hcju/proofwikiqa huggingface dataset:
From Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers we have that $\struct {\R_{>0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the set of real numbers without zero: :$\R_{\ne 0} = \R \setminus \set 0$ From Subgroup of Abelian Group is Abelian it also follows that $\struct {\R_{>0}, \times}$ is an abelian group. Its infinite nature follows from the nature of real numbers. {{Handwaving|Strictly state how the positive real numbers are uncountable. Straightforward, but we have no page for it.}} {{qed}}
{{MissingLinks}} {{improve|The justification for the derivation of $(1)$ from $(2)$ is handwavey and clumsy. Better to do $(1)$ first and then derive $(2)$ from it, to save having to solve a second order DE whose solution we have not yet put on {{ProofWiki}}. As the separate proofs for the separate parts have now been extracted into their own pages, complete with justifying links throughout (and no handwvery), it is suggested that this proof may be deleted.}} $B$ has acceleration $\mathbf a$. Let $\mathbf x$ be the vector corresponding to the position of $B$ at time $t$. Then: :$\dfrac {\d^2 \mathbf x} {\d t^2} = \mathbf a$ Solving this differential equation: :$\mathbf x = \mathbf c_0 + \mathbf c_1 t + \frac 1 2 \mathbf a t^2$ with $\mathbf c_0$ and $\mathbf c_1$ constant vectors. Evaluating $\mathbf x$ at $t = 0$ shows that $\mathbf c_0$ is the value $\mathbf x_0$ of $\mathbf x$ at time $t=0$. Taking the derivative of $\mathbf x$ at $t = 0$ shows that $\mathbf c_1$ corresponds to $\mathbf u$. Therefore, since $\mathbf s = \mathbf x - \mathbf x_0$, we have: :$\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$ and by taking the derivative of $\mathbf x$, we have: :$\mathbf v = \mathbf u + \mathbf a t$ Next, we dot $\mathbf v$ into itself using the previous statement. From the linearity and commutativity of the dot product: {{begin-eqn}} {{eqn | l = \mathbf v \cdot \mathbf v       | r = \paren {\mathbf u + \mathbf a t} \cdot \paren {\mathbf u + \mathbf a t}       | c =  }} {{eqn | r = \mathbf u \cdot \mathbf u + 2 \mathbf u \cdot \mathbf a t + t^2 \mathbf a \cdot \mathbf a       | c =  }} {{eqn | r = \mathbf u \cdot \mathbf u + \mathbf a \cdot \paren {2 \mathbf u t + t^2 \mathbf a}       | c =  }} {{end-eqn}} The expression in parentheses is $2 \mathbf s$, so: :$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf a \cdot \mathbf s$ and the proof is complete. {{qed}}
From Picard's Existence Theorem, every point in a given rectangle is passed through by some curve which is the solution of a given integral curve of a differential equation. The equation of this family can be written as: :$y = \map y {x, c}$ where different values of $c$ give different curves. The integral curve which passes through $\tuple {x_0, y_0}$ corresponds to the value of $c$ such that: :$y_0 = \map y {x_0, c}$ Conversely, consider the one-parameter family of curves described by: :$\map f {x, y, c} = 0$ Differentiate $f$ {{WRT|Differentiation}} $x$ to get a relation in the form: :$\map g {x, y, \dfrac {\d y} {\d x}, c} = 0$ Then eliminate $c$ between these expressions for $f$ and $g$ and get: :$\map F {x, y, \dfrac {\d y} {\d x} } = 0$ which is a first order ordinary differential equation. {{qed}}
Let $\map f {x, y, z}$ define a one-parameter family of curves $\FF$. From One-Parameter Family of Curves for First Order ODE, there is a corresponding first order ODE: :$\map F {x, y, \dfrac {\d y} {\d x} }$ whose solution is $\FF$. From Slope of Orthogonal Curves, the slope of one curve is the negative reciprocal of any curve orthogonal to it. So take the equation: :$\map F {x, y, \dfrac {\d y} {\d x} }$ and from it create the equation: :$\map F {x, y, -\dfrac {\d x} {\d y} }$ that is, replace $\dfrac {\d y} {\d x}$ with $-\dfrac {\d x} {\d y}$. This is also a first order ODE, which corresponds with a one-parameter family of curves $\GG$ defined by the implicit function $\map f {x, y, z}$. There is clearly one way of doing the above. Hence the result. {{qed}}
Let the circle have rolled so that the radius to the point $P = \tuple {x, y}$ is at angle $\theta$ to the vertical. :620px The center of the circle is at $\tuple {a \theta, a}$. Then it follows from the definition of sine and cosine that: :$x = a \theta - a \sin \theta$ :$y = a - a \cos \theta$ whence the result. {{qed}}
Let $L$ be the length of one arc of the cycloid. From Arc Length for Parametric Equations: :$\displaystyle L = \int_0^{2 \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$ where, from Equation of Cycloid: :$x = a \paren {\theta - \sin \theta}$ :$y = a \paren {1 - \cos \theta}$ we have: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = a \paren {1 - \cos \theta}       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = a \sin \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2       | r = a^2 \paren {\paren {1 - \cos \theta}^2 + \sin^2 \theta}       | c =  }} {{eqn | r = a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}       | c =  }} {{eqn | r = 2 a^2 \paren {1 - \cos \theta}       | c =  }} {{eqn | r = 4 a^2 \map {\sin^2} {\theta / 2}       | c = Half Angle Formula for Sine }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = L       | r = \int_0^{2 \pi} 2 a \, \map \sin {\theta / 2} \rd \theta       | c =  }} {{eqn | r = \bigintlimits {-4 a \, \map \cos {\theta / 2} } 0 {2 \pi}       | c =  }} {{eqn | r = 8 a       | c =  }} {{end-eqn}} So $L = 8 a$ where $a$ is the radius of the generating circle. {{qed}}
Consider the tangent $PQ$ to both the generating circle and the cycloid itself. :500px By {{wtd|still to be done}}: :$PR = 2 PQ$ In the limit, where $P$ 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. {{finish}}
Let $A$ be the area under of one arc of the cycloid.  Then $A$ 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}}
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 huudk9xbz4ybvfejj1bisse0czjilny
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. === Proof of Existence === Proof by Complete Induction: Let $\nu$ be the Euclidean valuation function on $D$. For all $n \in \Z_{>0}$, let $\map P n$ be the proposition: :every $x \in D$ such that $\map \nu x = n$ is either a unit of $D$ or has a complete factorization in $D$. That is, it can be written as the product of a finite number of irreducible elements. ==== Basis for the Induction ==== If $\map \nu x = \map \nu 1$ then $x$ is a unit of $D$. So $\map P n$ is true for $n = \map \nu 1$. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P k$ is true for all values of $k < \map \nu x$, then it logically follows that $\map P n$ is true. So this is our induction hypothesis: :Every $x \in D$ such that $\map \nu x < n$ is either a unit of $D$ or can be written as the product of a finite number of irreducible elements. Then we need to show: :Every $x \in D$ such that $\map \nu x = n$ is either a unit of $D$ or can be written as the product of a finite number of irreducible elements. ==== Induction Step ==== This is our induction step: Suppose $x \in D$ such that $\map \nu x = n > \map \nu 1$. If $x$ is irreducible, then it is the product of a finite number (that is, $1$) of irreducible elements. If not, then $x = y z$, where neither $b$ nor $c$ are either unit of $D$ or irreducible. By Euclidean Valuation of Non-Unit is less than that of Product we have that: :$\map \nu y < \map \nu x$ and $\map \nu z < \map \nu x$ By the induction hypothesis, we assume the truth of $\map P n$ for all values of $n < \map \nu x$. Hence we know that both $y$ and $z$ can be written as the product of a finite number of irreducible elements. Thus we may deduce the same about $x$. So $\map P n$ has been shown to be true by the Second Principle of Mathematical Induction. {{qed|lemma}} === Proof of Equivalence === Now we need to show that any two complete factorizations of $x$ in $D$ are equivalent. Suppose $p_1 p_2 \ldots p_r = q_1 q_2 \ldots q_s$ where all the $p$s and $q$s are irreducible elements of $D$. Then: :$p_1 \divides q_1 q_2 \ldots q_s$ and by Euclid's Lemma for Irreducible Elements: :$p_1 \divides q_i$ for some $i$. Since each of these is irreducible, they must by definition either be associates or that $p_1 = q_i$. Thus we can write $q_i = u_i p_1$ where $u_i$ is a unit of $D$. Cancelling $p_1$ from both sides, we continue similarly with $p_2$, and so on. After a finite number of steps we determine that $r = s$ and that $q_1, q_2, \ldots, q_s$ are just associates of $p_1, p_2, \ldots p_r$ perhaps in a different order. It follows directly, by definition, that $p_1 p_2 \ldots p_r$ and $q_1 q_2 \ldots q_s$ are equivalent factorizations. {{qed}}
{{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}}
: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}}
From the Fundamental Theorem of Calculus: :$(1): \quad \displaystyle \int_0^b x^n \ \mathrm d x = \left[{F \left({x}\right)}\right]_0^b = F \left({b}\right) - F \left({0}\right)$ where $F \left({x}\right)$ 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 \mathrm d x       | r = \left[{\frac {x^{n+1} } {n+1} }\right]_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}}
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 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: :$\displaystyle \lim_{r \mathop \to 1} S = \frac {b^{n + 1} } {n + 1}$ That is: :$\displaystyle \int_0^b x^n \rd x = \frac {b^{n + 1} } {n + 1}$ 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}}
Let $n \ge 2$. Let: : $\displaystyle I_n := \int \sin^n x \ \mathrm d x$ Then: {{begin-eqn}} {{eqn | l = I_n       | r = \int \sin^n x \ \mathrm d x       | c =  }} {{eqn | r = \int \sin^{n-1} x \sin x \ \mathrm d x       | c =  }} {{eqn | r = \int \sin^{n-1} x \ \mathrm d \left({-\cos x}\right)       | c = Derivative of Cosine Function }} {{eqn | r = - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \mathrm d \left({\sin^{n-1} x }\right)       | c = Integration by Parts }} {{eqn | r = - \sin^{n-1} x \cos x - \int \left({-\cos x}\right) \left({n - 1}\right)\sin^{n-2} x \cos x \ \mathrm d x       | c = Derivative of Power and Chain Rule for Derivatives }} {{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \cos^2 x \ \mathrm d x       | c = Linear Combination of Integrals and rearranging }} {{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \left({1 - \sin^2 x}\right) \ \mathrm d x       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) \int \sin^{n-2} x \ \mathrm d x - \left({n - 1}\right) \int \sin^n x \ \mathrm d x       | c =  }} {{eqn | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2} - \left({n - 1}\right) I_n       | c =  }} {{eqn | ll= \implies       | l = n I_n       | r = - \sin^{n-1} x \cos x + \left({n - 1}\right) I_{n-2}       | c = adding $\left({n - 1}\right) I_n$ to both sides }} {{eqn | ll= \implies       | 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 \ \mathrm d x       | r = \dfrac 0 1 \int \frac 1 {\sin x} \ \mathrm d 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}}
With a view to expressing the problem in the form: :$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$ 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 Integrals }} {{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}}
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 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}}
=== 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 Cylinder. From the Method of Disks, the volume of this sphere can be found by the definite integral: :$\displaystyle V = \pi \int_{-r}^r y^2 \rd x$ 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 = -r} {x = r}       | c = Linear Combination of 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}}