avewright/proofwiki-math huggingface dataset:
Theorem: Length of Arc of Cycloid

Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the length of one arc of the cycloid is $8 a$.

Proof:
Consider the tangent line $PQ$ to the cycloid generated by a generating circle moving at '''constant''' angular speed of $\omega = \dfrac \theta t$.
:750px
From Angular Speed of Particle in Circular Motion at Constant Speed, point $P$ is moving at a speed of $\dfrac {\d s} {\d t} = r \omega = 2 a \map \sin {\dfrac \theta 2} \times \omega$.
The length of chord $PQ$ varies with respect to time as: $PQ = 2 a \map \cos {\dfrac \theta 2} = 2 a \map \cos {\dfrac {\omega t} 2}$
Therefore, the length of chord $PQ$ is changing at a rate of $\dfrac {\d \paren{P Q} } {\d t} = -a \map \sin {\dfrac \theta 2} \times \omega$.
Since they both reach zero at the same time at the summit $R$ (that is arc length $PR$ is zero and chord length $PQ$ is zero because $P$ and $Q$ meet) then arc length $PR$ must be twice the length of chord $PQ$.
Therefore:
:$PR = 2 PQ$
When $P$ is at the cusp, the tangent $PQ$ is perpendicular to the straight line along which the generating circle rolls.
At this point:
:$PQ = 2 a$
Thus at this point:
:$PR = 4 a$
But $4 a$ is half the length of one arc of $C$.
Hence the result.
{{qed}}
Theorem: Area under Arc of Cycloid

Let $C$ be a cycloid generated by the equations:
:$x = a \paren {\theta - \sin \theta}$
:$y = a \paren {1 - \cos \theta}$
Then the area under one arc of the cycloid is $3 \pi a^2$.
That is, the area under one arc of the cycloid is three times the area of the generating circle.

Proof:
Let $A$ be the area under of one arc of the cycloid. 
From Area under Curve, $A$ is defined by:
{{begin-eqn}}
{{eqn | l = A
      | r = \int_0^{2 \pi a} y \rd x
      | c = 
}}
{{eqn | r = \int_0^{2 \pi} a \paren {1 - \cos \theta} \frac {\d x} {\d \theta} \rd \theta
      | c = 
}}
{{end-eqn}}
But:
:$\dfrac {\d x} {\d \theta} = a \paren {1 - \cos \theta}$
and so:
{{begin-eqn}}
{{eqn | l = A
      | r = a^2 \int_0^{2 \pi} \paren {1 - \cos \theta}^2 \rd \theta
      | c = 
}}
{{eqn | r = a^2 \int_0^{2 \pi} \paren {1 - 2 \cos \theta + \cos^2 \theta} \rd \theta
      | c = 
}}
{{eqn | r = a^2 \int_0^{2 \pi} \paren {1 - 2 \cos \theta + \frac 1 2 + \frac {\cos 2 \theta} 2} \rd \theta
      | c = 
}}
{{eqn | r = a^2 \intlimits {\theta + 2 \sin \theta + \frac {\theta} 2 + \frac {\sin 2 \theta} 4} 0 {2 \pi}
      | c = 
}}
{{eqn | r = 3 \pi a^2
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Goldbach Conjecture implies Weak Goldbach Conjecture

The Goldbach Conjecture:
: Every even integer greater than $2$ is the sum of two primes
implies Goldbach's Weak Conjecture:
: Every odd integer greater than $7$ is the sum of three odd primes.

Proof:
Take any odd integer $n$ such that $n > 7$.
Then $m = n - 3$ is an even integer $n$ such that $m > 4$.
If the Goldbach Conjecture holds, then $m$ is the sum of two primes: $m = p_1 + p_2$.
If one of them were $2$, then $m - 2$ would have to be even, which if it is prime it can not be.
So if $m > 4$, both $p_1$ and $p_2$ must be odd.
So then we have that $n = p_1 + p_2 + 3$, that is, the sum of three odd primes.
So, if the Goldbach Conjecture holds, then so does Goldbach's Weak Conjecture.
{{qed}}
Category:Number Theory
Category:Goldbach Conjecture
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Theorem: Euclidean Domain is UFD

Let $\struct {D, +, \times}$ be a Euclidean domain.
Then $\struct {D, +, \times}$ is a unique factorization domain.

Proof:
By the definition of unique factorization domain, we need to show that:
For all $x \in D$ such that $x$ is non-zero and not a unit of $D$:
:$(1): \quad x$ has a complete factorization in $D$
:$(2): \quad$ Any two complete factorizations of $x$ in $D$ are equivalent.
Theorem: Quadrature of Parabola

Let $T$ be a parabola.
Consider the parabolic segment bounded by an arbitrary chord $AB$.
Let $C$ be the point on $T$ where the tangent to $T$ is parallel to $AB$.
Then the area $S$ of the parabolic segment $ABC$ of $T$ is given by:
:$S = \dfrac 4 3 \triangle ABC$

Proof:
{{WLOG}}, consider the parabola $y = a x^2$.
Let $A, B, C$ be the points:
{{begin-eqn}}
{{eqn | l = A
      | r = \tuple {x_0, a {x_0}^2}
}}
{{eqn | l = B
      | r = \tuple {x_2, a {x_2}^2}
}}
{{eqn | l = C
      | r = \tuple {x_1, a {x_1}^2}
}}
{{end-eqn}}
:500px
The slope of the tangent at $C$ is given by using:
:$\dfrac {\d y} {\d x} 2 a x_1$
which is parallel to $AB$.
Thus:
:$2 a x_1 = \dfrac {a {x_0}^2 - a {x_2}^2} {x_0 - x_2}$
which leads to:
:$x_1 = \dfrac {x_0 + x_2} 2$
So the vertical line through $C$ is a bisector of $AB$, at point $P$.
Complete the parallelogram $CPBQ$.
Also, find $E$ which is the point where the tangent to $T$ is parallel to $BC$.
By the same reasoning, the vertical line through $E$ is a bisector of $BC$, and so it also bisects $BP$ at $H$.
Next:
{{begin-eqn}}     
{{eqn | l = EF
      | r = a \paren {\frac {x_1 + x_2} 2}^2 - \paren {a x_1^2 + 2 a x_1 \frac {x_2 - x_1} 2}
      | c = 
}}
{{eqn | r = \frac a 4 \paren {\paren {x_1 + x_2}^2 - 4 {x_1}^2 + 4 x_1 \paren {x_2 - x_1} }
      | c = 
}}
{{eqn | r = \frac a 4 \paren { {x_1}^2 - 2 x_1 x_2 + {x_2}^2}
      | c = 
}}
{{eqn | r = \frac a 4 \paren {x_2 - x_1}^2
      | c = 
}}
{{end-eqn}}
At the same time:
{{begin-eqn}}     
{{eqn | l = QB
      | r = a {x_2}^2 - \paren {a {x_1}^2 + 2 a x_1 \paren {x_2 - x_2} }
      | c = 
}}
{{eqn | r = a \paren { {x_1}^2 - 2 x_1 x_2 + {x_2}^2}
      | c = 
}}
{{eqn | r = a \paren {x_2 - x_1}^2
      | c = 
}}
{{end-eqn}}
So:
:$QB = 4 FE = FH$
and because $CB$ is the diagonal of a parallelogram:
:$2 FE = 2 EG = FG$
This implies that:
:$2 \triangle BEG = \triangle BGH$
and:
:$2 \triangle CEG = \triangle BGH$
So:
:$\triangle BCE = \triangle BGH$
and so as $\triangle BCP = 4 \triangle BGH$ we have that:
:$BCE = \dfrac {\triangle BCP} 4$
A similar relation holds for $\triangle APC$:
:500px
so it can be seen that:
:$\triangle ABC = 4 \paren {\triangle ADC + \triangle CEB}$
Similarly, we can create four more triangles underneath $\triangle ADC$ and $\triangle CEB$ which are $\dfrac 1 4$ the area of those combined, or $\dfrac 1 {4^2} \triangle ABC$.
This process can continue indefinitely.
So the area $S$ is given as:
:$S = \triangle ABC \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \cdots}$
But from Sum of Geometric Sequence it follows that:
:$S = \triangle ABC \paren {\dfrac 1 {1 - \dfrac 1 4} } = \dfrac 4 3 \triangle ABC$
{{qed}}
Theorem: Quadrature of Parabola

Let $T$ be a parabola.
Consider the parabolic segment bounded by an arbitrary chord $AB$.
Let $C$ be the point on $T$ where the tangent to $T$ is parallel to $AB$.
Then the area $S$ of the parabolic segment $ABC$ of $T$ is given by:
:$S = \dfrac 4 3 \triangle ABC$

Proof:
Let $T$ be the parabola which is the locus of points $\tuple {x, y}$ satisfying $y = x^2$.
By Area of Triangle Inscribed in Parabola:
:the point $C$ where the tangent to $T$ at $C$ is parallel to $AB$ has $x$-coordinate $\dfrac 1 2 \paren {x_0 + x_2}$.
Let $d$ be the horizontal distance between $A$ and $B$.
By Area of Triangle Inscribed in Parabola:
:$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$
Now construct two triangles.  
One is $BC$ bisected with point $E$ on the parabola, forming:
:$\triangle EBC$
The other is $AC$ bisected with point $D$ on the parabola, forming:
:$\triangle DAC$
For each triangle, the length along the $x$-coordinate $\Delta x$ is equal to $\dfrac d 2$.
The area of each triangle is then:
:$\AA = \dfrac 1 8 \paren {\dfrac d 2}^3 = \dfrac 1 8 \cdot \dfrac 1 8 d^3$
From above:
:$\map \AA {\triangle ABC} = \dfrac 1 8 d^3$
Thus
:$\map \AA {\triangle EBC} = \map \AA {\triangle DAC} = \dfrac 1 8 \map \AA {ABC}$
For two triangles, the area together is: 
:$\map \AA {\triangle DAC} + \map \AA {\triangle EBC}  = \dfrac 1 4 \map \AA {ABC}$
By a variant of the method of exhaustion, the total area over the parabola and under $AB$ is:
:$\map \AA {ABC} \cdot \paren {1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots}$
{{explain|If we are going to reference a method of construction, it is mandatory according to the site philosophy to link directly to it. So this "variant" needs its own page in order to make this proof tight.}}
By Sum of Infinite Geometric Sequence:
:$1 + \dfrac 1 4 + \dfrac 1 {4^2} + \dots = \dfrac 4 3$
Hence the total area is:
:$S = \map \AA {ABC} \cdot \dfrac 4 3$
{{qed}}
Theorem: Lune of Hippocrates

Take the circle whose center is $A$ and whose radius is $AB = AC = AD = AE$.
Let $C$ be the center of a circle whose radius is $CD = CF = CE$.
:400px
Consider the lune $DFEB$.
Its area is equal to that of the square $AEGC$.

Proof:
:400px
The chords $DB$ and $EB$ are tangent to the arc $DFE$. They divide the lune into three regions: yellow, green and blue.
From Pythagoras's Theorem, $CD = \sqrt 2 AD$.
The green and blue areas are of equal area as each subtend a right angle.
The orange area also subtends a right angle.
So the area of the orange area is $\paren {\sqrt 2}^2$ the area of either the green or blue areas.
That is, the orange area equals the sum of the green and blue areas together.
Thus the area of the lune $DFEB$ is equal to the area of $\triangle DEB$.
It is a simple matter then to show that the area of $\triangle DEB$ is the same as the area of the square $AEGC$.
{{qed}}
{{Namedfor|Hippocrates of Chios|cat = Hippocrates}}
Theorem: Integral of Power

:$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

Proof:
From the Fundamental Theorem of Calculus:
:$(1): \quad \ds \int_0^b x^n \rd x = \bigintlimits {\map F x} 0 b = \map F b - \map F 0$
where $\map F x$ is a primitive of $x^n$.
By Primitive of Power, $\dfrac {x^{n + 1} } {n + 1}$ is a primitive of $x^n$.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^b x^n \rd x
      | r = \intlimits {\frac {x^{n + 1} } {n + 1} } 0 b
      | c = substituting $\dfrac {x^{n + 1} } {n + 1}$ for $F$ in $(1)$
}}
{{eqn | r = \frac {b^{n + 1} } {n + 1} - \frac {0^{n + 1} } {n + 1}
      | c = 
}}
{{eqn | r = \frac {b^{n + 1} } {n + 1}
      | c =
}}
{{end-eqn}}
{{qed}}
Theorem: Integral of Power

:$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

Proof:
First let $n$ be a positive integer.
Take a real number $r \in \R$ such that $0 < r < 1$ but reasonably close to $1$.
Consider a subdivision $S$ of the closed interval $\closedint 0 b$ defined as:
:$S = \set {0, \ldots, r^2 b, r b, b}$
that is, by taking as the points of subdivision successive powers of $r$.
Now we take the upper Darboux sum $\map U S$ over $S$ (starting from the right):
{{begin-eqn}}
{{eqn | l = \map U S
      | r = b^n \paren {b - r b} + \paren {r b}^n \paren {r b - r^2 b} + \paren {r^2 b}^n \paren {r^2 b - r^3 b} + \cdots
      | c = 
}}
{{eqn | r = b^{n + 1} \paren {1 - r} + b^{n + 1} r^{n + 1} \paren {1 - r} + b^{n + 1} r^{2 n + 2} \paren {1 - r} + \cdots
      | c = 
}}
{{eqn | r = b^{n + 1} \paren {1 - r} \paren {1 + r^{n + 1} + r^{\paren {n + 1}^2} + \cdots}
      | c = 
}}
{{eqn | r = \frac {b^{n + 1} \paren {1 - r} } {1 - r^{n + 1} }
      | c = Sum of Geometric Sequence
}}
{{eqn | r = \frac {b^{n + 1} } {1 + r + r^2 + \cdots + r^n}
      | c = 
}}
{{end-eqn}}
Now we let $r \to 1$ and see that each of the terms on the bottom also approach $1$.
Thus:
:$\ds \lim_{r \mathop \to 1} S = \frac {b^{n + 1} } {n + 1}$
That is:
:$\ds \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$
for every positive integer $n$.
Now assume $n = \dfrac p q$ be a strictly positive rational number.
We set $s = r^{1/q}$ and proceed:
{{begin-eqn}}
{{eqn | l = \frac {1 - r} {1 - r^{n + 1} }
      | r = \frac {1 - s^q} {1 - \paren {s^q}^{p / q + 1} }
      | c = 
}}
{{eqn | r = \frac {1 - s^q} {1 - s^{p + q} }
      | c = 
}}
{{eqn | r = \frac {\paren {1 - s^q} / \paren {1 - s} } {\paren {1 - s^{p + q} } / \paren {1 - s} }
      | c = 
}}
{{eqn | r = \frac {1 + s + s^2 + \cdots + s^{q - 1} }{1 + s + s^2 + \cdots + s^{p + q - 1} }
      | c = 
}}
{{end-eqn}}
As $r \to 1$ we have $s \to 1$ and so that last expression shows:
{{begin-eqn}}
{{eqn | l = \frac {1 - r} {1 - r^{n + 1} }
      | o = \to
      | r = \frac q {p + q}
      | c = 
}}
{{eqn | r = \frac 1 {p / q + 1}
      | c = 
}}
{{eqn | r = \frac 1 {n + 1}
      | c = 
}}
{{end-eqn}}
So the expression for the main result still holds for rational $n$.
{{qed}}
Theorem: Integral of Power

:$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

Proof:
Let $n \ge 2$.
Let:
:$\ds I_n := \int \sin^n x \rd x$
Then:
{{begin-eqn}}
{{eqn | l = I_n
      | r = \int \sin^n x \rd x
      | c = 
}}
{{eqn | r = \int \sin^{n - 1} x \sin x \rd x
      | c = 
}}
{{eqn | r = \int \sin^{n - 1} x \map \rd {-\cos x}
      | c = Derivative of Cosine Function
}}
{{eqn | r = - \sin^{n - 1} x \cos x - \int \paren {-\cos x} \map \rd {\sin^{n - 1} x}
      | c = Integration by Parts
}}
{{eqn | r = - \sin^{n - 1} x \cos x - \int \paren {-\cos x} \paren {n - 1} \sin^{n - 2} x \cos x \rd x
      | c = Derivative of Power and Chain Rule for Derivatives
}}
{{eqn | r = - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \cos^2 x \rd x
      | c = Linear Combination of Primitives and rearranging
}}
{{eqn | r = - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \paren {1 - \sin^2 x} \rd x
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = - \sin^{n - 1} x \cos x + \paren {n - 1} \int \sin^{n - 2} x \rd x - \paren {n - 1} \int \sin^n x \rd x
      | c = 
}}
{{eqn | r = - \sin^{n - 1} x \cos x + \paren {n - 1} I_{n - 2} - \paren {n - 1} I_n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n I_n
      | r = - \sin^{n - 1} x \cos x + \paren {n - 1} I_{n - 2}
      | c = adding $\paren {n - 1} I_n$ to both sides
}}
{{eqn | ll= \leadsto
      | l = I_n
      | r = \dfrac {n - 1} n I_{n - 2} - \dfrac {\sin^{n - 1} x \cos x} n
      | c = dividing by $n$ and rearranging
}}
{{end-eqn}}
thus demonstrating the identity for all $n \ge 2$.
When $n = 1$ this degenerates to:
{{begin-eqn}}
{{eqn | l = \int \sin x \rd x
      | r = \dfrac 0 1 \int \frac 1 {\sin x} \rd x - \dfrac {\sin^0 x \cos x} 1
      | c = 
}}
{{eqn | r = 0 - 1 \cdot \cos x
      | c = 
}}
{{eqn | r = -\cos x
      | c =
}}
{{end-eqn}}
From Primitive of Sine Function this shows that the identity still holds for $n = 1$.
{{qed}}
Theorem: Integral of Power

:$\ds \forall n \in \R_{\ne -1}: \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$

Proof:
With a view to expressing the problem in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \sin^{n - 1} x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \paren {n - 1} \sin ^{n - 2} x \cos x
      | c = Chain Rule for Derivatives, Derivative of Sine Function, Derivative of Power
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \sin x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = -\cos x
      | c = Primitive of Sine Function
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \sin^n x \rd x
      | r = \int \sin^{n - 1} x \sin x \rd x
      | c = 
}}
{{eqn | r = \sin^{n - 1} x \paren {-\cos x} - \int \paren {-\cos x} \paren {\paren {n - 1} \sin^{n - 2} x \cos x} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \int \paren {n - 1} \sin^{n - 2} x \cos^2 x \rd x - \sin^{n - 1} x \cos x
      | c = rearranging
}}
{{eqn | r = \int \paren {n - 1} \sin^{n - 2} x \paren {1 - \sin^2 x} \rd x - \sin^{n - 1} x \cos x
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \paren {n - 1} \int \sin^{n - 2} x \rd x - \paren {n - 1} \int \sin^n x \rd x - \sin^{n - 1} x \cos x
      | c = Linear Combination of Primitives
}}
{{eqn | ll= \leadsto
      | l = n \int \sin^n x \rd x
      | r = \paren {n - 1} \int \sin^{n - 2} x \rd x - \sin^{n - 1} x \cos x
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \int \sin^n x \rd x
      | r = \dfrac {n - 1} n \int \sin^{n - 2} x \rd x - \dfrac {\sin^{n - 1} x \cos x} n
      | c = dividing both sides by $n$
}}
{{end-eqn}}
{{qed}}
Theorem: Volume of Sphere

The volume $V$ of a sphere of radius $r$ is given by:
:$V = \dfrac {4 \pi r^3} 3$

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.
:300px
On the {{LHS}} of equation $(3)$, the first two areas are added and multiplied by $2a$.
On the {{RHS}}, 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:
:{{AuthorRef|Democritus}}'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:
:$V = \dfrac 4 3 \pi a^3$
{{qed}}
Theorem: Volume of Sphere

The volume $V$ of a sphere of radius $r$ is given by:
:$V = \dfrac {4 \pi r^3} 3$

Proof:
:600px
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:
{{begin-eqn}}
{{eqn | l = V_C
      | r = \pi r^2 \times 2 r
      | c = 
}}
{{eqn | r = 2 \pi r^3
      | c = 
}}
{{end-eqn}}
$K$ 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:
{{begin-eqn}}
{{eqn | l = V_K
      | r = 2 \times \dfrac 1 3 \times \pi r^2 \times r
      | c = 
}}
{{eqn | r = \dfrac 2 3 \pi r^3
      | c = 
}}
{{end-eqn}}
Thus the volume $V_{C'}$ of $C'$ is given by:
{{begin-eqn}}
{{eqn | l = V_{C'}
      | r = 2 \pi r^3 - \dfrac 2 3 \pi r^3
      | c = 
}}
{{eqn | r = \dfrac {\left({6 - 2}\right) \pi r^3} 3
      | c = 
}}
{{eqn | r = \dfrac {4 \pi r^3} 3
      | c = 
}}
{{end-eqn}}
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.
{{qed}}
Theorem: Volume of Sphere

The volume $V$ of a sphere of radius $r$ is given by:
:$V = \dfrac {4 \pi r^3} 3$

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:
:$y = \sqrt {r^2 - x^2}$
:300px
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:
{{begin-eqn}}
{{eqn | l = V
      | r = \pi \int_{-r}^r \paren {\sqrt {r^2 - x^2} }^2 \rd x
      | c = 
}}
{{eqn | r = \pi \int_{-r}^r \paren {r^2 - x^2} \rd x
      | c = 
}}
{{eqn | r = \intlimits {\pi r^2 x - \pi \frac {x^3} 3} {x \mathop = -r} {x \mathop = r}
      | c = Linear Combination of Definite Integrals,  Integral of Constant, Power Rule
}}
{{eqn | r = \paren {\pi r^3 - \pi \frac {r^3} 3} - \paren {\pi \paren {-r^3} + \pi \frac {-r^3} 3}
}}
{{eqn | r = 2 \pi r^3 - \frac 2 3 \pi r^3
}}
{{eqn | r = \frac {4 \pi r^3} 3
}}
{{end-eqn}}
{{qed}}