Volume of the Tricylinder

## **Problem**

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**

### **Single Integral Solution**

Dissect the tricylinder into seven pieces (Figure 2): 1 cube of side length

2

r

{\displaystyle \sqrt{2} r }

 and 6 congruent caps. The integral for calculating the volume of the cap involves integrating the square cross-sections

4
(

r
2

−

x
2

)

{\displaystyle 4(r^2 - x^2) }

 on the interval

r

/

2

<
x
<
r

{\displaystyle r/\sqrt{2} < x < r}

. Thus, the volume of the tricylinder is

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} }

 swept across the azimuthal angle

0
≤
ϕ
≤

π
4

{\displaystyle 0 \le \phi \le \frac{\pi}{4} }

. Hence, the region to integrate is

# 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}

, we get

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. }