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Calculus 1 Problems & Solutions – Chapter 7 – Section 7.3.1 |
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7.3.1
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Review |
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 x-axis), x = a
(the vertical line at x
= a),
and x
= b
(the vertical line at x
= b).
When it's revolved (rotated) about the x-axis, it
generates a 3-dimensional space region, as shown in Fig. 1.2. This 3-D 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 x-axis,
the x-axis
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.
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Fig. 1.1Plane Region To Be Revolved About x-Axis. |
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Fig. 1.2Solid Of
Revolution.
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Fig. 1.3Gray piece is a slice. |
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Fig. 1.4Approximate Volume. |
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Fig. 1.5Approximate volume is a Riemann sum. |
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Fig. 1.6Value of volume of solid is same as
value of area of this colored
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Differential Volume – Volume Element
Let V(x) be the function defined by:
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Fig. 1.7
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Example 1.1
Solution
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Fig. 1.8Ball For Example 1.1. |
Refer to Fig. 1.8. Consider the ball centered at the origin
of the xyz-coordinate
system and with radius r.
It can be
regarded as being generated by revolving the upper half disk on the xy-plane
about the x-axis.
That half disk is the plane
region bounded by the upper semi-circle on the xy-plane and the x-axis. The
equation of that semi-circle is:
EOS
Remark that we use the symmetry of the ball about the yz-plane 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
Solution
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Fig. 2.1Region To Be Revolved About x-Axis. |
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Fig. 2.2Solid Obtained By Revolution In Fig.
2.1.
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EOS
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
= x2, y = 9, and x = 0 about the
y-axis.
Solution
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Fig. 3.1Colored plane region is revolved about y-axis. |
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Fig. 3.2Solid Obtained By Revolution Of Plane Region In Fig. 3.1 About y-Axis. |
EOS
Here the solid is obtained by revolving a plane region about
the y-axis;
the axis of revolution is the y-axis; the plane cross
section is perpendicular to the y-axis 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 y-axis.
As seen earlier, when the solid is obtained by revolving a
plane region about the x-axis,
the axis of revolution is the
x-axis;
the cross section is perpendicular to the x-axis 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 x-axis.
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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 = x2 and y
=
3 about the line y
= –1.
Solution
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Fig. 4.1Plane region bounded by y = x2 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 x-axis.
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 x-axis. Thus,
the increment is dx.
Hence, the integration is along the x-axis 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 x-axis.
Calculate the volume
of the generated solid.
Solution
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Fig. 5.1An 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 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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6. More-General 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 x-axis 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 x-axis 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 x-axis 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/h2)x2.
The volume of the slice obtained from the cross section, which is the volume
element, is dV
= A(x) dx =
(A/h2)x2
dx.
Therefore the volume V
of the pyramid is:
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Fig. 6.1Pyramid 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 x-axis.
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 45o
wooden wedge has a semi-circular base of radius r. The cross section on any plane
perpendicular to the
diameter of the semi-circle is a right isosceles triangle
with the right angle on the semi-circle. Calculate the volume of
the wedge.
Solution
5. The plane region
bounded by x
= y2 and y = – x + 2 is revolved about the line y = 1.
Compute the volume of the
generated solid.
Solution
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Let's find the points of intersection of x = y2 and y = – x + 2 by solving the system:
Solution
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
doughnut-shaped solid). Calculate the volume of the torus
displayed in the figure below by using the slice method.
Solution
Note
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