roman script formula is complicated.

### overall curve

A Clelia curve is a curve on a sphere whose longitude ${\displaystyle \varphi }$ and correlation ${\displaystyle \theta }$ satisfy Eq.

${\displaystyle \varphi =c\;\theta ,\quad c>0}$

Special cases are: the Vivian curve (${\displaystyle c=1}$) and spherical spirals (${\displaystyle c>2}$) such as the Seyfert spiral. All curves approximate the path of satellites in polar orbit.

### spherical cones

The analog of a conic section on a sphere is a spherical conic, a quartic curve that can be defined in several equivalent ways, including:

• as the point of intersection of a sphere with a quadratic cone whose vertex is the center of the sphere.
• as the intersection of a sphere with an elliptical or hyperbolic cylinder whose axis passes through the center of the sphere.
• As the locus of points that are the sum or difference of great circle distances from a pair of fixed foci.

Many theorems related to planar conic sections are extended to spherical cones as well.

### Intersection of a sphere with a more general surface

If one sphere intersects another surface, more complex spherical curves may exist.

#### Example

Sphere-cylinder intersection

The intersection of the sphere with the equation${\displaystyle \;x^{2}+y^{2}+z^{2}=r^{2}\;}$and the cylinder with equation ${\displaystyle \;(y-y_{0})^{2}+z^{2}=a^{2},\;y_{0}\neq 0\;}$Not just one or two circles. Solving the nonlinear system of equations

${\displaystyle x^{2}+y^{2}+z^{2}-r^{2}=0}$
${\displaystyle (y-y_{0})^{2}+z^{2}-a^{2}=0\ .}$

## Volume

The volume of the sphere is written based on this relationship. ${\displaystyle \!V={\frac {4}{3}}\pi r^{3}}$In this relation, r is the radius of the sphere and π is the pi number. This relation was first obtained by Archimedes. He showed that the volume of a sphere is 2/3 of the volume of its peripheral cylinder. The volume of the sphere is written based on this relationship.

### Prove the volume of sphere

We enclose a sphere in a cylinder whose diameter and height are equal. First, we divide the butter into two hemispheres. If we melt the hemisphere three times and pour it into the cylinder, the volume of the cylinder will be filled. So the volume of the hemisphere is one third of the volume of the cylinder and the volume of the sphere is two thirds of the volume of the cylinder.

Cylinder volume=

${\displaystyle V=(\pi r^{2})(2r)=2\pi r^{3}}$

If we multiply the ratio of the volume of the sphere to the volume of the cylinder by the volume of the cylinder, the volume of the sphere is obtained

${\displaystyle V={\frac {2}{3}}2\pi r^{3}={\frac {4}{3}}\pi r^{3}}$

where r is the radius and d is the diameter of the sphere. Archimedes first derived this formula by showing that the volume inside a sphere is twice the volume between the sphere and its enclosing cylinder (with the same height and diameter as the sphere). This can be proved by inscribing a cone upside down in the hemisphere, noting that the cross-sectional area of the cone plus the cross-sectional area of the sphere equals the cross-sectional area of the sphere. Limiting cylinder, and using the Cavalieri principle. This formula can also be obtained using integral calculus, i.e. disc integration for the sum of infinitesimal volumes of circular discs of infinitesimal thickness stacked side by side along the x-axis of x. = - r to x = r, assuming that the sphere of radius r is centered. Lineage.

### Proving the volume of the sphere by integral and trigonometric methods

First, we get the volume of the hemisphere, and since the sphere is symmetrical, the volume of a complete sphere is twice the volume of the hemisphere. Suppose this sphere is made of countless circular disks of very thin thickness. The sum (integral) of the volume of these disks makes the volume of the desired sphere. The axis of all these disks is located on the "y" axis, as a result, the disk that is located on the point "h = 0" has a radius equal to "r" ("s = r"). And the disk located at the point "h = r" has a radius equal to zero ("s = 0").

If the thickness of the discs at any desired point "h" is equal to "δh", then the volume of the disc will be equal to the cross-sectional area of the disc at its thickness:

${\displaystyle \!\delta V\approx \pi s^{2}\cdot \delta h.}$

So the total volume of the hemisphere is equal to the total volume of the disks:

${\displaystyle \!V\approx \sum \pi s^{2}\cdot \delta h.}$

Above the sphere, the radius of the disks is very small and close to zero. As a result, in order to obtain the total volume of the disks, one must take the integral from the above relation:

${\displaystyle \!V=\int _{0}^{r}\pi s^{2}dh.}$

According to the Pythagorean theorem, we know that at any point on the vertical axis we have:

${\displaystyle \!r^{2}=s^{2}+h^{2}}$

So, instead of ${\displaystyle s^{2}}$, we use the following relation which is a function of the value of ${\displaystyle h^{2}}$:

${\displaystyle \!r^{2}-h^{2}=s^{2}}$

We put the new value of ${\displaystyle s^{2}}$ in the integral:

${\displaystyle \!V=\int _{0}^{r}\pi (r^{2}-h^{2})dh.}$

The value of the integral is equal to:

${\displaystyle \!V=\pi \left[r^{2}h-{\frac {h^{3}}{3}}\right]_{0}^{r}=\pi \left(r^{3}-{\frac {r^{3}}{3}}\right)-\pi \left(-0^{3}+{\frac {0^{3}}{3}}\right)={\frac {2}{3}}\pi r^{3}.}$

The volume of half of the sphere is equal to ${\displaystyle \!V={\frac {2}{3}}\pi r^{3}}$, so the volume of the whole sphere is:

${\displaystyle \!V={\frac {4}{3}}\pi r^{3}.}$

The volume of the sphere can also be calculated in the polar coordinate system, in which case the following relationship should be used:

${\displaystyle \mathrm {d} V=r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi }$

Therefore we have

${\displaystyle V=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{r}r'^{2}\sin \theta \,dr'\,d\theta \,d\varphi =2\pi \int _{0}^{\pi }\int _{0}^{r}r'^{2}\sin \theta \,dr'\,d\theta =4\pi \int _{0}^{r}r'^{2}\,dr'\ ={\frac {4}{3}}\pi r^{3}.}$

For most practical purposes, the volume inside a sphere inscribed in a cube can be approximated as 52.4% of the volume of the cube, since V =π/6 d 3 where d The diameter of the sphere and the length of one side of the cube are π/6 ≈ 0.5236. For example, a sphere with a diameter of 1 meter has 52.4% of the volume of a cube with an edge length of 1 meter or about 0.524 cubic meters.

### Another way [1]

Based on a reflection of mathematicians Archimedes of Greece, for There is a hemisphere of radius ${\displaystyle r}$ of the reference field whose volume corresponds to the hemisphere, but the calculation is easy. This body compares to the transformation of a cylinder (more precisely: a right circular cylinder) with base surface radius ${\displaystyle r}$ and height ${\displaystyle r}$ is omitted to a cone (more precisely: right circular cone) with base radius ${\displaystyle r}$ and height ${\displaystyle r}$. I think it is wrong, that's why I didn't continue

Cavalieri's principle can be used to prove that the hemisphere and the comparison object have the same volume. This principle is based on the idea of dividing the considered objects into infinite slices of infinitesimal thickness (infinitely small). (An alternative to this method would be to use integral calculus). According to the mentioned principle, the intersection surfaces of both objects with plane which parallel and a distance ${\displaystyle h}$ is given from it. In the case of a hemisphere, the intersection is a circular area.

The radius ${\displaystyle s}$ of this circular area is obtained from the Pythagorean Theorem:

${\displaystyle s^{2}+h^{2}\,=\,r^{2}}$.

This gives the content of the interface

${\displaystyle A_{1}\,=\,\pi s^{2}=\pi (r^{2}-h^{2})=\pi r^{2}-\pi h^{2}}$.

On the other hand, in the case of the reference body, the intersection is a circular ring with an outer radius ${\displaystyle r}$ and an inner radius ${\displaystyle h}$. The area of this intersection is therefore:

${\displaystyle A_{2}\,=\,\pi r^{2}-\pi h^{2}}$.

For an arbitrary distance ${\displaystyle h}$ from the base region, the two intersection regions agree in region content. This follows from Cavalieri's principle that the hemisphere and the comparison body have the same volume.

Now it is possible to compare the volume of the body and thus calculate the hemisphere easily:

One less than the volume of the cylinder and one more than the volume of the cone.

${\displaystyle V_{\text{Zylinder}}=\pi r^{2}\cdot r=\pi r^{3}}$
${\displaystyle V_{\text{Kegel}}={\frac {1}{3}}\pi r^{2}\cdot r={\frac {1}{3}}\pi r^{3}}$
${\displaystyle V_{\text{Halbkugel}}=V_{\text{Vergleichskörper}}\,=\pi r^{3}-{\frac {1}{3}}\pi r^{3}={\frac {2}{3}}\pi r^{3}}$

Therefore, the following applies to the volume of the (solid) sphere:

${\displaystyle V_{\text{Kugel}}\,=\,2\cdot V_{\text{Halbkugel}}={\frac {4}{3}}\pi r^{3}}$.

#### Alternate derivation

The sphere can be divided into an infinite number of pyramids of height ${\displaystyle r}$ (vertice at the center of the sphere) whose entire base coincides with the surface of the sphere (see below). Therefore, the total volume of all pyramids is equal to:

${\displaystyle V={\frac {O\,r}{3}}={\frac {(4\pi r^{2})r}{3}}={\frac {4}{3}}\pi r^{3}}$.

#### Derivation using integral calculus

Radius at distance ${\displaystyle x}$:

${\displaystyle s={\sqrt {r^{2}-x^{2}}}}$.

Circular area at distance ${\displaystyle x}$:

${\displaystyle A_{x}=\pi s^{2}}$.

Volume of sphere ${\displaystyle V}$:

${\displaystyle V=\int _{-r}^{r}{A_{x}\,\mathrm {d} x}=\int _{-r}^{r}{\pi s^{2}\,\mathrm {d} x}=\int _{-r}^{r}{\left({r^{2}-x^{2}}\right)}\pi \,\mathrm {d} x=\int _{-r}^{r}\pi {r^{2}}\,\mathrm {d} x-\int _{-r}^{r}\pi {x^{2}}\,\mathrm {d} x}$
${\displaystyle V=\pi r^{2}\left[x\right]_{-r}^{r}-{\frac {1}{3}}\pi \left[{x^{3}}\right]_{-r}^{r}}$
${\displaystyle V=\pi r^{2}\left[r-(-r)\right]-{\frac {1}{3}}\pi \left[r^{3}-(-r)^{3}\right]=2\pi r^{3}-{\frac {2}{3}}\pi r^{3}={\frac {4}{3}}\pi r^{3}}$.

The volume of a spherical piece ${\displaystyle V_{\mathrm {KS} }}$ of height ${\displaystyle h}$ can be calculated in the same way:

${\displaystyle V_{\mathrm {KS} }=\int _{r-h}^{r}{A_{x}\,\mathrm {d} x}=\pi r^{2}\left[x\right]_{r-h}^{r}-{\frac {1}{3}}\pi \left[{x^{3}}\right]_{r-h}^{r}}$
${\displaystyle V_{\mathrm {KS} }=\pi r^{2}\left[r-(r-h)\right]-{\frac {1}{3}}\pi \left[r^{3}-(r-h)^{3}\right]=\pi r^{2}h-{\frac {1}{3}}\pi \left[r^{3}-(r^{3}-3r^{2}h+3rh^{2}-h^{3})\right]}$
${\displaystyle V_{\mathrm {KS} }=\pi r^{2}h-\pi r^{2}h+\pi rh^{2}-{\frac {1}{3}}\pi h^{3}={\frac {\pi h^{2}}{3}}(3r-h)}$.

#### More derivatives

A sphere with radius ${\displaystyle R}$ centered at the origin can be represented by the equation:

${\displaystyle K:~x^{2}+y^{2}+z^{2}\leq R^{2}}$

It is explained where ${\displaystyle x,y,z}$ are the distance coordinates.

This problem can be solved in two ways using integral calculus:

We parametrize the sphere into a smooth null set${\displaystyle {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}r~\sin \vartheta ~\cos \varphi \ \ r~\sin \vartheta ~\sin \varphi \\r~\cos \vartheta \end{pmatrix}}\qquad (0\leq r\leq R,\ 0\leq \vartheta \leq \pi ,\ 0\ leq\varphi \leq 2\pi )}$.

With functional specification:

${\displaystyle \det {\frac {\partial (x,y,z)}{\partial (r,\vartheta ,\varphi )}}=r^{2}\sin \vartheta }$

The required volume element is ${\displaystyle \mathrm {d} V}$ as a result:

${\displaystyle \mathrm {d} V=r^{2}\sin \vartheta \;\mathrm {d} r\,\mathrm {d} \varphi \,\mathrm {d} \vartheta }$.

Therefore, the volume of the sphere is given as:

{\displaystyle {\begin{aligned}\int _{K}\mathrm {d} V&=\int _{0}^{\pi }\int _{0}^{2\pi }\int _{0}^{R}r^{2}\sin \vartheta \;\mathrm {d} r\,\mathrm {d} \varphi \,\mathrm {d} \vartheta \\&=\int _{0}^{R}r^{2}\mathrm {d} r\int _{0}^{2\pi }\mathrm {d} \varphi \int _{0}^{\pi }\sin \vartheta \;\mathrm {d} \vartheta \\&={\frac {R^{3}}{3}}\cdot 2\pi \cdot 2\\&={\frac {4}{3}}\pi R^{3}.\\\end{aligned}}}

Another possibility is through polar coordinates:

{\displaystyle {\begin{aligned}\int _{K}\!\mathrm {d} V&=\iint \limits _{x^{2}+y^{2}\leq R^{2}}\left(\int \limits _{-{\sqrt {R^{2}-x^{2}-y^{2}}}}^{\ sqrt{R^{2}-x^{2}-y^{2}}}\mathrm {d} z\right)\mathrm {d} y\,\mathrm {d} x\\&=\iint \limits _{x^{2}+y^{2}\leq R^{2}}2{\sqrt {R^{2}-x^{2}-y^{2}}}\,\mathrm {d} y\,\mathrm {d} x.\end{aligned}}}

Cartesian coordinate system now converts to polar coordinate system, which means integration after changing the coordinate system using ${\displaystyle \,\!\varphi }$ and ${\displaystyle variables>\,\!r}$ instead of ${\displaystyle \,\!x}$ and ${\displaystyle \,\!y}$. The motivation of this development is the significant simplification of the calculation in the next period. For differential this means: ${\displaystyle \mathrm {d} y\,\mathrm {d} x\;\xrightarrow {\text{becomes}} \;r\,\mathrm {d} r\,\ mathrmd\varphi }$

{\displaystyle {\begin{aligned}\int _{K}\!\mathrm {d} V&=\int \limits _{0}^{2\pi }\int \limits _{0}^{R}2{\sqrt {R^{2}-r^{2}}}\;r\,\mathrm {d} r\,\mathrm {d} \varphi \\&=2\pi \int \limits _{0}^{R}2{\sqrt {R^{2}-r^{2}}}\;r\,\mathrm {d} r\\&=2\pi (-1){\frac {2}{3}}\left[{\sqrt {(R^{2}-r^{2})^{3}}}\right]_{r=0}^{R}\\&={\frac {4}{3}}\pi R^{3}.\end{aligned}}}

Another way with the help of the formula of bodies of revolution:

If you rotate a piece of surface around a fixed spatial axis, you get a body with a certain volume. In the case of a circular area, a sphere is formed. This can be visualized as a flipping coin.

The general formula for the rotation of a body of revolution around the x-axis gives:

${\displaystyle V=\pi \int _{a}^{b}[f(x)]^{2}\mathrm {d} x=\pi \int _{a}^{b}y^{2}\,\mathrm {d} x[/itex].Theequationiscircular::(x-x_{M})^{2}+(y-y_{M})^{2}\,=\,r^{2}}$ with center: ${\displaystyle M={\begin{pmatrix}x_{M}\\y_{M}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}}$. Plugging in the equation of the circle, we get ${\displaystyle x^{2}+y^{2}=r^{2}\Leftrightarrow y^{2}=r^{2}-x^{2}}$. Substituting the formula for rotating bodies around the x axis, we get: {\displaystyle {\begin{aligned}V_{\text{Kugel}}&=\pi \int _{-r}^{r}\left(r^{2}-x^{2}\right)\,\mathrm {d} x\\&=\pi \left[r^{2}x-{\frac {1}{3}}x^{3}\right]_{-r}^{r}\\&=\pi \left(r^{3}-{\frac {1}{3}}r^{3}\right)-\pi \left(r^{2}\cdot (-r)-{\frac {1}{3}}(-r)^{3}\right)\\&=\pi \left[\left({\frac {2}{3}}r^{3}\right)-\left(-{\frac {2}{3}}r^{3}\right)\right]\\&={\frac {4}{3}}\pi r^{3}.\\\end{aligned}}} ## Area The area of the sphere comes from the following relationship: ${\displaystyle \!A=4\pi r^{2}.}$ Archimedes was the first to calculate the area of the sphere. The derivative of the volume of the sphere with respect to "r", the radius of the sphere, gives the area of the sphere. Mazwan Qane Gnhhur Khare Kh Hajj Hajj one one one one k k k k arearaber e 'essathahahahahahahahahahahahahahahahahr kehnehahahahahaha Mohammad metatat bart. As a result, if you represent the volume of each sphere with "dV", the thickness of each shell with "r" and the area of each spherical shell with radius "r" with (A(r), the following relationship between these variations will be: ${\displaystyle \delta V\approx A(r)\cdot \delta r\,}$ The total volume is equal to the sum of the volume of each of these shells: ${\displaystyle V\approx \sum A(r)\cdot \delta r}$ When "r" tends to zero, we should use integral instead of sigma: ${\displaystyle V=\int _{0}^{r}A(r)\,dr}$ Because he had previously given the formula for the volume of butter, he had: ${\displaystyle {\frac {4}{3}}\pi r^{3}=\int _{0}^{r}A(r)\,dr}$ We derive from the two ends of the above relationship: ${\displaystyle \!4\pi r^{2}=A(r)}$ which is generally written as follows: ${\displaystyle \!A=4\pi r^{2}}$ In the polar coordinate system, the surface component is as ${\displaystyle dA=r^{2}\sin \theta \,d\theta \,d\phi }$ and in the Cartesian coordinate system as ${\displaystyle dS={\frac {r}{\sqrt {r^{2}-\sum _{i\neq k}x_{i}^{2}}}}\Pi _{i\neq k}dx_{i},\;\forall k}$ comes The total area of the sphere comes from the surface integral over the entire surface of the sphere: ${\displaystyle A=\int _{0}^{2\pi }\int _{0}^{\pi }r^{2}\sin \theta \,d\theta \,d\phi =4\pi r^{2}.}$ A sphere has the least surface area of all surfaces that enclose a given volume and the largest volume of all surfaces with a given surface area. This is how spheres appear in nature: for example, bubbles and small water droplets are nearly spherical because surface tension locally minimizes surface area. The surface area relative to the mass of a ball is called the specific surface area and it can be expressed from the values mentioned above. ${\displaystyle \mathrm {SSA} ={\frac {A}{V\rho }}={\frac {3}{r\rho }},}$ which is the density (ratio of mass to volume). ### Surface The sphere's surface is the two-dimensional surface that forms the edge of the sphere. It is therefore the set of all points whose distance from the center of the sphere has a fixed value ${\displaystyle r}$. It is a closed two-dimensional manifold. Its area is ${\displaystyle {A}=4\pi {r^{2}}}$ and is therefore the same as that of the surface area of the circular cylinder that envelops the sphere . For a given volume, the sphere has the smallest surface area – and thus the smallest A/V ratio – of all possible bodies. #### Rationale If you divide a sphere into: • Layers each with a height of ${\displaystyle d}$ and • "Meridians", which are also ${\displaystyle d}$ apart at the equator and let ${\displaystyle d}$ strive for ${\displaystyle 0}$, • so the length ${\displaystyle c}$ of each cell is inversely proportional to ${\displaystyle x}$—that is, to its distance from the central axis. This is clear from the upper drawing on the right: ${\displaystyle x}$ is the distance from the tangent point to the central axis. The tangent is perpendicular to the "spoke" ${\displaystyle r}$ and the two (right) triangles are similar. Accordingly: ${\displaystyle c={\frac {r}{x}}{d}}$. • However, the width of each field is proportional to ${\displaystyle x}$. This follows directly from the drawing below, "top view". The length multiplied by the width is therefore always the same, i.e. all square fields have the same area. The area at the equator is ${\displaystyle d^{2}}$ (${\displaystyle c\cdot d}$ where ${\displaystyle c}$ tends to ${\displaystyle d}$ because ${\displaystyle {\frac {r}{x}}}$ at the equator tends to ${\displaystyle 1}$ faster than ${\displaystyle d}$ tends to ${\displaystyle 0}$). Since all fields have the content ${\displaystyle d^{2}}$ and there is a total of (number of fields in the horizontal direction multiplied by the number of fields in the vertical direction, i.e.) ${\displaystyle {\frac {\text{size}}{d}}\cdot {\frac {\text{diameter}}{d}}={\frac {2\pi r\cdot 2r}{d^{2}}}}$ fields there is the total area of all fields: ${\displaystyle {A}={4}\pi {r^{2}}}$. #### Alternative derivation using the volume of the sphere A sphere can be imagined composed of an infinite number of infinitesimals (infinitely small) pyramids. The bases of these pyramids together form the surface of the sphere; the heights of the pyramids are equal to the sphere radius ${\displaystyle r}$. Since the pyramid volume is given by the formula ${\displaystyle V_{P}={\tfrac {1}{3}}Gh}$, a corresponding relationship applies to the total volume of all pyramids, i.e. the volume of the sphere: ${\displaystyle V={\frac {1}{3}}A_{O}r}$ (${\displaystyle A_{O}}$ = total surface area of the sphere) Because of ${\displaystyle V={\tfrac {4}{3}}\pi r^{3}}$ we get: ${\displaystyle {\frac {4}{3}}\pi r^{3}={\frac {1}{3}}A_{O}r}$ ${\displaystyle A_{O}=4\pi r^{2}}$ #### Alternative derivation using the volume of the sphere and the differential calculus Since the sphere volume with ${\displaystyle V={\frac {4}{3}}\pi r^{3}}$ is defined and, on the other hand, the surface according to a change in volume ${\displaystyle A_{O}={\frac {\mathrm {d} V}{\mathrm {d} r}}=4\pi r^{2}}$ is, the surface formula follows immediately from the derivation of the volume formula. #### Derivation using integral calculus From the first Guldin’s rule ${\displaystyle A_{O}=2\pi \int \limits _{a}^{b}f(x){\sqrt {1+(f'(x))^{2}}}\,\mathrm {d} x}$ for the lateral surface of a body of revolution: {\displaystyle {\begin{aligned}A_{O}&=2\pi \int \limits _{-r}^{r}{\sqrt {r^{2}-x^{2}}}{\sqrt {1+\left({\frac {-x}{\sqrt {r^{2}-x^{2}}}}\right)^{2}}}\,\mathrm {d} x\\&=2\pi \int \limits _{-r}^{r}{\sqrt {r^{2}-x^{2}}}{\sqrt {\frac {r^{2}}{r^{2}-x^{2}}}}\,\mathrm {d} x\\&=2\pi \int \limits _{-r}^{r}r\,\mathrm {d} x\\&=2\pi r\int \limits _{-r}^{r}1\,\mathrm {d} x\\&=4\pi r^{2}\end{aligned}}} #### Derivation using the integral calculus in spherical coordinates For the surface element on surfaces ${\displaystyle r}$ = constant applies in spherical coordinates: ${\displaystyle \mathrm {d} A=r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi }$. This makes it easy to calculate the surface: {\displaystyle {\begin{aligned}A_{O}&=\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }1\,\mathrm {d} A\\&=\int \limits _{0}^{2\pi }\int \limits _{0}^{\pi }r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \\&=2\pi r^{2}\int \limits _{0}^{\pi }\sin \theta \,\mathrm {d} \theta \\&=4\pi r^{2}\end{aligned}}} ## Generalization ### Higher-dimensional Euclidean spaces The concept of the sphere can be transferred to spaces of other dimension. Analogously to the three-dimensional solid sphere, for a natural number ${\displaystyle n}$ a ${\displaystyle n}$‑dimensional sphere is defined as the set of all points of ${\displaystyle n}$‑dimensional Euclidean space whose distance to a given point (the center) is less than or equal to a positive real number ${\displaystyle r}$ (the radius). The boundary of the ${\displaystyle n}$‑dimensional sphere, i.e. the set of all points whose distance from the center is equal to ${\displaystyle r}$, is denoted as ${\displaystyle (n-1)}$‑dimensional sphere or ${\displaystyle (n-1)}$‑sphere for short. If one speaks of the ${\displaystyle n}$‑dimensional sphere without further information, one usually means the ${\displaystyle n}$‑dimensional unit sphere; in this case the center is at the origin of the coordinate system and the radius is equal to 1. So, by this definition, a three-dimensional sphere is an ordinary sphere; its surface corresponds to a 2‑sphere. A two-dimensional sphere is a circular area, the associated edge of a circle is a 1‑sphere. Finally, a one-dimensional sphere is a line, where the two line end points can be understood as a 0‑sphere. Note: These terms are not used consistently. Spheres in the sense of the definition given here are sometimes called balls. Also, some authors speak of ${\displaystyle n}$ spheres when they mean ${\displaystyle (n-1)}$ dimensional spheres in ${\displaystyle n}$ dimensional space. The ${\displaystyle n}$-dimensional volume of a ${\displaystyle n}$-dimensional sphere of radius ${\displaystyle r}$ is ${\displaystyle V_{n}(r)=r^{n}{\frac {\pi ^{n/2}}{\Gamma ({\frac {n}{2}}+1)}}}$. Here ${\displaystyle \Gamma }$ is the gamma function, a continuous extension of the faculty. The intersection of a ${\displaystyle n}$ dimensional sphere in ${\displaystyle n}$ dimensional Euclidean space with a ${\displaystyle (n-1)}$ dimensional one hyperplane is a ${\displaystyle (n-1)}$‑dimensional sphere of radius ${\displaystyle {\sqrt {r^{2}-x^{2}}}}$, where is the distance of the hyperplane from the center of the sphere. The volume of the ${\displaystyle x,n}$‑dimensional sphere is therefore the integral over all parallel interfaces: ${\displaystyle V_{n}(r)=\int _{-r}^{r}V_{n-1}\left({\sqrt {r^{2}-x^{2}}}\right)\mathrm {d} x=r^{n-1}\int _{-r}^{r}V_{n-1}\left({\sqrt {1-\left({\frac {x}{r}}\right)^{2}}}\right)\mathrm {d} x}$ From the Substitution ${\displaystyle t:={\tfrac {x}{r}}}$ follows ${\displaystyle V_{n}(r)=r^{n}\int _{-1}^{1}V_{n-1}\left({\sqrt {1-t^{2}}}\right)dt=r^{n}V_{n}(1)}$ So the volume ${\displaystyle V_{n}(r)}$ is proportional to ${\displaystyle r^{n}}$. By complete induction over ${\displaystyle n}$ it follows that the volume for all dimensions ${\displaystyle n}$ is proportional to ${\displaystyle r^{n}}$. Preserves the ${\displaystyle (n-1)}$ dimensional content of the ${\displaystyle (n-1)}$ dimensional surface, i.e. the ${\displaystyle (n-1)}$ sphere one by derivation of the volume after the radius: ${\displaystyle O_{n}(r)=nr^{n-1}{\frac {\pi ^{n/2}}{\Gamma ({\frac {n}{2}}+1)}}=2r^{n-1}{\frac {\pi ^{n/2}}{\Gamma ({\frac {n}{2}})}}}$. For a unit sphere in ${\displaystyle n}$ dimensions one finds the following volumes and surface areas: Dimensions 1 2 3 4 5 6 7 8 9 10 n=2m n=2m+1 volume 2 ${\displaystyle \pi }$ ${\displaystyle {\frac {4\pi }{3}}}$ ${\displaystyle {\frac {\pi ^{2}}{2}}}$ ${\displaystyle {\frac {8\pi ^{2}}{15}}}$ ${\displaystyle {\frac {\pi ^{3}}{6}}}$ ${\displaystyle {\frac {16\pi ^{3}}{105}}}$ ${\displaystyle {\frac {\pi ^{4}}{24}}}$ ${\displaystyle {\frac {32\pi ^{4}}{945}}}$ ${\displaystyle {\frac {\pi ^{5}}{120}}}$ ${\displaystyle {\frac {\pi ^{m}}{m!}}}$ ${\displaystyle {\frac {2^{m+1}\pi ^{m}}{1\cdot 3\cdot \ldots \cdot (2m+1)}}}$ surface 2 ${\displaystyle 2\pi }$ ${\displaystyle 4\pi }$ ${\displaystyle 2\pi ^{2}}$ ${\displaystyle {\frac {8\pi ^{2}}{3}}}$ ${\displaystyle \pi ^{3}}$ ${\displaystyle {\frac {16\pi ^{3}}{15}}}$ ${\displaystyle {\frac {\pi ^{4}}{3}}}$ ${\displaystyle {\frac {32\pi ^{4}}{105}}}$ ${\displaystyle {\frac {\pi ^{5}}{12}}}$ ${\displaystyle {\frac {2\pi ^{m}}{(m-1)!}}}$ ${\displaystyle {\frac {2^{m+1}\pi ^{m}}{1\cdot 3\cdot \ldots \cdot (2m-1)}}}$ A ${\displaystyle n}$ sphere is an example of a compact ${\displaystyle n}$manifold. ### Metric Spaces The concept of the sphere can be generalized to all spaces in which one has a concept of distance, these are the metric spaces. If ${\displaystyle (X,d)}$ is a metric space, ${\displaystyle a\in X}$ and ${\displaystyle r\in \mathbb {R} }$, ${\displaystyle r>0,suchathingntman:[itex]B(a,r)=\{x\in X\mid d(a,x) the open sphere with center ${\displaystyle a}$ and radius ${\displaystyle r}$.[2] The amount:

${\displaystyle {\overline {B}}(a,r)=\{x\in X\mid d(a,x)\leq r\}}$

means closed sphere.

Some authors also write ${\displaystyle U(a,r)}$ for the open and ${\displaystyle B(a,r)}$ for the closed spheres.[3] Other notations for the open spheres are ${\displaystyle B_{r}(a)}$ and ${\displaystyle U_{r}(a)}$.

1. German Wikipedia
2. Boto von Querenburg: Set theoretical topology. Springer-Verlag, Berlin/Heidelberg 1976 , definition 1.3. 3rd edition 2001, ISBN 3-540-67790-9.
3. Herbert Federer: ' 'Geometric Measure Theory. Springer-Verlag, Berlin/Heidelberg 1969, 2.8.1.