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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
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. 2. This 3-D
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 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. 2. The
line segment (x, 0)-(x, f(x)) sweeps out a circular disk, light-yellow 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

Plane region to be revolved about
x-axis.

 

Fig. 2

Solid of revolution.
Gray colored disk is a plane cross section.

 

Fig. 3

Gray colored piece is a slice.

 

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 = x0 < x1 < x2 < ... < xn = b, with all the
sub-intervals [xi1, xi], for i = 1, 2, ..., n, having an equal length of (b a)/n. Let Dx = (b a)/n. On
each ith sub-interval [xi1, xi], construct a rectangle of base width Dx and height | f(xi1)| (we select
the left endpoints of the sub-intervals to construct the rectangles). When the plane region R is revolved
about the x-axis, each rectangle sweeps out a solid wheel. The radius of the ith wheel (the one on the
ith sub-interval) is | f(xi1)|. So its volume is Vi = p| f(xi1)|2Dx = p( f(xi1))2Dx. For n large enough,
we can approximate V by the sum V1 + V2 + ... + Vn. Thus, it's natural to define V as the limit of this
sum as n :



That's what we want to show.

Fig. 4

Approximate volume.

 

Fig. 5

Approximate volume is a Riemann sum.

 

Fig. 6

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


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 Vslice of the gray colored slice in Fig. 8. As (1) Vslice = 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 Vslice 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 x-axis 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

Area element
da = A(x) dx.

 

Fig. 8

Volume element
dV = p( f(x))2 dx = A(x) dx.


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Example 1

Show that the volume of a ball of radius r is V = (4/3)p r3.

Solution
Refer to Fig. 9. 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

Fig. 9

Volume of a ball of radius
r is V = (4/3)p r3.


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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3. Solids With A Hole

Example 2


EOS

Fig. 10

Region to be revolved about
x-axis.

 

Fig. 11

Solid obtained by revolution in Fig. 10.
It has a cylindrical hole.

 

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 y-Axis

Example 3

Find 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

EOS

Fig. 12

Colored plane region is revolved about
y-axis.

 

Fig. 13

Solid obtained by revolution of plane region in Fig. 10
about
y-axis.


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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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 = x2 and y = 3 about the line y = 1.

Solution


Fig. 14

Plane region bounded by
y = x2 and y = 3
is revolved about line
y = 1.

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 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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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 x-axis.
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/x2. So the volume of the slice having that cross section as one face is dV =
(p/x2) dx. Thus, the volume of the solid is:



Fig. 15

An infinitely long horn.

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. More-General 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 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
values 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 is dV = A(x) dx =
(A/h2)x2 dx. Therefore, the volume V of the pyramid is:



Fig. 16

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 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 x-axis.

Solution



Let x be an arbitrary point in [1, 3]. The area of the plane cross section on the plane perpendicular to
the x-axis 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 y-axis. The area of the plane cross section perpendicular to the y-axis at a
point y in [0, h] is A( y) = px2 = p(( y h)/(h/r))2 = pr2( y2/h2 2y/h + 1). So the volume of the
cone is:



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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. Find the volume of the wedge.

Solution





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5. The plane region bounded by x = y2 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 = ex, above y = 0, and
to the right of x = 0 about the y-axis. You may use the formula limx0+ x lnn 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 = ex. The
horizontal rectangle at y and of width dy sweeps out a slice. The interval of integration is [0, 1] on the
y-axis. From y = ex we obtain x = ln y. The volume of the slice is dV = p(x 0)2 dy = p(ln y)2 dy
= p ln2 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 doughnut-shaped 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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