Prismoidal Formula

## **Problem**

A *prismatoid* is a polyhedron whose vertices all lie in one or the other of two parallel planes. The perpendicular distance between the two planes is called the *height* or *altitude* of the prismatoid. The faces that lie in the parallel planes are called the *bases* of the prismatoid. The *midsection* is the polygon formed by cutting the prismatoid by a plane parallel to the bases halfway between them.

The cross-sections of a prismatoid follow a quadratic function along its height. The general formula for computing the volume of a prismoid is

# V

h
6

[

A
T

4

A
M

A
B

]

{\displaystyle V = \frac{h}{6}\left[A\_T + 4A\_M + A\_B \right] }

where

h

{\displaystyle h}

 is the height,

A
T

{\displaystyle A\_T}

 is the area of the top base,

A
M

{\displaystyle A\_M }

 is the area of the midsection, and

A
B

{\displaystyle A\_B }

 is the area of the bottom base. Use integration to prove why this volumetric formula is exact.

# h
b
−
a

{\displaystyle h = b-a }

 be the height of the prismatoid. For a solid with cross-sections that follow a quadratic function along its height,

f
(
x
# )
A

x
2

B
x
C

{\displaystyle f(x) = Ax^2 + Bx + C }

 and

# V

∫

a

b

f
(
x
)

d
x
.

{\displaystyle V = \int\_{a}^{b} f(x) \, dx. }

Let

A
T

=
f
(
b
)

{\displaystyle A\_T = f(b) }

,

A
M

=
f

(

a
b

2

)

{\displaystyle A\_M = f\left(\frac{a+b}{2}\right) }

, and

A
B

=
f
(
a
)

{\displaystyle A\_B = f(a) }

. Then it must be shown that

∫

a

b

A

x
2

B
x
C

d
# x

b
−
a

6

[

f
(
b
)
4
×
f

(

a
b

2

)

f
(
a
)

]

.

{\displaystyle \int\_{a}^{b} Ax^2 + Bx + C \, dx = \frac{b-a}{6}\left[f(b) + 4 \times f\left(\frac{a+b}{2}\right) + f(a) \right].}

Working out the algebra on the right-hand side yields

b
−
a

6

[

f
(
b
)
f

(

a
b

2

)

f
(
a
)

]

=

b
−
a

6

[

A

b
2

B
b
C
4
×

[

A

(

a
b

2

)

2

B

(

a
b

2

)

C

]

A

a
2

B
a
C

]

=

b
−
a

6

[

2
A
(

b
2

a
b

a
2

)
3
B
(
b
a
)
6
C

]

.

{\displaystyle \begin{align}
\frac{b-a}{6}\left[f(b) + f\left(\frac{a+b}{2}\right) + f(a) \right] &= \frac{b-a}{6}\left[Ab^2 + Bb + C + 4 \times \left[A\left(\frac{a+b}{2}\right)^2 + B\left(\frac{a+b}{2}\right) + C \right] + Aa^2 + Ba + C \right] \\[5 pt]
&= \frac{b-a}{6}\left[2A(b^2+ab+a^2) + 3B(b+a) + 6C \right].
\end{align}}

The algebra is simple but tedious.

Now for the left-hand side (which requires more tricks).

∫

a

b

A

x
2

B
x
C

d
x

=

A

x
3

3

B

x
2

2

C
x

a
b

=

A
3

(

b
3

−

a
3

)

B
2

(

b
2

−

a
2

)
C
(
b
−
a
)

{\displaystyle \begin{align}
\int\_{a}^{b} Ax^2 + Bx + C \, dx &= \frac{Ax^3}{3} + \frac{Bx^2}{2} + Cx \Biggr|\_a^b \\[5 pt]
 &= \frac{A}{3}(b^3-a^3) + \frac{B}{2}(b^2-a^2) + C(b-a)
\end{align}}

Use the following factorizations:

1. b
2

−

a
2

=
(
b
−
a
)
(
b
a
)

{\displaystyle b^2 - a^2 = (b-a)(b+a) }
2. b
3

−

a
3

=
(
b
−
a
)
(

b
2

a
b

a
2

)

{\displaystyle b^3 - a^3 = (b-a)(b^2 + ab + a^2) }

∫

a

b

A

x
2

B
x
C

d
x

=
(
b
−
a
)

[

A
3

(

b
2

a
b

a
2

)

B
2

(
b
a
)
C

]

=

b
−
a

6

[

2
A
(

b
2

a
b

a
2

)
3
B
(
b
a
)
6
C

]

.

{\displaystyle \begin{align}
\int\_{a}^{b} Ax^2 + Bx + C \,dx &= (b-a)\left[\frac{A}{3}(b^2+ab+a^2) + \frac{B}{2}(b+a) + C \right] \\[5 pt]
&= \frac{b-a}{6}\left[2A(b^2+ab+a^2) + 3B(b+a) + 6C \right].
\end{align} }

Since the left-hand and right-hand sides for evaluating the volume are equal,

# V

h
6

[

A
T

4

A
M

A
B

]

.

{\displaystyle V = \frac{h}{6}\left[A\_T + 4A\_M + A\_B \right]. }