Proof

Consider a cylinder ${\displaystyle x^{2}+y^{2}=r^{2}}$ bounded below by plane ${\displaystyle z=0}$ and above by plane ${\displaystyle z=ky}$ where k is the slope of the slanted roof:

${\displaystyle k={h \over r}}$.

Cutting up the volume into slices parallel to the y-axis, then a differential slice, shaped like a triangular prism, has volume

${\displaystyle A(x)\,dx}$

where

${\displaystyle A(x)={1 \over 2}{\sqrt {r^{2}-x^{2}}}\cdot k{\sqrt {r^{2}-x^{2}}}={1 \over 2}k(r^{2}-x^{2})}$

is the area of a right triangle whose vertices are, ${\displaystyle (x,0,0)}$, ${\displaystyle (x,{\sqrt {r^{2}-x^{2}}},0)}$, and ${\displaystyle (x,{\sqrt {r^{2}-x^{2}}},k{\sqrt {r^{2}-x^{2}}})}$, and whose base and height are thereby ${\displaystyle {\sqrt {r^{2}-x^{2}}}}$ and ${\displaystyle k{\sqrt {r^{2}-x^{2}}}}$, respectively. Then the volume of the whole cylindrical ungula is

${\displaystyle V=\int _{-r}^{r}A(x)\,dx=\int _{-r}^{r}{1 \over 2}k(r^{2}-x^{2})\,dx}$

${\displaystyle \qquad ={1 \over 2}k{\Big (}[r^{2}x]_{-r}^{r}-{\Big [}{1 \over 3}x^{3}{\Big ]}_{-r}^{r}{\Big )}={1 \over 2}k(2r^{3}-{2 \over 3}r^{3})={2 \over 3}kr^{3}}$

which equals

${\displaystyle V={2 \over 3}r^{2}h}$

after substituting ${\displaystyle rk=h}$.

A differential surface area of the curved side wall is

${\displaystyle dA_{s}=kr(\sin \theta )\cdot r\,d\theta =kr^{2}(\sin \theta )\,d\theta }$,

which area belongs to a nearly flat rectangle bounded by vertices ${\displaystyle (r\cos \theta ,r\sin \theta ,0)}$, ${\displaystyle (r\cos \theta ,r\sin \theta ,kr\sin \theta )}$, ${\displaystyle (r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),0)}$, and ${\displaystyle (r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),kr\sin(\theta +d\theta ))}$, and whose width and height are thereby ${\displaystyle r\,d\theta }$ and (close enough to) ${\displaystyle kr\sin \theta }$, respectively. Then the surface area of the wall is

${\displaystyle A_{s}=\int _{0}^{\pi }dA_{s}=\int _{0}^{\pi }kr^{2}(\sin \theta )\,d\theta =kr^{2}\int _{0}^{\pi }\sin \theta \,d\theta }$

where the integral yields ${\displaystyle -[\cos \theta ]_{0}^{\pi }=-[-1-1]=2}$, so that the area of the wall is

${\displaystyle A_{s}=2kr^{2}}$,

and substituting ${\displaystyle rk=h}$ yields

${\displaystyle A_{s}=2rh}$.

The base of the cylindrical ungula has the surface area of half a circle of radius r: ${\displaystyle {1 \over 2}\pi r^{2}}$, and the slanted top of the said ungula is a half-ellipse with semi-minor axis of length r and semi-major axis of length ${\displaystyle r{\sqrt {1+k^{2}}}}$, so that its area is

${\displaystyle A_{t}={1 \over 2}\pi r\cdot r{\sqrt {1+k^{2}}}={1 \over 2}\pi r{\sqrt {r^{2}+(kr)^{2}}}}$

and substituting ${\displaystyle kr=h}$ yields

${\displaystyle A_{t}={1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}}$. ∎

Note how the surface area of the side wall is related to the volume: such surface area being ${\displaystyle 2kr^{2}}$, multiplying it by ${\displaystyle dr}$ gives the volume of a differential half-shell, whose integral is ${\displaystyle {2 \over 3}kr^{3}}$, the volume.

When the slope k equals 1 then such ungula is precisely one eighth of a bicylinder, whose volume is ${\displaystyle {16 \over 3}r^{3}}$. One eighth of this is ${\displaystyle {2 \over 3}r^{3}}$.

Proof

Let a cone be described by

${\displaystyle 1-{\rho \over r}={z \over H}}$

where r and H are constants and z and ρ are variables, with

${\displaystyle \rho ={\sqrt {x^{2}+y^{2}}},\qquad 0\leq \rho \leq r}$

and

${\displaystyle x=\rho \cos \theta ,\qquad y=\rho \sin \theta }$.

Let the cone be cut by a plane

${\displaystyle z=ky=k\rho \sin \theta }$.

Substituting this z into the cone's equation, and solving for ρ yields

${\displaystyle \rho _{0}={1 \over {1 \over r}+{k\sin \theta \over H}}}$

which for a given value of θ is the radial coordinate of the point common to both the plane and the cone that is farthest from the cone's axis along an angle θ from the x-axis. The cylindrical height coordinate of this point is

${\displaystyle z_{0}=H{\Big (}1-{\rho _{0} \over r}{\Big )}}$.

So along the direction of angle θ, a cross-section of the conical ungula looks like the triangle

${\displaystyle (0,0,0)-(\rho _{0}\cos \theta ,\rho _{0}\sin \theta ,z_{0})-(r\cos \theta ,r\sin \theta ,0)}$.

Rotating this triangle by an angle ${\displaystyle d\theta }$ about the z-axis yields another triangle with ${\displaystyle \theta +d\theta }$, ${\displaystyle \rho _{1}}$, ${\displaystyle z_{1}}$ substituted for ${\displaystyle \theta }$, ${\displaystyle \rho _{0}}$, and ${\displaystyle z_{0}}$ respectively, where ${\displaystyle \rho _{1}}$ and ${\displaystyle z_{1}}$ are functions of ${\displaystyle \theta +d\theta }$ instead of ${\displaystyle \theta }$. Since ${\displaystyle d\theta }$ is infinitesimal then ${\displaystyle \rho _{1}}$ and ${\displaystyle z_{1}}$ also vary infinitesimally from ${\displaystyle \rho _{0}}$ and ${\displaystyle z_{0}}$, so for purposes of considering the volume of the differential trapezoidal pyramid, they may be considered equal.

The differential trapezoidal pyramid has a trapezoidal base with a length at the base (of the cone) of ${\displaystyle rd\theta }$, a length at the top of ${\displaystyle {\Big (}{H-z_{0} \over H}{\Big )}rd\theta }$, and altitude ${\displaystyle {z_{0} \over H}{\sqrt {r^{2}+H^{2}}}}$, so the trapezoid has area

${\displaystyle A_{T}={r\,d\theta +{\Big (}{H-z_{0} \over H}{\Big )}r\,d\theta \over 2}{z_{0} \over H}{\sqrt {r^{2}+H^{2}}}=r\,d\theta {(2H-z_{0})z_{0} \over 2H^{2}}{\sqrt {r^{2}+H^{2}}}}$.

An altitude from the trapezoidal base to the point ${\displaystyle (0,0,0)}$ has length differentially close to

${\displaystyle {rH \over {\sqrt {r^{2}+H^{2}}}}}$.

(This is an altitude of one of the side triangles of the trapezoidal pyramid.) The volume of the pyramid is one-third its base area times its altitudinal length, so the volume of the conical ungula is the integral of that:

${\displaystyle V=\int _{0}^{\pi }{1 \over 3}{rH \over {\sqrt {r^{2}+H^{2}}}}{(2H-z_{0})z_{0} \over 2H^{2}}{\sqrt {r^{2}+H^{2}}}r\,d\theta =\int _{0}^{\pi }{1 \over 3}r^{2}{(2H-z_{0})z_{0} \over 2H}d\theta ={r^{2}k \over 6H}\int _{0}^{\pi }(2H-ky_{0})y_{0}\,d\theta }$

where

${\displaystyle y_{0}=\rho _{0}\sin \theta ={\sin \theta \over {1 \over r}+{k\sin \theta \over H}}={1 \over {1 \over r\sin \theta }+{k \over H}}}$

Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven.

For the sidewall:

${\displaystyle A_{s}=\int _{0}^{\pi }A_{T}=\int _{0}^{\pi }{(2H-z_{0})z_{0} \over 2H^{2}}r{\sqrt {r^{2}+H^{2}}}\,d\theta ={kr{\sqrt {r^{2}+H^{2}}} \over 2H^{2}}\int _{0}^{\pi }(2H-z_{0})y_{0}\,d\theta }$

and the integral on the rightmost-hand-side simplifies to ${\displaystyle H^{2}rI}$. ∎

As a consistency check, consider what happens when k goes to infinity; then the conical ungula should become a semi-cone.

${\displaystyle \lim _{k\rightarrow \infty }{\Big (}I-{\pi \over kr}{\Big )}=0}$

${\displaystyle \lim _{k\rightarrow \infty }V={r^{3}kH \over 6}\cdot {\pi \over kr}={1 \over 2}{\Big (}{1 \over 3}\pi r^{2}H{\Big )}}$

which is half of the volume of a cone.

${\displaystyle \lim _{k\rightarrow \infty }A_{s}={kr^{2}{\sqrt {r^{2}+H^{2}}} \over 2}\cdot {\pi \over kr}={1 \over 2}\pi r{\sqrt {r^{2}+H^{2}}}}$

which is half of the surface area of the curved wall of a cone.

Surface area of top part

When ${\displaystyle k=H/r}$, the "top part" (i.e., the flat face that is not semicircular like the base) has a parabolic shape and its surface area is

${\displaystyle A_{t}={2 \over 3}r{\sqrt {r^{2}+H^{2}}}}$.

When ${\displaystyle k<H/r}$ then the top part has an elliptic shape (i.e., it is less than one-half of an ellipse) and its surface area is

${\displaystyle A_{t}={1 \over 2}\pi x_{max}(y_{1}-y_{m}){\sqrt {1+k^{2}}}\Lambda }$

where

${\displaystyle x_{max}={\sqrt {{k^{2}r^{4}H^{2}-k^{4}r^{6} \over (k^{2}r^{2}-H^{2})^{2}}+r^{2}}}}$,

${\displaystyle y_{1}={1 \over {1 \over r}+{k \over H}}}$,

${\displaystyle y_{m}={kr^{2}H \over k^{2}r^{2}-H^{2}}}$,

${\displaystyle \Lambda ={\pi \over 4}-{1 \over 2}\arcsin(1-\lambda )-{1 \over 4}\sin(2\arcsin(1-\lambda ))}$, and

${\displaystyle \lambda ={y_{1} \over y_{1}-y_{m}}}$.

When ${\displaystyle k>H/r}$ then the top part is a section of a hyperbola and its surface area is

${\displaystyle A_{t}={\sqrt {1+k^{2}}}(2Cr-aJ)}$

where

${\displaystyle C={y_{1}+y_{2} \over 2}=y_{m}}$,

${\displaystyle y_{1}}$ is as above,

${\displaystyle y_{2}={1 \over {k \over H}-{1 \over r}}}$,

${\displaystyle a={r \over {\sqrt {C^{2}-\Delta ^{2}}}}}$,

${\displaystyle \Delta ={y_{2}-y_{1} \over 2}}$,

${\displaystyle J={r \over a}B+{\Delta ^{2} \over 2}\log {\Biggr |}{{r \over a}+B \over {-r \over a}+B}{\Biggr |}}$,

where the logarithm is natural, and

${\displaystyle B={\sqrt {\Delta ^{2}+{r^{2} \over a^{2}}}}}$.