Claim: The volume of a conic solid whose base has area b and whose height is h is
1
3
b
h
{\displaystyle {1 \over 3}bh}
Proof: Let
α
→
(
t
)
{\displaystyle {\vec {\alpha }}(t)}
R
3
{\displaystyle \mathbb {R} ^{3}}
v
→
{\displaystyle {\vec {v}}}
α
→
{\displaystyle {\vec {\alpha }}}
Let the conic solid be parametrized by
σ
→
(
λ
,
t
)
=
(
1
−
λ
)
v
→
+
λ
α
→
(
t
)
{\displaystyle {\vec {\sigma }}(\lambda ,t)=(1-\lambda ){\vec {v}}+\lambda \,{\vec {\alpha }}(t)}
where
λ
,
t
∈
[
0
,
1
]
{\displaystyle \lambda ,t\in [0,1]}
For a fixed
λ
=
λ
0
{\displaystyle \lambda =\lambda _{0}}
σ
→
(
λ
0
,
t
)
=
(
1
−
λ
0
)
v
→
+
λ
0
α
→
(
t
)
{\displaystyle {\vec {\sigma }}(\lambda _{0},t)=(1-\lambda _{0}){\vec {v}}+\lambda _{0}\,{\vec {\alpha }}(t)}
α
→
(
t
)
{\displaystyle {\vec {\alpha }}(t)}
λ
0
α
→
(
t
)
{\displaystyle \lambda _{0}\,{\vec {\alpha }}(t)}
α
→
(
t
)
{\displaystyle {\vec {\alpha }}(t)}
(
1
−
λ
0
)
v
→
+
λ
0
α
→
(
t
)
{\displaystyle (1-\lambda _{0}){\vec {v}}+\lambda _{0}\,{\vec {\alpha }}(t)}
λ
0
α
→
(
t
)
{\displaystyle \lambda _{0}\,{\vec {\alpha }}(t)}
Moreover, the shape of
σ
→
(
λ
0
,
t
)
{\displaystyle {\vec {\sigma }}(\lambda _{0},t)}
α
(
t
)
{\displaystyle \alpha (t)}
σ
→
(
λ
0
,
t
)
{\displaystyle {\vec {\sigma }}(\lambda _{0},t)}
λ
0
2
{\displaystyle \lambda _{0}^{2}}
α
→
(
t
)
{\displaystyle {\vec {\alpha }}(t)}
b .
If the perpendiculars distance from the vertex to the plane of the base is h , then the distance between two slices
λ
=
λ
0
{\displaystyle \lambda =\lambda _{0}}
λ
=
λ
1
{\displaystyle \lambda =\lambda _{1}}
d
λ
=
λ
1
−
λ
0
{\displaystyle d\lambda =\lambda _{1}-\lambda _{0}}
h
d
λ
{\displaystyle h\,d\lambda }
d
V
=
(
λ
2
b
)
(
h
d
λ
)
{\displaystyle dV=(\lambda ^{2}b)(h\,d\lambda )}
Now integrate the volume:
V
=
∫
0
1
d
V
=
∫
0
1
b
h
λ
2
d
λ
=
b
h
[
1
3
λ
3
]
0
1
=
1
3
b
h
,
{\displaystyle V=\int _{0}^{1}dV=\int _{0}^{1}bh\lambda ^{2}\,d\lambda =bh\left[{1 \over 3}\lambda ^{3}\right]_{0}^{1}={1 \over 3}bh,}
Claim: the center of mass of a conic solid lies at one-fourth of the way from the center of mass of the base to the vertex.Proof: Let
M
=
ρ
V
{\displaystyle M=\rho V}
ρ is the uniform density and V is the volume (as given above).
A differential slice enclosed by the curve
σ
→
(
λ
0
,
t
)
{\displaystyle {\vec {\sigma }}(\lambda _{0},t)}
λ
=
λ
0
{\displaystyle \lambda =\lambda _{0}}
d
M
=
ρ
d
V
=
ρ
b
h
λ
2
d
λ
{\displaystyle dM=\rho \,dV=\rho bh\lambda ^{2}\,d\lambda }
Let us say that the base of the cone has center of mass
c
→
B
{\displaystyle {\vec {c}}_{B}}
λ
=
λ
0
{\displaystyle \lambda =\lambda _{0}}
c
→
S
(
λ
0
)
=
(
1
−
λ
0
)
v
→
+
λ
0
c
→
B
{\displaystyle {\vec {c}}_{S}(\lambda _{0})=(1-\lambda _{0}){\vec {v}}+\lambda _{0}{\vec {c}}_{B}}
Thus, the center of mass of the cone should be
c
→
c
o
n
e
=
1
M
∫
0
1
c
→
S
(
λ
)
d
M
{\displaystyle {\vec {c}}_{cone}={1 \over M}\int _{0}^{1}{\vec {c}}_{S}(\lambda )\,dM}
=
1
M
∫
0
1
[
(
1
−
λ
)
v
→
+
λ
c
→
B
]
ρ
b
h
λ
2
d
λ
{\displaystyle \qquad ={1 \over M}\int _{0}^{1}[(1-\lambda ){\vec {v}}+\lambda {\vec {c}}_{B}]\rho bh\lambda ^{2}\,d\lambda }
=
ρ
b
h
M
∫
0
1
[
v
→
λ
2
+
(
c
→
B
−
v
→
)
λ
3
]
d
λ
{\displaystyle \qquad ={\rho bh \over M}\int _{0}^{1}[{\vec {v}}\lambda ^{2}+({\vec {c}}_{B}-{\vec {v}})\lambda ^{3}]\,d\lambda }
=
ρ
b
h
M
[
v
→
∫
0
1
λ
2
d
λ
+
(
c
→
B
−
v
→
)
∫
0
1
λ
3
d
λ
]
{\displaystyle \qquad ={\rho bh \over M}\left[{\vec {v}}\int _{0}^{1}\lambda ^{2}\,d\lambda +({\vec {c}}_{B}-{\vec {v}})\int _{0}^{1}\lambda ^{3}\,d\lambda \right]}
=
ρ
b
h
1
3
ρ
b
h
[
1
3
v
→
+
1
4
(
c
→
B
−
v
→
)
]
{\displaystyle \qquad ={\rho bh \over {1 \over 3}\rho bh}\left[{1 \over 3}{\vec {v}}+{1 \over 4}({\vec {c}}_{B}-{\vec {v}})\right]}
=
3
(
v
→
12
+
c
→
B
4
)
{\displaystyle \qquad =3\left({{\vec {v}} \over 12}+{{\vec {c}}_{B} \over 4}\right)}
∴
c
→
c
o
n
e
=
v
→
4
+
3
4
c
→
B
{\displaystyle {\vec {c}}_{cone}={{\vec {v}} \over 4}+{3 \over 4}{\vec {c}}_{B}}
which is to say, that
c
→
c
o
n
e
{\displaystyle {\vec {c}}_{cone}}
c
→
B
{\displaystyle {\vec {c}}_{B}}
v
→
{\displaystyle {\vec {v}}}
Dimensional Comparison
edit
Note that the cone is, in a sense, a higher-dimensional version of a triangle, and that for the case of the triangle, the area is
1
2
b
h
{\displaystyle {1 \over 2}bh}
and the centroid lies 1/3 of the way from the center of mass of the base to the vertex.
A tetrahedron is a special type of cone, and it is also a stricter generalization of the triangle.
Claim: The Surface Area of a right circular cone is equal to
π
r
s
+
π
r
2
{\displaystyle \pi rs+\pi r^{2}}
r
{\displaystyle r}
s
{\displaystyle s}
r
2
+
h
2
{\displaystyle {\sqrt {r^{2}+h^{2}}}}
Proof: The
π
r
2
{\displaystyle \pi r^{2}}
r
{\displaystyle r}
Cut
n
{\displaystyle n}
n
{\displaystyle n}
d
C
{\displaystyle dC}
s
{\displaystyle s}
The number of triangles multiplied by
d
C
{\displaystyle dC}
C
=
2
π
r
{\displaystyle C=2\pi r}
d
C
{\displaystyle dC}
A
=
∫
0
2
π
r
1
2
s
d
C
{\displaystyle A=\int _{0}^{2\pi r}{\frac {1}{2}}sdC}
A
=
[
1
2
s
C
]
0
2
π
r
{\displaystyle A=\left[{\frac {1}{2}}sC\right]_{0}^{2\pi r}}
A
=
π
r
s
{\displaystyle A=\pi rs\!}
A
=
π
r
r
2
+
h
2
{\displaystyle A=\pi r{\sqrt {r^{2}+h^{2}}}}
Thus, the total surface area of the cone is equal to
π
r
2
+
π
r
r
2
+
h
2
{\displaystyle \pi r^{2}+\pi r{\sqrt {r^{2}+h^{2}}}}
![{\displaystyle \lambda ,t\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d41953a670f816b61086a7e8487791ca0d4e1617)
![{\displaystyle V=\int _{0}^{1}dV=\int _{0}^{1}bh\lambda ^{2}\,d\lambda =bh\left[{1 \over 3}\lambda ^{3}\right]_{0}^{1}={1 \over 3}bh,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edf32a64378a9d0550324bae7db4347f914cee5b)
![{\displaystyle \qquad ={1 \over M}\int _{0}^{1}[(1-\lambda ){\vec {v}}+\lambda {\vec {c}}_{B}]\rho bh\lambda ^{2}\,d\lambda }](https://wikimedia.org/api/rest_v1/media/math/render/svg/64d0d78042524e547ae3749bf5302fbb9b178d1b)
![{\displaystyle \qquad ={\rho bh \over M}\int _{0}^{1}[{\vec {v}}\lambda ^{2}+({\vec {c}}_{B}-{\vec {v}})\lambda ^{3}]\,d\lambda }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2da0cb4e937fa7db72f71785827c7831ce9c335e)
![{\displaystyle \qquad ={\rho bh \over M}\left[{\vec {v}}\int _{0}^{1}\lambda ^{2}\,d\lambda +({\vec {c}}_{B}-{\vec {v}})\int _{0}^{1}\lambda ^{3}\,d\lambda \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bdd6bbee05e12c584cf182fdc86e25118a4b1fa)
![{\displaystyle \qquad ={\rho bh \over {1 \over 3}\rho bh}\left[{1 \over 3}{\vec {v}}+{1 \over 4}({\vec {c}}_{B}-{\vec {v}})\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc7c46ea318dc2d6a7cbb5837790e6ea102a452d)
![{\displaystyle A=\left[{\frac {1}{2}}sC\right]_{0}^{2\pi r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cebf760ef0435bd491b2e3cae7a0b96d4418bf25)
