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6.2. USING DEFINITE INTEGRALS TO FIND VOLUME
by a single curve is often called the washer method.
If y = R(x) and y = r(x) are nonnegative continuous functions on [a, b] that satisfy
R(x) ≥ r(x) for all x in [a, b], then the volume of the solid of revolution generated by
revolving the region between them about the x-axis over this interval is given by
V =
b
a
π[R(x)
2 − r(x)
2 ] dx.
Activity 6.4.
In each of the following questions, draw a careful, labeled sketch of the region described,
as well as the resulting solid that results from revolving the region about the stated axis.
In addition, draw a representative slice and state the volume of that slice, along with a
definite integral whose value is the volume of the entire solid. It is not necessary to
evaluate the integrals you find.
(a) The region S bounded by the x-axis, the curve y =
√
x, and the line x = 4;
revolve S about the x-axis.
(b) The region S bounded by the y-axis, the curve y =
√
x, and the line y = 2;
revolve S about the x-axis.
(c) The finite region S bounded by the curves y =
√
x and y = x 3 ; revolve S about
the x-axis.
(d) The finite region S bounded by the curves y = 2x 2 + 1 and y = x 2 + 4; revolve
S about the x-axis
(e) The region S bounded by the y-axis, the curve y =
√
x, and the line y = 2;
revolve S about the y-axis. How does the problem change considerably when
we revolve about the y-axis?
⊳
Revolving about the y-axis
As seen in Activity 6.4, problem (e), the problem changes considerably when we revolve a
given region about the y-axis. Foremost, this is due to the fact that representative slices
now have thickness △y, which means that it becomes necessary to integrate with respect
to y. Let’s consider a particular example to demonstrate some of the key issues.
6.2. USING DEFINITE INTEGRALS TO FIND VOLUME
by a single curve is often called the washer method.
If y = R(x) and y = r(x) are nonnegative continuous functions on [a, b] that satisfy
R(x) ≥ r(x) for all x in [a, b], then the volume of the solid of revolution generated by
revolving the region between them about the x-axis over this interval is given by
V =
b
a
π[R(x)
2 − r(x)
2 ] dx.
Activity 6.4.
In each of the following questions, draw a careful, labeled sketch of the region described,
as well as the resulting solid that results from revolving the region about the stated axis.
In addition, draw a representative slice and state the volume of that slice, along with a
definite integral whose value is the volume of the entire solid. It is not necessary to
evaluate the integrals you find.
(a) The region S bounded by the x-axis, the curve y =
√
x, and the line x = 4;
revolve S about the x-axis.
(b) The region S bounded by the y-axis, the curve y =
√
x, and the line y = 2;
revolve S about the x-axis.
(c) The finite region S bounded by the curves y =
√
x and y = x 3 ; revolve S about
the x-axis.
(d) The finite region S bounded by the curves y = 2x 2 + 1 and y = x 2 + 4; revolve
S about the x-axis
(e) The region S bounded by the y-axis, the curve y =
√
x, and the line y = 2;
revolve S about the y-axis. How does the problem change considerably when
we revolve about the y-axis?
⊳
Revolving about the y-axis
As seen in Activity 6.4, problem (e), the problem changes considerably when we revolve a
given region about the y-axis. Foremost, this is due to the fact that representative slices
now have thickness △y, which means that it becomes necessary to integrate with respect
to y. Let’s consider a particular example to demonstrate some of the key issues.