=========================preview======================
(MATH024)[2012](s)final~=vfemuek^_21512.pdf
Back to MATH024 Login to download
======================================================
Solutions
1pp
sin2(n +1 . n)
1. (15 Marks) Determine whether the series X con-
log2(n +1)
n=1
verges.
1pp
sin2(n +1 . n)
Solution We shall show that the series X converges.
log2(n + 1)
n=1
pp1
Since n +1 . n = pp! 0, we have
n +1+ n
pp
sin2(n +1 . n)
lim pp=1.
n!+1 (n +1 . n)2
It is clear that
pp4n

lim 4n (n +1 . n)2 =lim pp
n!+1 n!+1 (n +1+ n)2 4
=lim =1.
n!+1
(p1+1/n + 1)2 log(n + 1)
We claim that lim = 1, since
n!+1 log n
.log(n + 1) . log(n + 1) . log n
lim . 1 =lim
n!+1 log n n!+1 log n log(1 + 1/n)
=lim =0.
n!+1 log n
Hence
pp
sin2(n+1n)
log2(n+1)

lim
1
n!+1
4n log2 n
pp
sin2(n +1 . n) pplog2 n
=lim pp 4n(n +1 . n)2
n!+1 (n +1 . n)2 log2(n + 1) 1
=1 1 =1.
12
It follows that there is N such that n . N implies
pp

sin2(n +1 . n)1
< 2 .
log2(n +1) 4n log2 n
Hence, by the Comparison Test, if we can show that the series X
n log2 n converges, then we can conclude that the original series converges.
1
To show that X converges, we notice that the function x log2 x is
n log2 n
increasing for suciently large x,since

(x log2 x)0 = log2 x + x 2 log x 1/x = (1+2log x) log x . 0.
1
Thus, f(x)= is decreasing for suciently large x. It is clear
x log2 x that f(x) ! 0 as x ! +1. Hence, by the Integral Test, we know that
X 1 converges, since, for a> 1,
n log2 n Z 1 11
dx = < +1.
a x log2 x log a
2. (15 Marks) Let R denote the region in the .rst quadrant enclosed by x = 0, y = c and y = x2,where c> 0. A solid is obtained by rotating the region R about y-axis.
(a)
Use two di.erent ways to .nd the volume of the solid.

(b)
Find the total surface area of the solid.

(c)
Where is the center of mass of the solid? Assume that the mass is evenly distributed.


p
Solution (a) The intersection of y = c and y = x2 is (c, c). The volume can be found by integrating a family of rings with in.nitesimal thickness. This gives
Z c p.c2
V = .(y)2 dy = .
2
0
The same volume can be found by integrating a family of cylindrical shells:
Z pc 22
. 1 c .c
V =2.x(c . x 2)dx =2. c . = .
24 2
0
(b) The total surface area consists of two parts: the top surface and the surface obtained from the rotation of the parabola. Hence
pZ c
S = .(c)2 +2. x(y)ds
0
Z c
= .c +2. x(y)px0(y)2 +1 dy
0
Z c ps.1 2
= .c +2. y y1/2 +1dy
2
0
Z c .1 1/2
= .c +2. + y dy
4
0 y=c
2 .1
3/2 .
= .c +2. + y
34
y=0
= .c + . h(1 + 4c)3/2 . 1i .
6
(c) Assume the density is .. The cross section made by the solid with the plane y = t is a circular disk. For each disk, the center of mass is located at (0,t). Thus
Z c p
0 ..(t)2dt x= 0Z c p=0, ..(t)2dt
0
Z c p
t ..(t)2dt
2c
0
y= = .
Z c pt)2dt 3
..(
0
3. (15 Marks) Expand the cosine function f(x) = cos x on (0, .) as a series of sine functions:
1
cos x = X bn sin nx, 0 <x< . . n=1
Use the series to .nd the su