nyuuzyou/wikis huggingface dataset:
Volume of Displaced Fluid equals Volume of Submerged Object/Historical Note

== Historical Note on Volume of Displaced Fluid equals Volume of Submerged Object ==
The story goes that was asked by to determine whether his new crown was made truly of gold, or whether the goldsmith had cheated him by stealing some of the gold and mixing the remainder with a metal of a lesser value, supposedly silver.
 was puzzling over this problem when he lay back in the bath and watched the water flow over the sides.
It was then that he suddenly realized that as gold was more dense than silver, it would have a smaller volume, and so displace less water, than a piece of silver of the same mass.
Legend has it that he jumped out of the bath and ran home without bothering to dress, shouting:
:Eureka! Eureka!
which means:
:I have found it! I have found it!
(It needs to be pointed out that public nakedness was not a forbidden phenomenon in that social milieu, hence this behaviour was not quite as outrageous as it would be if embedded in conventional early twenty-first century Western culture.)
He measured the mass of Hiero's new crown, and created a block of gold and one of silver of the same mass.
By measuring the volume of water that was displaced by each block, he determined that the crown was made of metal which was denser than silver, but less dense than gold, proving that the gold from which the crown was made was impure.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
Category:Historical Notes
Volume of Ellipsoid

== Theorem ==
Let $\EE$ be an ellipsoid such that:
:the semimajor axis of $\EE$ is $a$
:the semimean axis of $\EE$ is $b$
:the semiminor axis of $\EE$ is $c$.
The volume $\VV$ of $\EE$ is given by:
:$\VV = \dfrac {4 \pi a b c} 3$
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Ellipsoid of Semi-axes $a, b, c$: $4.47$
* 
* 
* : $\S 7$: Geometric Formulas: Ellipsoid of Semi-axes $a, b, c$: $7.47.$
Category:Ellipsoids
Category:Volume Formulas
Volume of Frustum of Cone or Pyramid

== Theorem ==
Let $F$ be a frustum of a cone or a pyramid.
The volume $V$ of $F$ is given as:
:$V = \dfrac {h \paren {A_1 + A_2 + \sqrt {A_1 A_2} } } 3$
where:
:$A_1$ and $A_2$ are the areas of the bases of $F$
:$h$ is the altitude of $F$.
== Proof ==
== Sources ==
* 
* 
Category:Frustums
Category:Volume Formulas
Volume of Frustum of Right Circular Cone

== Theorem ==
Let $F$ be a frustum of a right circular cone.
The volume $\VV$ of $F$ is given as:
:$\VV = \dfrac {\pi h \paren {a^2 + a b + b^2} } 3$
where:
:$a$ and $b$ are the radii of the bases of $F$
:$h$ is the altitude of $F$.
== Proof ==
From Volume of Frustum of Cone or Pyramid:
:$\VV = \dfrac {h \paren {A_1 + A_2 + \sqrt {A_1 A_2} } } 3$
where:
:$A_1$ and $A_2$ are the areas of the bases of $F$
:$h$ is the altitude of $F$.
Here we have that $F$ be a frustum of a right circular cone.
Hence the bases of $F$ are circles.
From Area of Circle, the areas of the bases of $F$ are therefore:
:$A_1 = \pi a^2$
:$A_2 = \pi b^2$
Hence:
Hence the result.
== Sources ==
* : $\S 4$: Geometric Formulas: Frustum of Right Circular Cone of Radii $a, b$ and Height $h$: $4.42$
* : $\S 7$: Geometric Formulas: Frustum of Right Circular Cone of Radii $a, b$ and Height $h$: $7.42.$
Category:Frustums
Category:Volume Formulas
Volume of Gabriel's Horn

== Theorem ==
Consider Gabriel's horn, the solid of revolution formed by rotating about the $x$-axis the curve:
:$y = \dfrac 1 x$
Consider the volume $V$ of the space enclosed by the planes $x = 1$, $x = a$ and the portion of Gabriel's horn where $1 \le x \le a$.
Then:
:$V = \pi \paren {1 - \dfrac 1 a}$
=== Corollary ===
== Proof ==
From Volume of Solid of Revolution:
== Historical Note ==
Category:Gabriel's Horn
Volume of Gabriel's Horn/Corollary

== Corollary to Volume of Gabriel's Horn ==
Consider Gabriel's horn, the solid of revolution formed by rotating about the $x$-axis the curve:
:$y = \dfrac 1 x$
Consider the volume $V$ of the space enclosed by the plane $x = 1$ and the portion of Gabriel's horn where $x \ge 1$.
Then:
:$V = \pi$
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Gabriel's Horn
Volume of Gabriel's Horn/Historical Note

== Historical Note to the Volume of Gabriel's Horn ==
The volume of Gabriel's horn was first demonstrated by in $1643$.
The result caused great astonishment at the time, as it was the first example of a solid figure demonstrated to be of infinite extent while also having finite volume.
As put it:
:To understand this for sense, it is not required that a man should be a geometrician or a logician, but that he should be mad.
== Sources ==
* : Chapter $\text {A}.15$: Torricelli ()
Category:Gabriel's Horn
Category:Historical Notes
Volume of Paraboloid

== Theorem ==
The volume of paraboloid is half the volume of its circumscribing cylinder.
== Proof ==
Consider a cylinder of radius $r$ and height $h$, circumscribing a paraboloid $y = h \paren {\dfrac x r}^2$ whose apex is at the center of the bottom base of the cylinder and whose base is the top base of the cylinder.
Also consider the paraboloid $y = h - h \paren {\dfrac x r}^2$, with equal dimensions but with its apex and base flipped.
For every height $0 \le y \le h$, the disk-shaped cross-sectional area $\pi \paren {r \sqrt {1 - \dfrac y h} }^2$ of the flipped paraboloid is equal to the ring-shaped cross-sectional area $\pi r^2 - \pi r^2 \paren {\sqrt {\dfrac y h} }^2$ of the cylinder part outside the inscribed paraboloid.
Indeed:
Therefore, the volume of the flipped paraboloid is equal to the volume of the cylinder part outside the inscribed paraboloid.
In other words, the volume of the paraboloid is $\dfrac \pi 2 r^2 h$, half the volume of its circumscribing cylinder.
Category:Paraboloids
Category:Volume Formulas
Volume of Parallelepiped

== Theorem ==
Let $P$ be a parallelepiped.
The volume $V_P$ of $P$ is given by:
:$V_P = A h$
where:
:$A$ is the area of the base of $P$
:$h$ is the height of $P$.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Parallelepiped of Cross-sectional Area $A$ and Height $h$: $4.28$
* 
* 
* : $\S 7$: Geometric Formulas: Parallelepiped of Cross-sectional Area $A$ and Height $h$: $7.28.$
Category:Parallelepipeds
Category:Volume Formulas
Volume of Parallelepiped/Mistake

== Source Work ==
:
:Chapter $4$: Geometric Formulas
::Parallelepiped of Cross-sectional Area $A$ and Height $h$: $4.28$
:
:Chapter $7$: Geometric Formulas
::Parallelepiped of Cross-sectional Area $A$ and Height $h$: $7.28.$
== Mistake ==
:Volume $= A h = a b c \sin \theta$
== Correction ==
The formula 'Volume $= a b c \sin \theta$ applies only when the base is rectangular.
== Sources ==
* : $\S 4$: Geometric Formulas: Parallelepiped of Cross-sectional Area $A$ and Height $h$: $4.28$
* : $\S 7$: Geometric Formulas: Parallelepiped of Cross-sectional Area $A$ and Height $h$: $7.28.$
Category:Mistakes/Mathematical Handbook of Formulas and Tables
Volume of Pyramid

== Theorem ==
The volume $V$ of a pyramid $P$ is given by:
:$V = \dfrac {A h} 3$
where:
:$A$ is the area of the base of $P$
:$h$ is the height of $P$.
== Proof ==
== Sources ==
* : $\S 4$: Geometric Formulas: Pyramid of Base Area $A$ and Height $h$: $4.39$
* 
* 
* : $\S 7$: Geometric Formulas: Pyramid of Base Area $A$ and Height $h$: $7.39.$
Category:Pyramids
Category:Volume Formulas
Volume of Right Circular Cone

== Theorem ==
The volume $V$ of a right circular cone is given by:
:$V = \dfrac 1 3 \pi r^2 h$
where:
:$r$ is the radius of the base
:$h$ is the height of the cone, that is, the distance between the apex and the center of the base.
== Construction ==
Construct the following triangle:
:
Let $A$ be located at the origin of the $xy$-plane.
By definition of tangent as opposite over adjacent, the line segment $\overline {AB}$ can be described by the equation $y = \dfrac {BC} {AC} x$.
By Euclid's definition of a cone, the solid of revolution generated by rotating $\triangle ABC$ about the $x$-axis is a right circular cone:
:whose axis is $\overline {AC}$
:whose base consists of the circle whose center is $C$, whose radius is $BC$ and whose plane is perpendicular to $\overline {AC}$.
As $\overline {AC}$ is perpendicular to the base of the cone, the height of the cone is $AC$.
Let $h = AC$ denote the height and $r = BC$ denote the radius of the base.
== Proof ==
This proof utilizes the Method of Disks and thus is dependent on Volume of Right Circular Cylinder.
From the Method of Disks, the volume of the cone can be found by the definite integral:
:$\ds (1): \quad V = \pi \int_0^{AC} \paren {\map R x}^2 \rd x$
where $\map R x$ is the function describing the line which is to be rotated about the $x$-axis in order to create the required solid of revolution.
In this example, $\map R x$ describes the line segment $\overline {AB}$, and so:
:$\map R x = \dfrac r h x$
We have also defined:
:$\overline {AC}$ as the axis of the cone, whose length is $h$
:$A$ as the origin.
So the equation $(1)$ is interpreted as:
== Sources ==
* : $\S 4$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $4.37$
* : $\S 7$: Geometric Formulas: Right Circular Cone of Radius $r$ and Height $h$: $7.37.$
* 
* 
Category:Right Circular Cones
Category:Volume Formulas
Category:Analytic Geometry
Category:Integral Calculus
Volume of Right Circular Cylinder

== Theorem ==
The volume $V_C$ of a right circular cylinder whose bases are circles of radius $r$ and whose height is $h$ is given by the formula:
:$V_C = \pi r^2 h$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$: Geometric Formulas: Right Circular Cylinder of Radius $r$ and Height $h$: $4.31$
* : $\S 7$: Geometric Formulas: Right Circular Cylinder of Radius $r$ and Height $h$: $7.31.$
Category:Volume of Right Circular Cylinder
Category:Right Circular Cylinders
Category:Volume Formulas
Volume of Right Circular Cylinder/Proof 1

== Theorem ==
== Proof ==
Consider a right circular cylinder $C$ whose base is a circle of radius $r$ and whose height is $h$.
Let $V_C$ denote the volume of $C$.
From Volume of Cylinder in terms of Height and Base Area:
:$V_C = A h$
where $A$ is the area of the base of $C$.
From Area of Circle, the area of each base is:
:$A = \pi r^2$
Hence:
:$V_C = \pi r^2 h$
Category:Volume of Right Circular Cylinder
Volume of Right Circular Cylinder/Proof 2

== Theorem ==
== Proof ==
:
Consider a right circular cylinder $C$ whose base is a circle of radius $r$ and whose height is $h$.
Consider a cuboid $K$ whose height is $h$ and whose base has the same area as the base of $C$.
Let the area of those bases be $A$.
Let $C$ be positioned with its base in the same plane as the base of $K$.
By Cavalieri's Principle $C$ and $K$ have the same volume.
The bases of $C$ are circles of radius $r$.
From Area of Circle, the area of each base therefore gives:
:$A = \pi r^2$
From Volume of Cuboid, $K$ has volume given by:
:$V_K = A h = \pi r^2 h$
Hence the result.
Category:Volume of Right Circular Cylinder
Volume of Smallest Rational Tetrahedron

== Theorem ==
The only rational tetrahedron whose edge lengths are less than $157$ has:
: edges of length $117$, $80$, $53$, $52$, $51$, $84$
: faces of area $1800$, $1890$, $2016$, $1170$
: volume of $18 \, 144$.
== Proof ==
== Sources ==
* 
* 
* : $18,144$
* 
Category:Tetrahedra
Volume of Smallest Tetrahedron with Integer Edges and Integer Volume

== Theorem ==
The volume of the smallest tetrahedron with integer edges and integer volume is $3$.
There are $2$ possible sets of edges:
:$32, 33, 35, 44, 70, 76$
:$21, 32, 47, 56, 58, 76$
== Proof ==
Tartaglia's Formula gives us that:
:$V_T^2 = \dfrac {1} {288} \det \ \begin{vmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{vmatrix}$
where:
:$V_T$ denotes the volume of a tetrahedron
:$d_{ij}$ denotes the distance between vertex $\mathbf d_i$ and vertex $\mathbf d_j$
So the volume of the set lengths $32, 33, 35, 40, 70, 76$ is:
Hence the result.
== Sources ==
* 
* : $3$
Category:Tetrahedra
Volume of Solid of Revolution

== Theorem ==
Let $f: \R \to \R$ be a real function which is integrable on the interval $\closedint a b$.
Let the points be defined:
:$A = \tuple {a, \map f a}$
:$B = \tuple {b, \map f b}$
:$C = \tuple {b, 0}$
:$D = \tuple {a, 0}$
Let the figure $ABCD$ be defined as being bounded by the straight lines $y = 0$, $x = a$, $x = b$ and the curve defined by $\set {\map f x: a \le x \le b}$.
Let the solid of revolution $S$ be generated by rotating $ABCD$ around the $x$-axis (that is, $y = 0$).
Then the volume $V$ of $S$ is given by:
:$\ds V = \pi \int_a^b \paren {\map f x}^2 \rd x$
=== Parametric Form ===
== Proof ==
:
Consider a rectangle bounded by the lines:
:$y = 0$
:$x = \xi$
:$x = \xi + \delta x$
:$y = \map f x$
Consider the right circular cylinder generated by revolving it about the $x$-axis.
By Volume of Right Circular Cylinder, the volume of this cylinder is:
:$V_\xi = \pi \paren {\map f x}^2 \delta x$
== Historical Note ==
== Sources ==
* : $\text {II}$. Calculus: Area, Volume and Centre of Gravity: Volume of Rotation about the $x$-axis
* 
* 
* 
* 
* 
Category:Integral Calculus
Category:Solids of Revolution
Volume of Solid of Revolution/Historical Note

== Theorem ==
The technique of finding the Volume of Solid of Revolution by dividing up the solid of revolution into many thin disks and approximating them to cylinders was devised by sometime around or after $1612$, reportedly on the occasion of his wedding in $1613$.
His inspiration was in the problem of finding the volume of wine barrels accurately.
He published his technique in his $1615$ work (New Stereometry of Wine Barrels).
 redefined the problem by applying the techniques of integral calculus around $1680$.
== Sources ==
* : Chapter $\text {A}.10$: Kepler ()
* : Chapter $8$: The System of the World: Leibniz
Category:Historical Notes
Volume of Solid of Revolution/Parametric Form

== Theorem ==
Let $x: \R \to \R$ and $y: \R \to \R$ be real functions defined on the interval $\closedint a b$.
Let $y$ be integrable on the (closed) interval $\closedint a b$.
Let $x$ be differentiable on the (open) interval $\openint a b$.
Let the points be defined:
:$A = \tuple {\map x a, \map y a}$
:$B = \tuple {\map x b, \map y b}$
:$C = \tuple {\map x b, 0}$
:$D = \tuple {\map x a, 0}$
Let the figure $ABCD$ be defined as being bounded by the straight lines $y = 0$, $x = a$, $x = b$ and the curve defined by:
:$\set {\tuple {\map x t, \map y t}: a \le t \le b}$
Let the solid of revolution $S$ be generated by rotating $ABCD$ around the $x$-axis (that is, $y = 0$).
Then the volume $V$ of $S$ is given by:
:$\ds V = \pi \int_a^b \paren {\map y t}^2 \map {x'} t \rd t$
== Proof ==
== Historical Note ==
== Sources ==
* 
Category:Integral Calculus
Category:Solids of Revolution
Volume of Sphere

== Theorem ==
The volume $V$ of a sphere of radius $r$ is given by:
:$V = \dfrac {4 \pi r^3} 3$
== Proof by Archimedes ==
== Proof by Method of Disks ==
== Proof by Cavalieri ==
== Sources ==
* : $\S 4$: Geometric Formulas: Sphere of Radius $r$: $4.29$
* 
* 
* : $\S 7$: Geometric Formulas: Sphere of Radius $r$: $7.29.$
* 
Category:Volume of Sphere
Category:Spheres
Category:Volume Formulas
Volume of Sphere/Proof by Archimedes

== Theorem ==
== Proof ==
Consider the circle in the cartesian plane whose center is at $\tuple {a, 0}$ and whose radius is $a$.
From Equation of Circle, its equation is:
:$(1): \quad x^2 + y^2 = 2 a x$
Consider this circle as the cross-section through the center of a sphere which has the x-axis passing through its center, which is at $\tuple {a, 0}$.
Consider the cross-section of this sphere formed by the plane $x$ units to the right of the origin.
The area of this cross-section is $\pi y^2$.
We write $(1)$ in the form:
:$(2): \quad \pi x^2 + \pi y^2 = 2 \pi a x$
We can likewise interpret $\pi x^2$ as the area of the cross-section of the cone generated by revolving the line $y = x$ about the x-axis.
Now, consider the $2 \pi a x$ in equation $(2)$.
We write equation $(2)$ in the following form:
:$(3): \quad 2 a \paren {\pi x^2 + \pi y^2} = \paren {2 a}^2 \pi x$
and we see that $\paren {2 a}^2 \pi$ is the area of the cross-section of the right circular cylinder which has the same height and base as the cone.
So, we have three circular discs whose areas are $\pi y^2$, $\pi x^2$ and $\paren {2 a}^2 \pi$.
These are all the intersections with the plane $x$ units from the right of the origin with the three solids of revolution described above.
Illustrated below is the intersections of these with the X-Y plane.
:
On the of equation $(3)$, the first two areas are added and multiplied by $2a$.
On the , the third area is multiplied by $x$.
Now, consider the discs themselves as bodies of constant density, so their mass (and hence weight) is proportional to their area.
Imagine the x-axis as the arm of a balance whose fulcrum is at the origin.
We leave the disc of radius $2 a$ where it is, $x$ units to the right of the origin.
We move the discs of radius $x$ and $y$ to a point $2 a$ units to the left of the origin, and imagine them hanging from the point $\tuple {-2 a, 0}$ on the balance.
Now from the Principle of Moments we see that the combined moments of the two discs on the left equals the moment of the disc on the left.
So the balance is at equilibrium.
Now for the last bit.
As $x$ increases from $0$ to $2 a$, the three cross-sections go through their respective solids and fill them.
Since, throughout this entire exercise, they are in equilibrium, so are the solids.
From:
:'s Formula for Volume of Cone:
::$\dfrac 1 3 \pi \paren {2 a}^2 2 a$
:The obvious position of the center of gravity of the cylinder, that is, at $\tuple {a, 0}$
we have:
:$2 a \paren {\frac 1 3 \pi \paren {2 a}^2 \paren {2 a} + V} = a \pi \paren {2 a}^2 \paren {2 a}$
from which drops out:
:$V = \dfrac 4 3 \pi a^3$
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.6$: How Archimedes Discovered Integration
* : Historical Introduction
* : Chapter $2$: The Logic of Shape: Archimedes
Category:Volume of Sphere
Volume of Sphere/Proof by Archimedes/Historical Note

== Historical Note on Archimedes' Proof of Volume of Sphere ==
The volume of the sphere was proved by in around $250$ BCE.
His proof appears as Proposition $33$ in his .
In the version given here, the notation has been brought up to date.
He also describes in detail how he arrived at this result in .
In his words:
:I am persuaded that this method will be of no little service to mathematics. For I can see that once it is understood and established, it will be used to discover other theorems which have not yet occurred to me, by other mathematicians, now living or yet unborn.
::::-- : 
It can be argued that this was the point integral calculus was first invented.
Tradition has it that he asked his friends to mark his tombstone with the representation of a sphere inscribed within a cylinder, to commemorate this achievement.
This was done at the order of , the Roman general who finally defeated the City of Syracuse in $212$ BCE.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $\text {A}.5$: Archimedes (ca. B.C.)
* : Chapter $\text {B}.6$: How Archimedes Discovered Integration
* : Historical Introduction
Category:Historical Notes
Volume of Sphere/Proof by Cavalieri

== Theorem ==
== Proof ==
:
Consider a sphere $S$ of radius $r$.
Consider a right circular cylinder $C$ whose bases are circles of radius $r$ and whose height is $2 r$.
Let $K$ be a double napped cone each of whose nappes has bases which coincide with the bases of $C$.
Let $K$ be removed from $C$ to leave a solid figure $C'$ described as a right circular cylinder with a conical hollow at either end.
The bases of $C$ are circles of radius $r$.
From Area of Circle, the area of each base is therefore $\pi r^2$.
From Volume of Right Circular Cylinder, $C$ has volume given by:
$K$ has twice the volume of each of its nappes.
Each of the nappes of $K$ is of height $r$ and has a base of area $\pi r^2$.
Hence from Volume of Cone, $K$ has volume given by:
Thus the volume $V_{C'}$ of $C'$ is given by:
Let the sphere $S$ be positioned so as to be tangent to the plane containing the base of $C$
Consider a plane parallel to the base of $C$ intersecting both $S$ and $C$ at a distance $x$ from the center of $S$.
Let $y$ be the radius of the circle which is the intersection of $S$ with that plane.
From Pythagoras's Theorem, $y$ is given by:
:$x^2 + y^2 = r^2$
and so:
:$y^2 = r^2 - x^2$
Thus the area of this circle is given by:
:$A = \pi \left({r^2 - x^2}\right)$
Similarly, consider the intersection of $C'$ with the same plane.
This consists of:
: the circle of radius $r$ which has an area equal to that of the base of $C$
with:
: the circle of radius $x$ which is the intersection of the plane with the cone
removed from it.
So the area of the remaining circle with the inner circle removed is:
:$A' = \pi r^2 - \pi x^2$
from Area of Circle.
Thus the intersection of the plane with both $S$ and $C'$ are the same.
Thus by Cavalieri's Principle $S$ and $C'$ have the same volume.
Hence the result.
== Historical Note ==
This proof was given by .
== Sources ==
* : Chapter $\text {A}.14$: Cavalieri ()
Category:Volume of Sphere
Volume of Sphere/Proof by Method of Disks

== Theorem ==
== Proof ==
=== Construction ===
Describe a circle on the $x y$-plane.
Let its center be the origin.
By Equation of Circle, this circle is the locus of:
:$x^2 + y^2 = r^2$
where $r$ is a constant radius.
Solving for $y$:
:$y = \pm \sqrt {r^2 - x^2}$
Considering only the upper half of the circle:
:$y = \sqrt {r^2 - x^2}$
:
This plane region is a semicircle whose radius is $r$ and whose extremes are at $x = -r$ and $x = r$.
By Euclid's definition of a sphere, the solid of revolution of this plane region about the $x$-axis is a sphere whose radius is $r$.
=== Proof ===
Note that this proof uses the Method of Disks and thus is dependent on Volume of Right Circular Cylinder.
From the Method of Disks, the volume of this sphere can be found by the definite integral:
:$\ds V = \pi \int_{-r}^r y^2 \rd x$
where $y$ is the function of $x$ describing the curve which is to be rotated about the $x$-axis in order to create the required solid of revolution.
By construction, $y = \sqrt {r^2 - x^2}$.
The volume, then, is given by:
Category:Volume of Sphere
Volume of Sphere from Surface Area

== Theorem ==
The volume $V$ of a sphere of radius $r$ is given by:
:$V = \dfrac {r A} 3$
where $A$ is the surface area of the sphere.
== Proof ==
Let the surface of the sphere of radius $r$ be divided into many small areas.
If they are made small enough, they can be approximated to plane figures.
Let the areas of these plane figures be denoted:
:$a_1, a_2, a_3, \ldots$
Let the sphere of radius $r$ be divided into as many pyramids whose apices are at the center and whose bases are these areas.
From Volume of Pyramid, their volumes are:
:$\dfrac {r a_1} 3, \dfrac {r a_2} 3, \dfrac {r a_3} 3, \ldots$
The volume $\VV$ of the sphere is given by the sum of the volumes of each of these pyramids:
But $a_1 + a_2 + a_3 + \cdots$ is the surface area of the sphere.
Hence:
It needs to be noted that this proof is intuitive and non-rigorous.
== Historical Note ==
This was the method used by as an offshoot of the proof he gave for the area of a circle
== Sources ==
* : Chapter $\text {A}.10$: Kepler ()
Category:Volume of Sphere
Category:Spheres
Category:Volume Formulas
Volume of Spherical Pyramid

== Theorem ==
Let $Q$ be a spherical pyramid constructed from a spherical polygon $P$ on the surface of a sphere $S$.
The volume $V$ enclosed by $Q$ is given by:
:$V = \dfrac {\pi r^3 E} {540}$
where:
:$r$ is the radius of $S$
:$E$ is the spherical excess of $P$.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Pyramids
Category:Volume Formulas
Volume of Spherical Sector

== Theorem ==
Let $Q$ be a spherical sector constructed on the surface of a sphere $S$.
The volume $V$ enclosed by $Q$ is given by:
:$V = \dfrac {2 \pi r^2 h} 3$
where:
:$r$ is the radius of $S$
:$h$ is the altitude of the zone forming the surface of $Q$.
== Proof ==
== Sources ==
* 
* 
Category:Spherical Sectors
Category:Volume Formulas