Calculus Of One Real Variable –
By Pheng Kim Ving 
12.3 
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Let f
be a continuous function on [a,
b]. See Fig. 1.1. Let R be the plane region bounded
by f, y = 0 (the xaxis), x = a
(the vertical line at x
= a), and x = b (the vertical line at x = b).
When it's revolved (rotated) about the xaxis,
it
generates a 3dimensional space region, as shown in Fig. 1.2. This 3D space
region is called a solid of revolution.
Let's label it S.
We wish to calculate its volume. Let V
be that volume.
First, let's examine S
a little more closely. When R
is revolved about the xaxis,
the xaxis is
called the axis of
revolution of S. A
point (x, f(x)) on the graph of f describes a circle, as shown in Fig. 1.2. The
line segment (x,
0)(x,
f(x)) sweeps out a circular
disk, gray colored in Fig. 1.2. This disk is perpendicular to the axis of
revolution at its centre.
It can also be obtained as the intersection of the plane perpendicular to the
axis of revolution at x
and the solid S.
It's
called a plane cross section of S.
A strip of R with x at its lower left corner,
gray colored in Fig. 1.1, sweeps out a
circular piece of S,
gray colored in Fig. 1.3. This piece is called a slice of S.
Fig. 1.1
Plane Region To Be Revolved About xAxis.

Fig. 1.2
Solid Of Revolution.

Fig. 1.3
Gray piece is a slice.

Fig. 1.4
Approximate Volume.

Fig. 1.5
Approximate volume is a Riemann sum.

Fig. 1.6
Value of volume of solid is same as
value of area of this colored

Let V(x) be the function defined by:
Fig. 1.7

Example 1.1
Solution
Fig. 1.8
Ball For Example 1.1.

Refer to Fig. 1.8. Consider the ball centered at the origin
of the xyzcoordinate
system and with radius r.
It can be
regarded as being generated by revolving the upper half disk on the xyplane about the xaxis. That half disk is the
plane
region bounded by the upper semicircle on the xyplane and the xaxis. The equation of that semicircle is:
EOS
Remark that we use the symmetry of the ball about the yzplane in calculating the
integral. We prefer to use symmetry
whenever it allows us to have 0 as a limit of integration, because then the
calculation is simpler.
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A washer slice is a slice with a hole.
Example 2.1
Fig. 2.1
Region To Be Revolved About xAxis.

Fig. 2.2
Solid Obtained By Revolution In Fig.
2.1.

When a solid has a hole in it, a plane cross section and thus
a slice have a hole in them. The volume of a washer slice is
equal to the volume of the disk slice (slice as if it had no hole) minus that
of the hole.
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Example 3.1
Compute the volume of the solid generated by revolving the
plane region bounded by y
= x^{2}, y = 9, and x = 0 about the
yaxis.
Solution
Fig. 3.1
Colored plane region is revolved about yaxis.

Fig. 3.2
Solid Obtained By Revolution Of Plane
Region In Fig. 3.1 About yAxis.

EOS
Here the solid is obtained by revolving a plane region about
the yaxis; the
axis of revolution is the yaxis;
the plane cross
section is perpendicular to the yaxis
at its centre; the thickness of the slice obtained from it is dy; its radius must be
expressed in terms of y;
and integration is along the yaxis.
As seen earlier, when the solid is obtained by revolving a
plane region about the xaxis,
the axis of revolution is the
xaxis; the cross section is
perpendicular to the xaxis
at its centre; the thickness of the slice obtained from it is dx; its
radius must be expressed in terms of x;
and integration is along the xaxis.
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Example 4.1
Use the slicing method to find the volume of the solid
generated by revolving the plane region bounded by y = x^{2} and y =
3 about the line y =
–1.
Solution
Fig. 4.1
Plane region bounded by y = x^{2}
and y = 3 is revolved about line y = –1.

EOS
Remarks 4.1
i. The slicing method can also be employed when the axis of revolution doesn't coincide with a coordinate axis.
ii. The axis of revolution is parallel to the xaxis.
iii. The plane cross section or the slice will be
perpendicular to the axis of revolution, so the rectangle must be
perpendicular to the axis of
revolution.
v. The rectangle is perpendicular to the xaxis. Thus, the increment is
dx. Hence, the integration is
along the xaxis and
we must express the integrand in
terms of x.
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5. Unbounded Solids 
Example 5.1
The plane region below y
= 1/x, above y = 0, and to the right of x = 1 is revolved about the xaxis. Calculate the volume
of the generated solid.
Solution
Fig. 5.1
An Infinitely Long Horn With A Finite
Volume.

EOS
Remarks 5.1
i. The solid is generated by revolving an unbounded
plane region and hence is itself unbounded. We use improper integral
to handle it. It's an infinitely long
horn.
ii. The area of the given unbounded plane region is
infinite; see
Section 11.2 Part 10. It's interesting that revolving a region
with an infinite area gives rise to a solid with a finite volume.
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6. MoreGeneral Solids 
Example 6.1
A pyramid has a triangular base of area A and has a height of h measured
perpendicular to the plane of the base. Show
that its volume is V
= (1/3)Ah.
Solution
The pyramid is shown in Fig. 6.1. Let the xaxis
be on the line passing thru the vertex and perpendicular to the plane of
the base, with the origin at the vertex, and oriented in the direction vertex
to base. Let x be an
arbitrary point in [0, h].
Let A(x) be the area of the plane
cross section on the plane perpendicular to the xaxis at x.
The cross section is a
triangle similar to the base triangle. So A(x)/A = (MP/NQ)^{2}. Let's use the digit and letters 0, x, and h
to also indicate
the points representing their positions on the xaxis respectively. Triangles 0MP and 0NQ are similar; thus MP/NQ =
0P/0Q. Triangles 0xP and 0hQ are similar; consequently 0P/0Q
= 0x/0h = x/h.
Hence A(x)/A
= (x/h)^{2}, yielding A(x)
= (A/h^{2})x^{2}. The volume of the slice obtained from the cross
section, which is the volume element, is dV
= A(x) dx =
(A/h^{2})x^{2} dx.
Therefore the volume V
of the pyramid is:
Fig. 6.1
Pyramid with a triangular base of area A and a height of h has volume V = (1/3)Ah.

EOS
The pyramid isn't a solid of revolution. It's of a more general
type of solid. However, the slicing method can still be used
to find its volume. The slicing method can often be used to find the volume of
a solid if that solid can be sliced up into
parallel cross sections whose faces have readily computed areas. A solid of
revolution and the pyramid are 2 such solids.
The proof is similar to that for the solid of revolution as discussed in Part 1 above.
Problems & Solutions 
1. Calculate the volume of the solid generated by
revolving the plane region bounded by y
= 1/x, x = 1, and x = 3 about
the xaxis.
Solution
2. A cylindrical hole of radius r is drilled thru the centre of a ball of radius R. Compute the volume of the
remaining part
of the ball.
Solution
3. Find the volume of a right circular cone of base radius r and height h.
Solution
4. A 45^{o} wooden wedge has a semicircular base of radius r. The cross section on any
plane perpendicular to the
diameter of the semicircle is a
right isosceles triangle with the right angle on the semicircle. Calculate the
volume of
the wedge.
Solution
5. The plane region
bounded by x = y^{2} and y = – x
+ 2 is revolved about the line y
= 1. Compute the volume of the
generated solid.
Solution
Let's find the points of intersection of x = y^{2} and y = – x + 2 by solving the system:
On [1, 4] it is:
Solution
Hence the volume is:
7. A solid generated by
revolving a disk about an axis that is on its plane and external to it is
called a torus (a
doughnutshaped solid). Calculate
the volume of the torus displayed in the figure below by using the slice
method.
Solution
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