Region between a revolved surface and oblique plane
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}}}}} .