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CALCULUS 1 PROBLEMS & SOLUTIONS
7.3.1
Finding Volumes By
Slicing
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Review
1. Volumes Of Solids Of
Revolution
Let f be a
continuous function on [a,
b]. See Fig. 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. 2. This 3D
space region is called a solid of revolution. Let's label it S. We wish to find 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. 2. The
line segment (x,
0)(x, f(x)) sweeps out a circular disk, lightyellow
colored in Fig. 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, sweeps out a circular piece of S,
gray colored in Fig. 3. The piece is called a slice of S.
Fig. 1

Fig. 2

Fig. 3

Now, let's find the volume V. Divide the interval [a, b]
into a regular partition of order n.
Refer to Fig.
4, where we choose n
=
5 as an example. The partition is: a
=
x_{0} < x_{1} < x_{2} < ... < x_{n} = b, with all the
subintervals [x_{i}_{–1}, x_{i}], for i
=
1, 2, ..., n,
having an equal length of (b
– a)/n. Let Dx
=
(b – a)/n. On
each ith
subinterval [x_{i}_{–1}, x_{i}], construct a rectangle of base width Dx and height  f(x_{i}_{–1}) (we select
the left endpoints of the subintervals to construct the rectangles). When the
plane region R is
revolved
about the xaxis,
each rectangle sweeps out a solid wheel. The radius of the ith wheel (the one on the
ith subinterval) is  f(x_{i}_{–1}). So its volume is V_{i} = p f(x_{i}_{–1})^{2}Dx = p(
f(x_{i}_{–1}))^{2}Dx. For n
large enough,
we can approximate V
by the sum V_{1} + V_{2} + ... + V_{n}.
Thus, it's natural to define V
as the limit of this
sum as n ® ¥:
That's what we want to show.
Fig. 4

Fig. 5

Fig. 6

Note that the value of the volume of the solid S in Fig. 2 is the same as the value of the area
under A
over [a, b] between x = a and x = b, which is shown in Fig. 6. Don't be confused by
these two areas:
A and the area under the graph of A over [a,
b]. Since A is a continuous function of x, it has an
integral on [a, b], and since A(x) > 0 for all x in [a,
b], that integral is the area under the graph of A
over [a, b].
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2. Volume Elements
The volume V of the
solid S is the
integral of A(x) over [a, b].
See Fig. 7. By Section
7.2 Review
Part 3, we can treat this integral as the integral of area elements under
the graph of A(x). Let a be
the area under A
over [a, b]. The area element is da
=
A(x) dx,
gray colored in Fig. 7. Now, A(x) dx
=
p( f(x))^{2} dx is the volume V_{slice} of the gray colored slice
in Fig. 8. As (1) V_{slice} =
da,
(2) da is the
area element of the area a, and
(3) the value of the volume V
is the same as that of the area a, we
can view V_{slice} as the volume
element dV of V. From now on we take this
point of view, that the
volume V is the
sum or integral of infinitely many infinitesimal volume elements dV.
Thus, to find the volume V
of a solid of revolution generated by revolving the graph of y = f(x) about
the xaxis and
bounded by x =
a and x = b, draw a slice of thickness dx, the volume of the slice being
dV = p( f(x))^{2} dx, which is the volume
element, and integrate the volume element on [a, b].
For
obvious reason, this method is called the slice method.
It's ok to draw just a plane cross section instead of a slice. This is
tremendously welcome because it
can occur that the slice is difficult to draw. The volume of the slice having
that cross section as one face
is the same: dV =
p( f(x))^{2 }dx.
Fig. 7

Fig. 8

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Example 1
Show that the volume of a ball of radius r
is V =
(4/3)p r^{3}.
Solution
Refer to Fig. 9. 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
Fig. 9

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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3. Solids With A Hole
Example 2
EOS
Fig. 10

Fig. 11

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 (a slice with a hole) 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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4. Revolution About The yAxis
Example 3
Find 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
EOS
Fig. 12

Fig. 13

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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5. Axis Of Revolution Not
Coinciding With A Coordinate Axis
Example 4
Use the slice 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. 14

EOS
Remarks 1
i. The slice 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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6. Unbounded Solids
Example 5
The plane region below y
=
1/x, above y = 0, and to the
right of x =
1 is revolved about the xaxis.
Find the volume of the generated solid.
Solution
The solid is sketched in Fig. 15. Let x
be any point in [1, ¥).
The area of the plane cross section at x
is
A(x) = p(1/x)^{2} = p/x^{2}. So the volume of the slice having that cross section as
one face is dV =
(p/x^{2}) dx.
Thus, the volume of the solid is:
Fig. 15

EOS
Remarks 2
i. The solid is generated by
revolving an unbounded plane region and 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 6.7.1 Review 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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7. MoreGeneral Solids
Example 6
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. 16. 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
values 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
is dV =
A(x) dx
=
(A/h^{2})x^{2} dx. Therefore, the volume V of the pyramid is:
Fig. 16

EOS
The pyramid isn't a solid of revolution. It's of a more general type of solid.
However, the slice method
can still be used to find its volume. The slice 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 two such solids. The proof is similar to that
for the solid of revolution
as discussed in Parts 1 and 2 above.
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Problems & Solutions
1. Find 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
Let x be an
arbitrary point in [1, 3]. The area of the plane cross section on the plane
perpendicular to
the xaxis at x is A(x)
=
p(1/x)^{2}. So the volume of the solid is:
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2. A cylindrical hole of radius r is drilled thru the centre
of a ball of radius R.
Find the volume of the
remaining part of the ball.
Solution
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3. Find the volume of a right
circular cone of base radius r
and height h.
Solution
The equation of the line joining the points (r, 0) and (0, h)
is (y – 0)/(x – r) = (h – 0)/(0 – r),
or y =
(–h/r)x + h.
The cone can be generated by revolving the plane region bounded by y = (–h/r)x
+ h,
y = 0, and x = 0 about the yaxis. The area of the plane
cross section perpendicular to the yaxis
at a
point y in [0, h] is A( y)
=
px^{2} = p((
y – h)/(–h/r))^{2} =
pr^{2}( y^{2}/h^{2} – 2y/h
+ 1). So the volume of the
cone is:
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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. Find the volume of
the wedge.
Solution
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5.
The plane region bounded by x
=
y^{2} and y = – x + 2 is revolved about the line y = 1. Find the
volume of the generated solid.
Solution
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6. Find the volume of the solid
generated by revolving the plane region under y = e^{–}^{x},
above y =
0, and
to
the right of x =
0 about the yaxis. You
may use the formula lim_{x}_{®}_{0+} x ln^{n}
x = 0 for any n in Z.
Solution
Let y be any
point in [0, 1) and x
be the point such that the point (x,
y) is on the curve y = e^{–}^{x}. The
horizontal rectangle at y
and of width dy sweeps
out a slice. The interval of integration is [0, 1] on the
yaxis. From y = e^{–}^{x} we obtain x = – ln y. The
volume of the slice is dV
=
p(x – 0)^{2} dy
=
p(–ln y)^{2} dy
=
p ln^{2}
y dy. So the volume of the solid is:
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7.
A solid generated by revolving a disk about an axis which is on its
plane and external to it is called
a torus (a doughnut  Wow, delicious! says Homer Simpson  Well, a
doughnutshaped solid 
Doh! ... Nuts! ... Hmmm, doughnuts ...). Find the volume
of the torus displayed in the figure below
by using the slice method.
Solution
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