(a)
GURE 6
GURE 7
(b)
Examples
Calculus
II, Fall the
2016
v Some
EXAMPLE
3 Find the volume of theMath
solid 142,
obtained
by rotating
region bounded
1
by There
y 苷 x ,are
y 苷three
8, andmethods
x 苷 0 about
the y-axis.
in this course to find the volume of a solid:
SOLUTION The region is shown in Figure 7(a) and the resulting solid is shown in Fig1. Slicing
ure 7(b). Because the region is rotated about the y-axis, it makes sense to slice the solid
2. Disks or
Washers
perpendicular
to the
y-axis and therefore to integrate with respect to y. If we slice at
3
height y, we get a circular disk with radius x, where x 苷 s
y . So the area of a cross3. Shells
section through y is
3
Examples on Slicing Method
were
the
quiz
A共y兲
苷 included
␲ x 2 苷 ␲in(s
y )2last
苷␲
y 2兾3and the solutions I posted online of
3
problems 6.2-5, 6 & 7. Here are some examples on the Disks (Or Wahsers) Method:
and the volume of the approximating cylinder pictured in Figure 7(b) is
Ex. 1: Using disks (wahsers) method, find the volume of the solid obtained by rotating the region
A共y兲
␲ y 2兾3
bounded by y = x3 , y = 8 and
x =⌬y0 苷
about
the⌬y
y-axis.
Since the
solid lies between y 苷 0 and y 苷 8, its volume is
Solution:
The region bounded
by these curves
is shown in the 8left figure
8
8
96␲ and the resulting solid is
3 5兾3
2兾3
V
苷
A共
y兲
dy
苷
␲
y
dy
苷
␲
y
苷
0
5
shown in the right
figure.
0
0
5
y
[
y
y
]
y
y=8
8
x
(x, y)
Îy
x=0
y=˛
or
x=œ„
œ3 y
0
x
0
x
If we slice at an arbitrary
height y, we get a circular disk(b)
with radius x, where x = y 1/3 . So
(a)
the area of a cross section through y is
A(y) = πx2 = π(y 1/3 )2 = πy 2/3 .
Since the solid lies between y = 0 and y = 8, its volume
Z 8
Z 8
i8
h3
2/3
5/3
=
V =
A(y) dy = π
y dy = π y
5
0
0
0
is
3 √
3
96
3
π( 8)5 = π 25 = π.
5
5
5
J
Ex. 2: The region R enclosed by the curves y = x and y = x2 is rotated about the x-axis. Find
the volume of the resulting solid using the method of washers.
Solution:
We graph the region first (See the figure below), then we find the intersection points by
setting y = x equal to y = x2 to have x2 = x or x2 − x = 0 or x(x − 1) = 0 and thus x = 0
and x = 1.
冋
苷␲
ON
Some Examples
Math 142, Calculus II, Fall 2016
x
x
⫺
3
5
册
苷
0
2␲
15
2
2
E 4 The region ᏾ enclosed by the
y curves y 苷 x and y 苷 x isyrotated about the
ind the volume of the resulting solid.
(1, 1)
y=x
N The curves y 苷 x and y 苷 x 2 intersect
at the points 共0, 0兲 and 共1, 1兲. The
y=≈
etween them, the solid of rotation, and a cross-section
perpendicular to the
e shown in Figure 8. A cross-section in the plane Px has the shape of a washer
lar ring) with inner radius x 2 and outer radius x, so we find the cross-sectional
x
0
(0, 0)
subtracting the area of the inner circle from the area of the outer circle:
A共x兲 苷 ␲ x 2 ⫺ ␲ 共x 2 兲2 苷 ␲ 共x 2 ⫺ x 4 兲
e we have
Using the washers method the volume of the resulting solid (See the figure below) is
Z 1
V =π
[R2 (x) − r2 (x)] dx,
1
1
0
V 苷 A共x兲 dx 苷
共x 2 ⫺ x 4 兲 dx
FIGURE
8 0radius is the distance
where
R the larger
(a) a random x on the x−axis to the line y =
(b)x
0 from
and it is in fact equal to y = x, and r the smaller radius is the distance from that random
x to the curve y = x2 and3 it is just5y =1x2 . Thus, R = x, r = x2 and thus
x
x5 Find
2
Z 1 the volume
solid2 obtained
h 1 of the
苷 Z 1 EXAMPLE
⫺
苷
1 5 i1
2
4
3
2
2 2
[x 15
[(x)
) ] dx
= πy 苷
V =π
3− (xthe
5line
about
2−.x ] dx = π 3 x − 5 x 0 = 15 π.
0 0
0
y
y
冋
␲
册
␲
␲
by rotat
SOLUTION
The solid and a cross-section are shown in Figu
y
y
a washer, but this time the inner radius is 2 ⫺ x and the ou
(1, 1)
A(x)
y
y=x
y=≈
4
≈
x
(0, 0)
x
x
0
y=2
(a)
( b)
(c)
2-≈
J
y=x
E 5 Find the volume of the solid obtained
by rotatingy=≈
the region in Example 4
≈
ea by subtracting the area of the inner circle from the area of the outer circle:
A共x兲 苷 ␲ x 2 ⫺ ␲ 共x 2 兲2 苷 ␲ 共x 2 ⫺ x 4 兲
Some Examples
3
Math 142, Calculus II, Fall 2016
erefore we have
FIGURE 8
(a)
(c)
(b)
1
1
Ex. 3: Find the volume
of the solid
obtained
by
rotating
the
region
in
Example
2
about
the
line
V 苷 y A共x兲 dx 苷 y ␲ 共x 2 ⫺ x 4 兲 dx
0
0
y = 2.
EXAMPLE 5 Find the volume of the solid obtained by rotating the region in Exam
冋
册
Solution:
x 3 about
x 5 the line
2␲ y 苷 2.
苷␲
⫺
苷
After rotating the
bounded by y = x and y = x2 about the horizontal line y = 2,
3 region
5 0that15
SOLUTION The solid and a cross-section are shown in Figure 9. Again the cross-se
the resulting solid will look like this
2
1
ya
y
washer, but this time the inner radius is 2 ⫺ x and the outer radius is 2 ⫺ x .
y
(1, 1)
A(x)
y=x
4
y=≈
≈
x
(0, 0)
x
x
0
y=2
y=2
(a)
(c)
( b)
2-x
2-≈
XAMPLE 5 Find the volume of the solid obtained by rotating the region in Example 4
out the line y 苷 2.
y=≈
y=x
x
≈
x
0 9. xAgain1 the cross-section
LUTION The solid and a cross-section are shown in Figure
is
9 is 2 ⫺ x and the outer radius is 2 ⫺ x .
washer, but this time the FIGURE
inner radius
x
2
y
The washer in purple will look like this:
4
y=2
y=2
2-x
2-≈
y=≈
y=x
x
≈
0
x
1
x
x
x
The larger radius R will be the distance from the line y = 2 (at some random point) to the
curve y = x2 (the bottom curve). Since the distance from the x−axis to the the line y = 2
is just 2 and the distance from the x−axis (at some random point x) to the curve y = x2 is
just y which is x2 , then R is the remaining distance, i.e., R = 2 − x2 . The smaller radius r
will be the distance from the line y = 2 (at some random point) to the curve y = x (the top
x
Some Examples
Math 142, Calculus II, Fall 2016
4
curve) and that will be r = 2 − x. Now this purple washer can only move along the x−axis
and it moves from x = 0 to x = 1, and thus
Z 1
Z 1
2
2
[R − r ] dx = π
[(2 − x2 )2 − (2 − x)2 ] dx
V =π
0
Z0 1
[(4 − 4x2 + x4 ) − (4 − 4x + x2 )] dx
=π
Z0 1
[4 − 4x2 + x4 −4+4x−x2 ] dx
=π
Z0 1
[4x − 5x2 + x4 ] dx
=π
0
h
5 3 1 5 i1
8
2
= π 2x − x + x
= π.
3
5
15
0
J