DEO
PAT-
ET
RIE
Sa
n
U
6.2 Selected Solutions
of
versity
ni
a
Find the volume of the solid obtained by rotating the region bounded by the given curves s k a
tc he w
about the specified line. Sketch the region, the solid, and a typical disk or “washer”.
(14)
y=
1
, y = 0, x = 1, x = 3, about y = −1.
x
Solution:
The outer radius is O(x)
=
I(x) = 0 − (−1) = 1, so
3
2
1
− (−1)
x
3
2
=
1
+ 1, and
x
2
1
2
+ 1 − [1] dx =
x
V =π
[O(x)] − [I(x)] dx = π
1
3
3 1
1
1
1
=
dx
=
−
+
2
ln
x
+
2
π
2
x
x
1 x
1
1
2
1
1
− + 2 ln 3 − − + 2 ln 1 = − + 2 ln 3 + 1 =
+ 2 ln 3
3
1
3
3
y
1
0
x
0
1
2
-1
-2
-3
1
3
4
U
of
√
x; about x = 2.
sk
Solution:
Rewrite the equations of the curves as functions of y: x = y and x = y 2 .
Then the outer radius, as a function of y, is O(y) = 2 − y 2 , and the inner radius is
I(y) = 2 − y.
π
1
0

0
2
4 − 2y + y
4
− 4 − 2y + y
2
dy = π

y
π
y
y

π
=
−3
+2
5
3
2 0
5
5
3
1
y 4 − 3y 2 + 2y dy =
0
y
2 1
2
1
0
-1
x
0
-1
2
1
2
3
4
5
PAT-
ET
RIE
atc he w
Therefore the volume of the solid is
1
1
2 2
2
2 2
V =π
2−y
− 2 − y dy =
O(y) − I(y) dy = π
0
DEO
n
y = x, y =
Sa
(16)
versity
ni
a
Sa
sk
Solution:
y
1
0
-1
x
0
1
2
3
-2
-3
-4
V =
π (sin x − (−2))2 − (0 − (−2))2 dx =
0
π
V =π
0
-5
π (sin x + 2)2 − 22 dx =
0
sin2 x + 4 sin xdx
(36)
3
y and
2
x = 4−(y −1)2 . Then the outer radius, as a function of y, is O(y) = 4−(y −1)2 −(−5) =
3
3
9 − (y − 1)2 = 8 + 2y − y 2 , and the inner radius is I(y) = 3 − 2 y − (−5) = 8 − 2 y. The
9 7
curves intersect at (3,0) and − ,
, so the volume of the solid is
4 2
7
2 V =π
O(y) − I(y) O(y) + I(y) dy =
Solution:
Rewrite the equations of the curves as functions of y: x = 3 −
0
π
7 2
π
8 + 2y − y
0
7
2
0
7
y − y2
2
2
3
− 8− y
2
8 + 2y − y
2
3
+ 8− y
2
dy =
7
1
π 2
2
y[2y − 7][2y 2 − y − 32]dy
16 + y − y dy =
2
4 0
3
DEO
PAT-
ET
RIE
n
U
y = 0, y = sin x, 0 ≤ x ≤ π ; about y = −2.
of
(34)
versity
ni
atc he w
a
x =3−
3
y
2
y
4
3
x = 4 − (y − 1)2
2
1
0
-6 -5 -4 -3 -2 -1 0
-1
4
x
1
2
3
4
5
radius r .
Sa
(46)
A frustum of a right circular cone with height h, lower base radius R, and top s k a
y
A
Solution:
z
r
h
R
O
B
x
z+h
rh
z
=
or Rz = r z + r h. Therefore z =
and
r
R
R
−r
r
R
the height of the larger triangle is |OA| = h + z = h 1 +
=
h. Thus
R−r
R−r
R
A has coordinates 0,
h and B has coordinates (0, R). The line AB has slope
R−r
R
h
h
h
R−r
−
=−
and therefore its equation is y = −
(x − R). Solving for x, we
R
R−r
R−r
get:
By similar triangles we have
R−r
R−r
y or x = R −
y, which is the radius of the circular cross-section
h
h
at height y.
y=h 2
R−r
R−r
Thus we get V = π
R−
y dy = letting u = R −
y
h
h
y=0


u=r
3 u=r
3 u=R
−hdu
−hπ
u
u
R3 r 3
π
h
πh
2


π
=
−
=
u
=
=
R−r
R−r
3 R−r
3 R−r
3
3
u=R
x−R = −
u=R
π h (R − r )(R 2 + Rr + r 2 )
=
R−r
3
π
u=r
R 2 + Rr + r 2
h
3
In the forest industry, this formula is might be stated in the following form:
The volume of a log equals its length times the average area of the small and large ends,
and the area of the ellipse whose axes equal the diameters of the top and bottom.
5
DEO
PAT-
ET
RIE
n
U
Find the volume of the described solid S.
of
versity
ni
tc he w
a
Sa
height h.
sk
Solution:
DEO
PAT-
ET
RIE
n
U
A frustum of a pyramid with square base of side b, square top of side a, and
of
(48)
versity
ni
atc he w
Using essentially the same diagram as above, we get
b−a
b/2 − a/2
y = b/2 −
y,
h
2h
which is one-half the side of the square cross-section at height y.
Thus the area of the cross-section at height y is
x = b/2 −
A(y) = (2x)2 = 4 b/2 −
4 b2 /4 − 2b/2
b2 − 2b
b−a
y
2h
A
2
2
=
(b − a) 2
b−a
y+
y
2h
4h2
z
=
a/2
(b − a)2 2
b−a
y+
y ,
h
h2
h
h
b/2
(b − a)2 2
b−a
y+
y dy =
h
h2
0
2
b − a y 2 (b − a)2 y 3 2
=
b y − 2b
+
h
2
h2
3 0
so the volume is V =
b2 h − 2b
b2 − 2b
b − a h2 (b − a)2 h3
+
=
h 2
h2
3
(b − a)2 h
=
3
2
(b
−
a)
h b2 − b(b − a) +
=
3
b2 − 2ab + a2
h ab +
=
3
b2 h − b(b − a)h +
h
y
a2 + ab + b2
3
6
O
B
x
a
U
of
(54)
versity
ni
DEO
PAT-
ET
RIE
Thus the volume of S is
V =
1
0
A(y)dy =
1
0
1.0
√
3
2
3ydy =
0.5
a
0.0
x
-1.0 -0.5 0.0 0.5 1.0
a/2
(56)
The base of S is the triangular region with vertices (0,0), (3,0), and (0,2). Crosssections perpendicular to the y-axis are semicircles.
3
3
2
The line AB has equation y = − (x − 3) or x − 3 = − y or x = 3 − y,
3
2
2
x
3 3
the diameter of the semicircle at height y. The radius is therefore
= − y and the
2
2 4
2
2
1
3 3
3
1
9π
area of the cross-section is A(y) = π
− y = π
(2 − y)2
(2 − y)2 =
2
2 4
2
4
32
Solution:
The volume is thus
2
y=2
9π
9π y=2
2
V =
(2 − y) dy =
A(y)dy =
(2 − y)2 dy = letting u = 2 − y
32 y=0
0
y=0 32

u=2 
9π u=0 2
9π u=2 2
9π  u3  = 9π 8 = 3π
u (−du) =
u du =
32 u=2
32 u=0
32
3 32 3
4
u=0
y
A
O
P
7
B
x
n
Sa
The base of S is the parabolic region {(x, y)|x ≤ y ≤ 1}. Cross-Sections
perpendicular to the y-axis are equilateral triangles.
sk
a
atc he w
√
Solution: The height of an equilateral triangle of side a is 3 a, so the area of
2
√
1 3 2
such a triangle is A = a 32a =
a .
2
4
√
For 0 ≤ y ≤ 1, the side of the equilateral triangle based on
√ it is a = 2 y, so the area of
3 2 the cross-section perpendicular to the y-axis is A(y) =
(2 y) = 3y.
4
y