In solid geometry , an ungula is a region of a solid of revolution , cut off by a plane oblique to its base.[ 1] A common instance is the spherical wedge . The term ungula refers to the hoof of a horse , an anatomical feature that defines a class of mammals called ungulates .
The volume of an ungula of a cylinder was calculated by Grégoire de Saint Vincent .[ 2] Two cylinders with equal radii and perpendicular axes intersect in four double ungulae.[ 3] The bicylinder formed by the intersection had been measured by Archimedes in The Method of Mechanical Theorems , but the manuscript was lost until 1906.
A historian of calculus described the role of the ungula in integral calculus :
Grégoire himself was primarily concerned to illustrate by reference to the ungula that volumetric integration could be reduced, through the ductus in planum , to a consideration of geometric relations between the lies of plane figures. The ungula , however, proved a valuable source of inspiration for those who followed him, and who saw in it a means of representing and transforming integrals in many ingenious ways.[ 4] : 146
Ungula of a right circular cylinder.
A cylindrical ungula of base radius r and height h has volume
V
=
2
3
r
2
h
{\displaystyle V={2 \over 3}r^{2}h}
,.[ 5]
Its total surface area is
A
=
1
2
π
r
2
+
1
2
π
r
r
2
+
h
2
+
2
r
h
{\displaystyle A={1 \over 2}\pi r^{2}+{1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}+2rh}
,
the surface area of its curved sidewall is
A
s
=
2
r
h
{\displaystyle A_{s}=2rh}
,
and the surface area of its top (slanted roof) is
A
t
=
1
2
π
r
r
2
+
h
2
{\displaystyle A_{t}={1 \over 2}\pi r{\sqrt {r^{2}+h^{2}}}}
.
Consider a cylinder
x
2
+
y
2
=
r
2
{\displaystyle x^{2}+y^{2}=r^{2}}
bounded below by plane
z
=
0
{\displaystyle z=0}
and above by plane
z
=
k
y
{\displaystyle z=ky}
where k is the slope of the slanted roof:
k
=
h
r
{\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
A
(
x
)
d
x
{\displaystyle A(x)\,dx}
where
A
(
x
)
=
1
2
r
2
−
x
2
⋅
k
r
2
−
x
2
=
1
2
k
(
r
2
−
x
2
)
{\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,
(
x
,
0
,
0
)
{\displaystyle (x,0,0)}
,
(
x
,
r
2
−
x
2
,
0
)
{\displaystyle (x,{\sqrt {r^{2}-x^{2}}},0)}
, and
(
x
,
r
2
−
x
2
,
k
r
2
−
x
2
)
{\displaystyle (x,{\sqrt {r^{2}-x^{2}}},k{\sqrt {r^{2}-x^{2}}})}
,
and whose base and height are thereby
r
2
−
x
2
{\displaystyle {\sqrt {r^{2}-x^{2}}}}
and
k
r
2
−
x
2
{\displaystyle k{\sqrt {r^{2}-x^{2}}}}
, respectively.
Then the volume of the whole cylindrical ungula is
V
=
∫
−
r
r
A
(
x
)
d
x
=
∫
−
r
r
1
2
k
(
r
2
−
x
2
)
d
x
{\displaystyle V=\int _{-r}^{r}A(x)\,dx=\int _{-r}^{r}{1 \over 2}k(r^{2}-x^{2})\,dx}
=
1
2
k
(
[
r
2
x
]
−
r
r
−
[
1
3
x
3
]
−
r
r
)
=
1
2
k
(
2
r
3
−
2
3
r
3
)
=
2
3
k
r
3
{\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
V
=
2
3
r
2
h
{\displaystyle V={2 \over 3}r^{2}h}
after substituting
r
k
=
h
{\displaystyle rk=h}
.
A differential surface area of the curved side wall is
d
A
s
=
k
r
(
sin
θ
)
⋅
r
d
θ
=
k
r
2
(
sin
θ
)
d
θ
{\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
(
r
cos
θ
,
r
sin
θ
,
0
)
{\displaystyle (r\cos \theta ,r\sin \theta ,0)}
,
(
r
cos
θ
,
r
sin
θ
,
k
r
sin
θ
)
{\displaystyle (r\cos \theta ,r\sin \theta ,kr\sin \theta )}
,
(
r
cos
(
θ
+
d
θ
)
,
r
sin
(
θ
+
d
θ
)
,
0
)
{\displaystyle (r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),0)}
, and
(
r
cos
(
θ
+
d
θ
)
,
r
sin
(
θ
+
d
θ
)
,
k
r
sin
(
θ
+
d
θ
)
)
{\displaystyle (r\cos(\theta +d\theta ),r\sin(\theta +d\theta ),kr\sin(\theta +d\theta ))}
, and whose width and height are thereby
r
d
θ
{\displaystyle r\,d\theta }
and (close enough to)
k
r
sin
θ
{\displaystyle kr\sin \theta }
, respectively.
Then the surface area of the wall is
A
s
=
∫
0
π
d
A
s
=
∫
0
π
k
r
2
(
sin
θ
)
d
θ
=
k
r
2
∫
0
π
sin
θ
d
θ
{\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
−
[
cos
θ
]
0
π
=
−
[
−
1
−
1
]
=
2
{\displaystyle -[\cos \theta ]_{0}^{\pi }=-[-1-1]=2}
, so that the area of the wall is
A
s
=
2
k
r
2
{\displaystyle A_{s}=2kr^{2}}
,
and substituting
r
k
=
h
{\displaystyle rk=h}
yields
A
s
=
2
r
h
{\displaystyle A_{s}=2rh}
.
The base of the cylindrical ungula has the surface area of half a circle of radius r :
1
2
π
r
2
{\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
r
1
+
k
2
{\displaystyle r{\sqrt {1+k^{2}}}}
, so that its area is
A
t
=
1
2
π
r
⋅
r
1
+
k
2
=
1
2
π
r
r
2
+
(
k
r
)
2
{\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
k
r
=
h
{\displaystyle kr=h}
yields
A
t
=
1
2
π
r
r
2
+
h
2
{\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
2
k
r
2
{\displaystyle 2kr^{2}}
, multiplying it by
d
r
{\displaystyle dr}
gives the volume of a differential half-shell , whose integral is
2
3
k
r
3
{\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
16
3
r
3
{\displaystyle {16 \over 3}r^{3}}
. One eighth of this is
2
3
r
3
{\displaystyle {2 \over 3}r^{3}}
.
Ungula of a right circular cone.
A conical ungula of height h , base radius r , and upper flat surface slope k (if the semicircular base is at the bottom, on the plane z = 0) has volume
V
=
r
3
k
H
I
6
{\displaystyle V={r^{3}kHI \over 6}}
where
H
=
1
1
h
−
1
r
k
{\displaystyle H={1 \over {1 \over h}-{1 \over rk}}}
is the height of the cone from which the ungula has been cut out, and
I
=
∫
0
π
2
H
+
k
r
sin
θ
(
H
+
k
r
sin
θ
)
2
sin
θ
d
θ
{\displaystyle I=\int _{0}^{\pi }{2H+kr\sin \theta \over (H+kr\sin \theta )^{2}}\sin \theta \,d\theta }
.
The surface area of the curved sidewall is
A
s
=
k
r
2
r
2
+
H
2
2
I
{\displaystyle A_{s}={kr^{2}{\sqrt {r^{2}+H^{2}}} \over 2}I}
.
As a consistency check, consider what happens when the height of the cone goes to infinity, so that the cone becomes a cylinder in the limit:
lim
H
→
∞
(
I
−
4
H
)
=
lim
H
→
∞
(
2
H
H
2
∫
0
π
sin
θ
d
θ
−
4
H
)
=
0
{\displaystyle \lim _{H\rightarrow \infty }{\Big (}I-{4 \over H}{\Big )}=\lim _{H\rightarrow \infty }{\Big (}{2H \over H^{2}}\int _{0}^{\pi }\sin \theta \,d\theta -{4 \over H}{\Big )}=0}
so that
lim
H
→
∞
V
=
r
3
k
H
6
⋅
4
H
=
2
3
k
r
3
{\displaystyle \lim _{H\rightarrow \infty }V={r^{3}kH \over 6}\cdot {4 \over H}={2 \over 3}kr^{3}}
,
lim
H
→
∞
A
s
=
k
r
2
H
2
⋅
4
H
=
2
k
r
2
{\displaystyle \lim _{H\rightarrow \infty }A_{s}={kr^{2}H \over 2}\cdot {4 \over H}=2kr^{2}}
, and
lim
H
→
∞
A
t
=
1
2
π
r
2
1
+
k
2
1
+
0
=
1
2
π
r
2
1
+
k
2
=
1
2
π
r
r
2
+
(
r
k
)
2
{\displaystyle \lim _{H\rightarrow \infty }A_{t}={1 \over 2}\pi r^{2}{{\sqrt {1+k^{2}}} \over 1+0}={1 \over 2}\pi r^{2}{\sqrt {1+k^{2}}}={1 \over 2}\pi r{\sqrt {r^{2}+(rk)^{2}}}}
,
which results agree with the cylindrical case.
Let a cone be described by
1
−
ρ
r
=
z
H
{\displaystyle 1-{\rho \over r}={z \over H}}
where r and H are constants and z and ρ are variables, with
ρ
=
x
2
+
y
2
,
0
≤
ρ
≤
r
{\displaystyle \rho ={\sqrt {x^{2}+y^{2}}},\qquad 0\leq \rho \leq r}
and
x
=
ρ
cos
θ
,
y
=
ρ
sin
θ
{\displaystyle x=\rho \cos \theta ,\qquad y=\rho \sin \theta }
.
Let the cone be cut by a plane
z
=
k
y
=
k
ρ
sin
θ
{\displaystyle z=ky=k\rho \sin \theta }
.
Substituting this z into the cone's equation, and solving for ρ yields
ρ
0
=
1
1
r
+
k
sin
θ
H
{\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
z
0
=
H
(
1
−
ρ
0
r
)
{\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
(
0
,
0
,
0
)
−
(
ρ
0
cos
θ
,
ρ
0
sin
θ
,
z
0
)
−
(
r
cos
θ
,
r
sin
θ
,
0
)
{\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
d
θ
{\displaystyle d\theta }
about the z -axis yields another triangle with
θ
+
d
θ
{\displaystyle \theta +d\theta }
,
ρ
1
{\displaystyle \rho _{1}}
,
z
1
{\displaystyle z_{1}}
substituted for
θ
{\displaystyle \theta }
,
ρ
0
{\displaystyle \rho _{0}}
, and
z
0
{\displaystyle z_{0}}
respectively, where
ρ
1
{\displaystyle \rho _{1}}
and
z
1
{\displaystyle z_{1}}
are functions of
θ
+
d
θ
{\displaystyle \theta +d\theta }
instead of
θ
{\displaystyle \theta }
. Since
d
θ
{\displaystyle d\theta }
is infinitesimal then
ρ
1
{\displaystyle \rho _{1}}
and
z
1
{\displaystyle z_{1}}
also vary infinitesimally from
ρ
0
{\displaystyle \rho _{0}}
and
z
0
{\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
r
d
θ
{\displaystyle rd\theta }
, a length at the top of
(
H
−
z
0
H
)
r
d
θ
{\displaystyle {\Big (}{H-z_{0} \over H}{\Big )}rd\theta }
, and altitude
z
0
H
r
2
+
H
2
{\displaystyle {z_{0} \over H}{\sqrt {r^{2}+H^{2}}}}
, so the trapezoid has area
A
T
=
r
d
θ
+
(
H
−
z
0
H
)
r
d
θ
2
z
0
H
r
2
+
H
2
=
r
d
θ
(
2
H
−
z
0
)
z
0
2
H
2
r
2
+
H
2
{\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
(
0
,
0
,
0
)
{\displaystyle (0,0,0)}
has length differentially close to
r
H
r
2
+
H
2
{\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:
V
=
∫
0
π
1
3
r
H
r
2
+
H
2
(
2
H
−
z
0
)
z
0
2
H
2
r
2
+
H
2
r
d
θ
=
∫
0
π
1
3
r
2
(
2
H
−
z
0
)
z
0
2
H
d
θ
=
r
2
k
6
H
∫
0
π
(
2
H
−
k
y
0
)
y
0
d
θ
{\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
y
0
=
ρ
0
sin
θ
=
sin
θ
1
r
+
k
sin
θ
H
=
1
1
r
sin
θ
+
k
H
{\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:
A
s
=
∫
0
π
A
T
=
∫
0
π
(
2
H
−
z
0
)
z
0
2
H
2
r
r
2
+
H
2
d
θ
=
k
r
r
2
+
H
2
2
H
2
∫
0
π
(
2
H
−
z
0
)
y
0
d
θ
{\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
H
2
r
I
{\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.
lim
k
→
∞
(
I
−
π
k
r
)
=
0
{\displaystyle \lim _{k\rightarrow \infty }{\Big (}I-{\pi \over kr}{\Big )}=0}
lim
k
→
∞
V
=
r
3
k
H
6
⋅
π
k
r
=
1
2
(
1
3
π
r
2
H
)
{\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.
lim
k
→
∞
A
s
=
k
r
2
r
2
+
H
2
2
⋅
π
k
r
=
1
2
π
r
r
2
+
H
2
{\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
edit
When
k
=
H
/
r
{\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
A
t
=
2
3
r
r
2
+
H
2
{\displaystyle A_{t}={2 \over 3}r{\sqrt {r^{2}+H^{2}}}}
.
When
k
<
H
/
r
{\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
A
t
=
1
2
π
x
m
a
x
(
y
1
−
y
m
)
1
+
k
2
Λ
{\displaystyle A_{t}={1 \over 2}\pi x_{max}(y_{1}-y_{m}){\sqrt {1+k^{2}}}\Lambda }
where
x
m
a
x
=
k
2
r
4
H
2
−
k
4
r
6
(
k
2
r
2
−
H
2
)
2
+
r
2
{\displaystyle x_{max}={\sqrt {{k^{2}r^{4}H^{2}-k^{4}r^{6} \over (k^{2}r^{2}-H^{2})^{2}}+r^{2}}}}
,
y
1
=
1
1
r
+
k
H
{\displaystyle y_{1}={1 \over {1 \over r}+{k \over H}}}
,
y
m
=
k
r
2
H
k
2
r
2
−
H
2
{\displaystyle y_{m}={kr^{2}H \over k^{2}r^{2}-H^{2}}}
,
Λ
=
π
4
−
1
2
arcsin
(
1
−
λ
)
−
1
4
sin
(
2
arcsin
(
1
−
λ
)
)
{\displaystyle \Lambda ={\pi \over 4}-{1 \over 2}\arcsin(1-\lambda )-{1 \over 4}\sin(2\arcsin(1-\lambda ))}
, and
λ
=
y
1
y
1
−
y
m
{\displaystyle \lambda ={y_{1} \over y_{1}-y_{m}}}
.
When
k
>
H
/
r
{\displaystyle k>H/r}
then the top part is a section of a hyperbola and its surface area is
A
t
=
1
+
k
2
(
2
C
r
−
a
J
)
{\displaystyle A_{t}={\sqrt {1+k^{2}}}(2Cr-aJ)}
where
C
=
y
1
+
y
2
2
=
y
m
{\displaystyle C={y_{1}+y_{2} \over 2}=y_{m}}
,
y
1
{\displaystyle y_{1}}
is as above,
y
2
=
1
k
H
−
1
r
{\displaystyle y_{2}={1 \over {k \over H}-{1 \over r}}}
,
a
=
r
C
2
−
Δ
2
{\displaystyle a={r \over {\sqrt {C^{2}-\Delta ^{2}}}}}
,
Δ
=
y
2
−
y
1
2
{\displaystyle \Delta ={y_{2}-y_{1} \over 2}}
,
J
=
r
a
B
+
Δ
2
2
log
|
r
a
+
B
−
r
a
+
B
|
{\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
B
=
Δ
2
+
r
2
a
2
{\displaystyle B={\sqrt {\Delta ^{2}+{r^{2} \over a^{2}}}}}
.