Solution to PSet 6
Xin Sun
Part II 1
a)
By second fundamental theorem,
Si0 (x) = sinc(x).
For x 6= 0,
Si00 = sinc0 (x) =
x cos x − sin x
.
x2
For x = 0,
sin x − x
= 0,
x2
where the last step is due to the second order approximation of sin x around
0.
Si00 = sinc0 (0) = lim
x→0
b)
Si0 (x) = sinc(x) = 0 gives
sin(x) = 0, x 6= 0.
Therefore x = kπ where k is a non-zero integer.
Since Si00 (kπ) = cos(kπ)
for k 6= 0,
kπ
1. if k is an non-zero positive even integer or a negative odd integer, kπ
is a local minimum;
2. if k is an non-zero negative even integer or a positive odd integer, , kπ
is a local maximum.
1
c)
See Figure Si0 and Figure Si00 .
Si0
d)
Please see the Figure Si in the last page.
e)
By second fundamental theorem,
h0 (x) = Si0 (xr )xr−1 =
1
(Si(xr ))0 .
r
Therefore
1
h(x) = Si(xr ) + C.
r
Since h(0) = 0 and Si(0) = 1, we have C = − 1r . Therefore
Si(xr ) − 1
h(x) =
r
2
Si00
f)
x2
lim
x→3 x − 3
Z
x
3
sinc(t)dt = lim x2
x→3
Si(x) − Si(3)
= 9Si0 (3) = 9sinc(3) = 3 sin 3.
x−3
7.3 22
a)
By chain rule,
dV
dV dh
=
.
dt
dh dt
= A(h). Therefore
By second fundamental theorem, dV
dh
−cA(h) = A(h)
which means
dh
= −c.
dt
3
dh
,
dt
b)
From a)
h = h0 − ct.
Therefore it will take time h0 /c.
Part II 3
Fix z and consider the horizontal slice, the area is given by 21 z 2 − 12 z 8 . By
the constraint we have, z < 1, the volume is
Z 1
1 2 1 8
1
z − z dz = .
2
9
0 2
Part II 4
a)
Let R − `/2 ≤ r ≤ R + `/2. Then for each r there is a cylinder shell for the
solid with radius r. For R − `/2 ≤ r ≤ R, the volume of the shell is
√
2πr × 3(r − R + `/2)dr.
For R ≤ r ≤ R + `/2, the volume of the shell is
√
2πr × 3(R − r + `/2)dr.
So the integral is
Z
R
Z
√
2πr × 3(r − R + `/2)dr +
R−`/2
R+`/2
R
4
2πr ×
√
3(R − r + `/2)dr
b)
The integral is
Z
Z R+`/2
√
√
2πr × 3 × `/2dr + 2 3π
R
r2 dr −
R−`/2
R−`/2
Z
!
R+`/2
r2 dr
R
√ √ `
2 3π
` 2
` 2
` 3
` 3
3
= 3π (R + ) − (R − ) +
2R − (R + ) − (R − )
2
2
2
3
2
2
√
`
`
− 3πR 2R2 − (R + )2 − (R − )2
2
2
√
2
3πR`
=
.
2
c)
√
By disk method, for 0 ≤ h ≤ 3`/2, each slice is a annulus with volume
√
√
`
3h 2
`
3h 2
π(R + −
) − π(R − +
) dr.
2
3
2
3
Therefore the total volume is
√
√
Z √3`/2
`
`
3h 2
3h 2
) − π(R − +
) dh
π(R + −
2
3
2
3
0
Z √3`/2
4πhR
=
2πR` − √ dh
3
√0
2
3πR`
.
=
2
7.4 12
The line passes (r, 0) and (0, h). For 0 ≤ x ≤ r, there is a shell with volume
x
2πx × h(1 − )dx.
r
Therefore the total volume of the cone is
Z r
πr2 h
x
2πx × h(1 − )dx =
r
3
0
5
7.4 13
Assume that the sphere has radius a, the cylinder has radius r and the height
is h. We want to first compute the volume of the hole removed. There is
cylinder of height h and radius r inside the hole separating the hole into an
upper piece and lower piece. This cylinder has volume
h2
πhr = πh(a − )
4
2
2
Now we use the disk method to compute the volume of the upper piece.
For h/2 ≤ y ≤ a, there is a disk of height y of volume
π(a2 − y 2 )dy
Therefore the volume is
Z a
1
1
π(a2 − y 2 )dy = πa2 (a − h/2) − πa3 + π(h/2)3 = .
3
3
h/2
Therefore the hole has volume
h2
1 3 1
π
2
2
3
πh(a − ) + 2 × πa (a − h/2) − πa + π(h/2) = (8a3 − h3 ).
4
3
3
6
Therefore the volume of the spherical ring is
4 3 π
π
πa − (8a3 − h3 ) = h3 .
3
6
6
6
Si
7