Problem [ ] Figure 1. Tricylinder
Consider the solid formed by intersecting three perpendicular cylinders:
{
x
2
+
y
2
=
r
2
x
2
+
z
2
=
r
2
y
2
+
z
2
=
r
2
{\displaystyle \begin{cases} x^2 + y^2 = r^2 \\ x^2 + z^2 = r^2 \\ y^2 + z^2 = r^2 \end{cases} }
This solid is called the tricylinder . Determine the volume of the tricylinder. Express the volume in terms of the radius of the cylinder. Then express the volume in terms of the diameter of the cylinder.
Solution [ ] Figure 2. Dissecting the tricylinder
Single Integral Solution [ ] Dissect the tricylinder into seven pieces (Figure 2): 1 cube of side length
2
r
{\displaystyle \sqrt{2} r }
4
(
r
2
−
x
2
)
{\displaystyle 4(r^2 - x^2) }
r
/
2
<
x
<
r
{\displaystyle r/\sqrt{2} < x < r}
V
t
r
i
c
y
l
i
n
d
e
r
=
V
c
u
b
e
+
6
×
V
c
a
p
.
{\displaystyle V_{tricylinder} = {V}_{cube} + 6 \times {V}_{cap}.}
The volume of the cube is
V
c
u
b
e
=
(
2
r
)
3
=
2
2
r
3
.
{\displaystyle \begin{align}
{V}_{cube} &= {\left(\sqrt{2} r \right)}^{3} \\[5 pt]
&= 2\sqrt{2} {r}^{3}.
\end{align} }
The volume of each cap is
V
c
a
p
=
∫
r
/
2
r
4
(
r
2
−
x
2
)
d
x
=
4
[
r
2
x
−
x
3
3
|
r
/
2
r
]
=
1
3
(
8
−
5
2
)
r
3
.
{\displaystyle \begin{align}
{V}_{cap} &= \int_{r/\sqrt{2}}^{r} 4(r^2 - x^2)\, dx \\[5 pt]
&= 4\left[r^2x-\frac{x^3}{3}\Biggr|_{r/\sqrt{2}}^{r}\right] \\[5 pt]
&= \frac{1}{3}\left(8-5\sqrt{2}\right){r}^{3}.
\end{align} }
Combining all the parts yields:
V
t
r
i
c
y
l
i
n
d
e
r
=
V
c
u
b
e
+
6
×
V
c
a
p
=
2
2
r
3
+
6
×
1
3
(
8
−
5
2
)
r
3
=
(
16
−
8
2
)
r
3
{\displaystyle \begin{align}
V_{tricylinder} &= {V}_{cube} + 6 \times {V}_{cap} \\[5 pt]
&= 2\sqrt{2} {r}^{3} + 6\times\frac{1}{3}\left(8-5\sqrt{2}\right){r}^{3} \\[5 pt]
&= \left(16-8\sqrt{2}\right){r}^{3} \\[5 pt]
\end{align}}
or
V
t
r
i
c
y
l
i
n
d
e
r
=
8
(
2
−
2
)
r
3
.
{\displaystyle V_{tricylinder} = 8\left(2-\sqrt{2}\right){r}^{3}. }
Triple Integral Solution [ ] Split the solid into 48 congruent segments. The segments are bounded by the circular arc
r
2
−
(
r
cos
ϕ
)
2
=
r
sin
ϕ
{\displaystyle \sqrt{r^2 - {(r\cos{\phi})}^{2}} = r \sin{\phi} }
0
≤
ϕ
≤
π
4
{\displaystyle 0 \le \phi \le \frac{\pi}{4} }
D
=
{
(
ρ
,
ϕ
,
z
)
|
0
≤
ρ
≤
r
,
0
≤
ϕ
≤
π
/
4
,
0
≤
z
≤
r
sin
ϕ
}
.
{\displaystyle D = \{(\rho, \phi, z)|\, 0 \le \rho \le r, \, 0 \le \phi \le \pi/4, \, 0 \le z \le r\sin{\phi} \}. }
V
s
e
g
m
e
n
t
=
∫
0
π
/
4
∫
0
r
∫
0
r
sin
ϕ
ρ
d
z
d
ρ
d
ϕ
=
∫
0
π
/
4
∫
0
r
ρ
[
z
|
0
r
sin
ϕ
]
d
ρ
d
ϕ
=
∫
0
π
/
4
∫
0
r
ρ
2
sin
ϕ
d
ρ
d
ϕ
=
∫
0
π
/
4
[
ρ
3
3
sin
ϕ
|
0
r
]
d
ϕ
=
r
3
3
∫
0
π
/
4
sin
ϕ
d
ϕ
=
r
3
3
[
−
cos
ϕ
|
0
π
/
4
]
=
(
2
−
2
)
r
3
6
{\displaystyle \begin{align}
{V}_{segment} &= \int_{0}^{\pi/4} \int_{0}^{r} \int_{0}^{r\sin \phi} \rho \,dz \,d\rho \,d\phi \\[5 pt]
&= \int_{0}^{\pi/4} \int_{0}^{r} \rho \left[z \Biggr|_{0}^{r\sin{\phi}}\right] d\rho \,d\phi \\[5 pt]
&= \int_{0}^{\pi/4} \int_{0}^{r} {\rho}^2 \sin{\phi} \,d\rho \,d\phi \\[5 pt]
&= \int_{0}^{\pi/4} \left[\frac{{\rho}^3}{3} \sin{\phi} \Biggr|_{0}^{r} \right] \,d\phi \\[5 pt]
&= \frac{r^3}{3} \int_{0}^{\pi/4} \sin{\phi} \,d\phi \\[5 pt]
&= \frac{r^3}{3} \left[-\cos{\phi} \Biggr|_{0}^{\pi/4} \right] \\[5 pt]
&= \frac{\left(2-\sqrt{2} \right)r^3}{6}
\end{align} }
Thus, the volume of the tricylinder is
V
t
r
i
c
y
l
i
n
d
e
r
=
48
×
V
s
e
g
m
e
n
t
=
48
×
(
2
−
2
)
r
3
6
=
8
(
2
−
2
)
r
3
.
{\displaystyle \begin{align}
V_{tricylinder} &= 48 \times {V}_{segment} \\[5 pt]
&= 48 \times \frac{\left(2-\sqrt{2} \right)r^3}{6} \\[5 pt]
&= 8 \left(2-\sqrt{2} \right)r^3.
\end{align} }
Since diameter is twice the radius,
d
=
2
r
{\displaystyle d = 2r}
V
t
r
i
c
y
l
i
n
d
e
r
=
(
2
−
2
)
d
3
.
{\displaystyle V_{tricylinder} = \left(2-\sqrt{2} \right)d^3. }
![{\displaystyle {\begin{aligned}{V}_{cube}&={\left({\sqrt {2}}r\right)}^{3}\\[5pt]&=2{\sqrt {2}}{r}^{3}.\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/ad64e36f339c02f0ed784c0baa669a47ecfb6f60)
![{\displaystyle {\begin{aligned}{V}_{cap}&=\int _{r/{\sqrt {2}}}^{r}4(r^{2}-x^{2})\,dx\\[5pt]&=4\left[r^{2}x-{\frac {x^{3}}{3}}{\Biggr |}_{r/{\sqrt {2}}}^{r}\right]\\[5pt]&={\frac {1}{3}}\left(8-5{\sqrt {2}}\right){r}^{3}.\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/e8af649147c3fe4e093719e47a195bef15cb501f)
![{\displaystyle {\begin{aligned}V_{tricylinder}&={V}_{cube}+6\times {V}_{cap}\\[5pt]&=2{\sqrt {2}}{r}^{3}+6\times {\frac {1}{3}}\left(8-5{\sqrt {2}}\right){r}^{3}\\[5pt]&=\left(16-8{\sqrt {2}}\right){r}^{3}\\[5pt]\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/1f0d43cf25aee9b9ec9c3c5127b2d7190dfccc6b)
![{\displaystyle {\begin{aligned}{V}_{segment}&=\int _{0}^{\pi /4}\int _{0}^{r}\int _{0}^{r\sin \phi }\rho \,dz\,d\rho \,d\phi \\[5pt]&=\int _{0}^{\pi /4}\int _{0}^{r}\rho \left[z{\Biggr |}_{0}^{r\sin {\phi }}\right]d\rho \,d\phi \\[5pt]&=\int _{0}^{\pi /4}\int _{0}^{r}{\rho }^{2}\sin {\phi }\,d\rho \,d\phi \\[5pt]&=\int _{0}^{\pi /4}\left[{\frac {{\rho }^{3}}{3}}\sin {\phi }{\Biggr |}_{0}^{r}\right]\,d\phi \\[5pt]&={\frac {r^{3}}{3}}\int _{0}^{\pi /4}\sin {\phi }\,d\phi \\[5pt]&={\frac {r^{3}}{3}}\left[-\cos {\phi }{\Biggr |}_{0}^{\pi /4}\right]\\[5pt]&={\frac {\left(2-{\sqrt {2}}\right)r^{3}}{6}}\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/c4862e328283b4c17ac1a353a98d132d46105e7c)
![{\displaystyle {\begin{aligned}V_{tricylinder}&=48\times {V}_{segment}\\[5pt]&=48\times {\frac {\left(2-{\sqrt {2}}\right)r^{3}}{6}}\\[5pt]&=8\left(2-{\sqrt {2}}\right)r^{3}.\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/5b00d6f88c2980d0a41daa65f667ae799c824ec9)
