CHAPTER VIII

THE GENERAL PRISMATOID

42. SECTION OF A GENERAL PRISMATOID

Definition. A general prismatoid is a solid such that the area of any section, say A y, parallel to and
distant y from a fixed plane can be expressed as a polynomial in y of degree not higher than the third.
That is, a solid is a general prismatoid if

A y =ay 2+ by 2+ cy +d ,
Where a , b , c , and d are constants which may be positive, negative, or zero.

Inasmuch as all the solids of Solid Mensuration, and many other considered in advanced mathermatics,
are such that A yis expressible as a polynomial in y of degree not higher than the second, we shall
confine our discussion to prismatoids for which

A y =ay 2+ by +c
For example, consider a prismatoid with its lower base L lying in the fixed
plane CD (see figure). Then the area of a section A parallel to the base L
y

and distant y above it will be

A y =ay 2+ by + L
Example 1. For the hemisphere shown in the figure, find the value of A
y

where y is the distance from the base of the hemisphere. Also find the area
of the mid-section parallel to the base.

Solution. Let the base of the hemisphere be in the fixed plane. Then A y is any section parallel to
and distant y from the plane. Denote by r the radius of this section. From right triangle OAB ,we have

r 2=R 2− y 2 .

Therefore

A y =πr 2 −πR 2−πy 2 . Ans .

R
Substituting in this equation y= , we find the mid-section
2

R 2 1 2
M = A R =πR 2−π
2
( ) 2
= πR . Ans .
4

PROBLEMS
1. For each of the following solids verify the value of A y given and find the area of the mid-section
parallel to the base. The letters U , L and h in these formulas denote upper base, lower base,
and altitude, respectively.
A
( a ) Prisms, cylinder. y =L.
(b) Right triangular prism in which and edge is taken as lower base and the opposite face as
U ( h− y )
upper base. A y = , where y is the distance form the upper base.
h
Ly 2
( c )Pyramid. A y = 2 , where y is the distance from the vertex.
h
2
πr 2
(d )Cone. A y = 2 y , where y is the distance from the vertex.
h
( e )Sphere. A y =2 πRy−π y 2 , where y is the distance from the plane parallel to the base and
tangent to the spherical surface.
2
L− U
( f )Frustum of pyramid. A y = √ √
h ( ) y −2( L−√h LU ) y+ L,
2
where y is the distance form

the lower base L .

2
y 2 y
A
( g ) Frustum of cone. y =r r 1
2
[+ 2r r
2( 2 −r 1)
h
+ r −r
h ]
( 2 1 ) 2 , where y is the distance from base
of radius r 1 .
( h ) Segment of sphere of two bases and altitude h .
( a 2+ h2−b 2)
[
A y =π b 2
h ]
y− y 2 , where y is the distance from the base of radius b .

2. Prove that, if in a general prismatoid y is measured from a plane of units from the fixed plane,
A y is still a polynomial in y of not higher than the second degree. Show also that the coefficent
of y 2 is unchanged.
3. If a solid has a square for upper base, a retangle for lower base, and isosceles trapezoids for
faces, find A y in terms of the distance from the lower base.
4. The bases of a solid are perpendicular to a leteral edge. The upper base in a right triangle, the
lower base is a sqaure, and the faces are triangles and trapezoids. Find A y in terms of the
distance from the lower base.
43. PRISMATOID THEOREM
The volume of a prismatoid is equal to the algebraic sum of the volumes of a pyramid, a wedge,
and a parallelepiped.
Analysis. Consider the prismatoid (see figure) of upper base U , lower base L , and
altituded h , for which section
A y =± ay 2 ± by+ L, (1)
Where y is the distance from the base L . Substituting y=h in (1), we get
U =A h=± a h2 ± bh+ L. (2)
 2
Now construct a pyramid (see figure) of base U 1=ah and of altitude h ; a wedge (right
traingular prism standing on lateral edge)

Of altitude h and of
base b by h ; and a
parallelepiped of
altitude h and of base

L . Place these solids so that the vertex of the pyramid, the edge of the wedge, and the lower
base of the parallelepiped lie in the plane containing the base L of the prismatoid and bases
U 1 ,U 2 and Llie in the plane of base U as shown. Denote by A1 , A 2 , and A3 the sections cut
from the pyramid, wedge, and parallelepiped, respectively, by the place containing section A y
of the prismatoid.
Considering the pyramid, we have from the theorem for similar figure 22,
A1 y 2
= ,
U 1 h2
Or
U 1 y 2 ah 2 y 2 2
A1= 2 = 2 =a y ; ( 3)
h h
And considering the wedge, we have from similar triangles
z b by
= , or z= .
y h h
Therefore
A1=zh−by . ( 4)
Also we note that
A3 =L. (5)
2
Substituting in ( 1 ) a y = A1 from ( 3 ) , by= A 2 from ( 4 ) , and L= A 3 (6)
Whence in accordance with Cavalieri’s theorem the volume of a prismatoid is equal to the
algebraic sum of the volumes of a pyramid, a wedge, and a parallelepiped, or
1 1
V =± ah 2 ± bh± Lh ,
3 2
1 2 1
Where ah is the volume of the pyramid, bh is the volume of the wedge, and Lh is the
3 2
volume of the parallelepiped.
44. PROOF OF THE PRISMOIDAL FORMULA
In the preceding acticle we found the volume of the prismatoid to be
1 1
V =± ah 2 ± bh± Lh , ( 1)
3 2
Where a and b are constants which are positve, negative, or zero. We shall now show that this
can be transformed into the familliar prismoidal formula of 30.
h
Factoring out from the right-hand member of (1) and transforming the resulting terms, we
6
have
h
V= [ 2 a h2+ 3 bh+ 6 L ]
6
h
¿
6
[ ( a h2 +bh+ L ) + ( a h 2+ 2bh+ 4 L ) + L ]
ah2 bh
¿
6 [
h ( 2
a h +bh+ L ) + L+ 4
4 (
+ +L
2 )]
2
h
( a h2 +bh+ L ) + L+ 4 a h + b h + L
¿
6 { 2
2 ( ( ) ( ) )}
2
But since A y =ay + by + L , we have
h 2 h
ah2 +bh+ L= Ah=U , and a () ()
2
+b
2
+ L= As =M .
Hence
h
V = ( U + L+4 M ) .
6
This remarkable formula may be used, in general, to find not only all the areas and volumes of
plane and solid geometry but a great many others besides. It can be used to get a close
approximation to most of the physical quantities that are used in engineering science.
Example 1. The solid shown in the figure has an elliptic base
perpendiicular to the axis CD and has an altitudeh . If the
area of any section parallel to the base is proportional to the
distance y from the vertex D , that is A y =ky , find the
volume of the solid. (This solid is called a segment of an
elliptic paraboloid.)
Solution. Since A y =ky , we have

L= A s=0 , M = A a =k
z
( h2 ) , and U =A =kh .
h

Substituting these values of L , M and U in the prismoidal formula, we have

h 1 1
V=
6 ( 2 4 )
kh+0+ kh = k h2 . Ans .

Example 2. A solid has a circular base of diameter R . The line AB is a diameter of the base.
Find the volume of the solid if every section perpendicular to AB is a right triangle of base c and
altitude mc .
Solution. Let y be the distance from A to section A y. From the figure we write
1 mc 2
A y = c ( mc )= ,
2 2
And from plane geometry (Reference 64, 50) we have
c 2
()
2
=Ry− y 2 ,
Or
c2 2
=2 Ry−2 y
2
But
m c2
A y=
2
Therefore
A y =2mRy −2 m y 2 . (a)
Therefore the solid is a prismatoid.
Consider verices A and B as bases of this solid. Then in the formula

V =h ( U + L+4
6
M
) (b )
U =0, and L=0
To find tje mid-section M substitute y=R /2 in (a) to obtain
m R2
M = AR= .
2
Whence, substituting these values of U , L and M in ( b ) , we get
1
V = m R3 . Ans .
8

PROBLEMS
1. Find the volume of a general prismatoid (a) whose altitude is 2, and for which
A y =15+5 y−3 y 2 ,(b) whose altitude is 3 , and for which A y =9 y+ 4 y 2 , where y is the
distance from the base of the prismatoid.
 2
2. Find the volume of a sphere of radius R for which A y =2 πRy−π y , where y is the
distance from a plane tangent to the sphere.
3. A segment of one base of a sphere of radius R has an altitude h . Find its volume, using the
value of A y in Prob.2.
4. Find the volume of a spherical segment of one base if its altitude is h and its base radius is b .
See Prob. 1 h , 42.
5. Find the volume of the spherical segment of altitude h and of base radii a and b . See Prob.
1 h , 42.
6. If R and r are respectively the radii of the larger and smaller bases of the frustum of a cone
of altitude h , show that A y =π ¿ where y is the distance from the smaller base. Find the
volume of the frustum.
7. The segment of a paraboloid of revolution (see figure) is a solid in
which every section parallel to the base is a circle the radius of which is
the mean proportional between the distance y from the vertex and
the radius of the base. Find the volume of the segment of altitude h
and base radius b if h=b.
8. Using the prismoidal formula, derive the formula for the volume of
each of the following prismatoids; (a) pyramid, (b) cone, (c ) sphere, (d ) segment of a
sphere, (e ) frustum of a pyramid.
9. For the solid shown in the figure every section perpendicular to the axis
DE is an ellipse of an area A y =M ¿ where M =πab is the mid-
1
section, y is the distance of A y from M , and c= DE . Find the
2
volume of the solid. (This solid is called an ellipeoid.)
10. In a cone of altitude h and elliptic base B , every section parallel to the
B y2
base has an area A y = , where y is the distance from the vertex to
h2
the section. Find the volume of the cone. (This solid is called an elliptic cone.)
11. For the solid shown in the figure every section perpendicular to the
2 y2 ( y
axis DE is an ellipse of area A y =2
B−2 M )+ ( 4 M −B ) ,
h h
where B=πab is the upper base, M is the mid-section, y is the
distance from D to A y , and h=DE is the altitude. Find the
volume of the solid. (This solid is called a segment of an elliptic
hyperboloid of two aboets.)
12. For the solid shown in the figure every section perpendicular to the axis
y2
DE is an ellipse of area A y =( B−M ) + M , where B=π a1 b1 , is the
h2
area of the upper base as wellas that of the lower base, M =πab is the
1
area of the mid-section, y is the distance from O to A y , and h= DE is
2
 one-half the altitude. Find the volume of the solid. (This solid is called a segment of an
elliptic hyperboloid of cone aboets.)
13. A variable rectangle generates a solid by moving from a fixed point and keeping parallel to
its first position. One sode of the rectangle is equal to the distance the rectangle moves
from the fixed point, and the other side is equal to twice the distance. What is the volume
generated while the rectangle moves a distance of 2 ft . ?
14. A wedge (see fig. (a)¿ is cut from a cylinder of radius 5∈. by two planes, one perpendicular
to the axis of the cylinder and the other passing through a diameter of the section made by
the first plane and inclined to this plane at an angle of 45 ° . Find the volume of the wedge.

15. A solid has a circular base of radius 20∈. Find the volume of the solid if every plane section
perpendicular to a certain diameter is
(a) an equilateral triangle (see fig. (b));
(b) an isosceles right triangle with its hypotenuse in the plane of the base;
(c ) an isosceles right triangle with one leg in the plane of the base;
(d ) an isosceles triangle with its altitude equal to its base;
(e ) a square.

16. The base (See figure) of a certain solid is a quarter circle of radius 10∈.
Every section parallel to one face is a right triangle , whose altitude is 1.6
times its base. Find its volume.
17. The base of a certain solid is a circle of radius r . If all sections perpendicular
to a fixed diameter of the base are (a) square, (b) isosceles right triangles,
find the volume of the solid. (These solids are called conoids.)
18. For the solid shown in Fig.(c ) every section perpendicular to edge CD is a circle of diameter
20 y−3 y 2 , where y is the distance of the section from C. If the altitude DC of the solid is
4 , find its volume.
19. For the solid shown in Fig.(d ) every section perpendicular to edge CD is a circle. If are CD
is a semicircle of diameter 2 R, find the volume of the solid.
20. The base of a certain solid is a triangle of base b and altitude h . If all sections perpendicular
to the altitude of the triangle are regular hexagons, find the volume of the solid.
21. Two cylinders (see figure) with circular bases have a common upper
base and tangent lower bases. If the radius of each base is 10∈., find
the volume of the part common to the two cylinders
22. The center of a square moves along a diameter of a given circle of
radius a , the plane of the square being perpendicular to this
diameter, and its magnitude varying in such a way that two opposite
vertices move along the circumference of the given circle. Find the
volume of the solid generated.
23. Two cylinders (see figure below) of equal radius r have their axis meeting at right angles.
Find the volume of the common part.

24. Show that the solid represented in the figure is a

prismatoid. MQ is a plane and RT is a plane
perpendicular to MQ . The warped surfaces MS and RQ
are generated by moving straight lines that remain parallel to the base plane MRN .

25. Find the volume of the prismatoid shown in the figure.

 26. The figure represents a solid cut out of a circular cylinder of radius a
and altitude h . O is on the axis of the cylinder, the semicircular base
ABD is perpendicular to the axis, and ABD is a plane section. Find the
volume of the solid.
27. By means of the prismoidal formula, derive a formula for the area of a
(a) square, (b) rectangle, (c ) parallelogram, (d ) triangle, (e ) trapezoid.
28. By means of the prismoidal formula, derive a formula for the area of the lateral surface of a
(a) cube, (b) rectangular, (c ) right cylinder, (d ) right prism.

45. APPLICATION OF PRISMATOID THEOREM

It is interesting to note that the volume of a prismatoid may be found by direct application of the
prismatoid theorem of 43.

Thus after finding

A y =ay 2+ by + L
We write

Ah =ah2 +bh+ L
We may think of the terms in the right-hand member of this equation as the areas of the bases of
pyramid, a wedge, and a parallelepiped, respectively, of common altitude h , and write for the sum of
the volumes of these solids

1 1
V = ah3 + bh+ Lh.
3 2
Example. If the area of section of a prismatoid at a distance y from the base is

A y =10−4 y +6 y 2 sq .∈.
And the altitude is 5∈. , find the volume.

Solution. Since

A y =10−4 y +6 y 2
The volume of the prismatoid may be thought of as the algebraic sum of the volumes of a
parallelepiped, a wedge, and a pyramid, with a common altitude h and with cases 10 , ( 4 )( 5 ) , and (6) ¿
respectively. Therefore

4 ( 5 )2 6 ( 5 ) 3
V = (10 )( 5 ) − + =250 cu .∈. Ans .
2 3
PROBLEMS

1. Using the method of this article solve Probs. 1 ¿ 27 of 44.

2. Prove that the volume of prismatoid of altitude h is equal to the sum of the volume of a frustum
of a pyramid and a prism each of alititude h . When is the volume of a prismatoid equal to the
volume of the frustum alone? At what distance from the base of the frustum is the vextex of the
pyramid?
b 2 b2
Hint 2
Write A y =ay + by + L=a y +
(2a )(
+ L−
4a
. )
3. By means of the calculus (a subject in advanced mathematics), it can be shows that the volume
2
of a solid whose altitude is h and for which A y =ay is equal to the volume of a prism of altitude
ahn+ 1
h and base . Using this fact, show that the volume of any solid whose altitude is h and for
n+ 1
which A y =ay 2+ by 2+ cy + L , is equal to the sum of the volumes of a prism, pyramid, wedge,
and parallelepiped. Also show that the volume of this solid can be found by the prismoidal
formula.