10.
x = cos t + sin t, y = cos t − sin t, (0 ≤ t ≤ 2π )
The circle x 2 + y 2 = 2, traversed clockwise, starting and ending at (1, 1).
17.
x = et − t, y = 4et/2 , (0 ≤ t ≤ 2). Length is
Z
2p
(et − 1)2 + 4et dt
Z 2p
Z 2
t
2
(e + 1) dt =
=
(et + 1) dt
0
0
2
= (et + t) = e2 + 1 units.
L=
0
0
25. Area of a large loop:
Z
1 π/3
(1 + 2 cos(2θ ))2 dθ
A =2×
2 0
Z π/3
=
[1 + 4 cos(2θ ) + 2(1 + cos(4θ ))] dθ
0
π/3
1
= 3θ + 2 sin(2θ ) + sin(4θ ) 2
0
√
3 3
=π+
sq. units.
4
5.
y
b
c
2 ft
2 ft
a
x
Fig. C-5
x2
y2
+
= 1, with a = 2 and foci at (0, ±2) so that c = 2 and
a2
b2
b2 = a 2 + c2 = 8. The volume of the barrel is
Let the ellipse be
V =2
Z
2
2
Z
2
y2
4 1−
dy
8
π x dy = 2π
0
2
3
y 40π 3
= 8π y −
=
ft .
24 0
3
0
16. If r 2 = cos 2θ , then
2r
dr
sin 2θ
dr
= −2 sin 2θ ⇒
= −√
dθ
dθ
cos 2θ
and
ds =
s
cos 2θ +
dθ
sin2 2θ
dθ = √
.
cos 2θ
cos 2θ
a) Area of the surface generated by rotation about the x- axis is
Sx = 2π
Z
π/4
r sin θ ds
0
Z
π/4 √
dθ
cos 2θ sin θ √
cos 2θ
0
π/4
√
= (2 − 2)π sq. units.
= −2π cos θ = 2π
0
b) Area of the surface generated by rotation about the y- axis is
S y = 2π
= 4π
Z
π/4
r cos θ ds
−π/4
Z π/4 √
cos 2θ cos θ √
dθ
cos 2θ
π/4
√
= 4π sin θ = 2 2π sq. units.
0
0
3. A strip along the slant wall of the dam between depths h and h + dh has area
dA =
26
200 dh
= 200 ×
dh.
cos θ
24
The force on this strip is
d F = 9, 800 h d A ≈ 2.12 × 106 h dh N.
Thus the total force on the dam is
F = 2.12 × 10
6
Z
24
h dh ≈ 6.12 × 108 N.
0
θ
h
h+dh
26
24
Fig. 6-3