6. The area of revolution of y =
Z
6
√
s
x, (0 ≤ x ≤ 6), about the x-axis is
dy
dx
2
dx
y 1+
r
Z 6
√
1
x 1+
dx
= 2π
4x
0
Z 6r
1
= 2π
x + dx
4
0
3/2 6
4π
1
1
4π
125
62π
=
=
−
sq. units.
x+
=
3
4
3
8
8
3
0
S = 2π
0
15.
y
x 2 +y 2 =a 2
ds
θ dθ
−a
s
xa
Fig. 4-15
Consider the area element which is the thin half-ring shown in the figure. We have
dm = ks π s ds = kπ s 2 ds.
kπ 3
Thus, m =
a .
3
Regard this area element as itself composed of smaller elements at positions given by the
angle θ as shown. Then
d M y=0 =
M y=0
Z
π
0
3
(s sin θ )s dθ ks ds
= 2ks ds,
Z a
ka 4
.
s 3 ds =
= 2k
2
0
ka 4
3
3a
Therefore, y =
·
=
. By symmetry, x = 0. Thus, the centre of mass of the
3
2π
2 kπ a
3a
plate is 0,
.
2π
1.
a) The nth bead extends from x = (n − 1)π to x = nπ , and has volume
Z
nπ
e−2kx sin2 x d x
Vn = π
Z (n−1)π
π nπ
e−2kx (1 − cos(2x)) d x
=
2 (n−1)π
Let x = u + (n − 1)π
Z d x = du
π π −2ku −2k(n−1)π =
e
e
1 − cos(2u + 2(n − 1)π ) du
2 0
Z
π −2k(n−1)π π −2ku
e
(1 − cos(2u)) du
= e
2
0
= e−2k(n−1)π V1 .
Vn+1
e−2knπ V1
= e−2kπ , which depends on k but not n.
= −2k(n−1)π
Vn
e
V1
b) Vn+1 / Vn = 1/2 if −2kπ = ln(1/2) = − ln 2, that is, if k = (ln 2)/(2π ).
Thus
c) Using the result of Example 4 in Section 7.1, we calculate the volume of the first bead:
V1 =
=
=
=
Z
π π −2kx
e
(1 − cos(2x)) d x
2 0
π e−2kx π π e−2kx (2 sin(2x) − 2k cos(2x)) π
−
−4k 0
2
4(1 + k 2 )
0
π
π
−2kπ
−2kπ
(k − ke
)
(1 − e
)−
4k
4(1 + k 2 )
π
(1 − e−2kπ ).
4k(1 + k 2 )
By part (a) and Theorem 1(d) of Section 6.1, the sum of the volumes of the first n beads
is
π
(1 − e−2kπ )
Sn =
2
4k(1 + k )
2
n−1 × 1 + e−2kπ + e−2kπ + · · · + e−2kπ
−2knπ
π
−2kπ 1 − e
(1
−
e
)
4k(1 + k 2 )
1 − e−2kπ
π
=
(1 − e−2knπ ).
4k(1 + k 2 )
=
Thus the total volume of all the beads is
V = lim Sn =
n→∞
π
cu. units..
4k(1 + k 2 )
12.
s=
Z
π/4 p
π/6
π/4
1 + tan2 x d x
π/4
sec x d x = ln | sec x + tan x|
=
π/6
π/6
√
1
2
= ln( 2 + 1) − ln √ + √
3
3
√
2+1
units.
= ln √
3
Z
28. The area of the cone obtained by rotating the line
y = (h/r )x, 0 ≤ x ≤ r , about the y-axis is
S = 2π
Z
0
r
√
2 + h 2 x 2 r
r
x 1 + (h/r )2 d x = 2π
r
2 0
p
= πr r 2 + h 2 sq. units.
q