#13 , 2014 14

#13
,
2014
14
F~ = 3x2 yx̂ + (x3 + 1)ŷ + 9z 2 ẑ
.x = 2
.x2 + y 2 > 1
,
x2 + y 2 + z 2 = 9
x2 + y 2 + z 2 = 4
S
.z
,F~ = (y, x, z)
.S
.1
.
.
.2
S
.3
.n̂ = (x, y, z)/2
F~ · n̂
=
=
=
.1
(x, y, z)
z2
(y, x, z) ·
=
2
2
¨
ˆ 2⇡ ˆ 5⇡/6
4 cos2 ✓
F~ · n̂dS =
4 sin ✓d✓d =
2
S
0
⇡/6
ˆ 5⇡/6
5⇡/6
p
cos3 ✓
16⇡
cos2 ✓ sin ✓d✓ = 16⇡
= 4 3⇡
3 ⇡/6
⇡/6
(1)
(2)
(3)
F~ · n = yx
.
~ · F~
r
=
=
.2
xy = 0
n̂ = (x, y, 0)
˜
˝
~ · F~ dV ,
, S F~ · n̂dS = V r
˚
˚
~ · F~ dV =
1)
r
dV = V =
V
V
¨
¨
p
F~ · n̂dS +
F~ · n̂dS = 4 3⇡
S
.3
(4)
(5)
cylinder
a, b ,F~ = (ax2 y + y 3 + 1)x̂ + (2x3 + bxy 2 + 2)ŷ
,
.
.F~ = rf
a, b
,
f (x, y)
.x = et cos t, y = et sin t, 0  t  ⇡
C
1
´
C
F~ ·d~r
.1
.2
.3
,@x Fy = @y Fx
@ x Fy
@ y Fx
=
=
.1
6x2 + by 2
2
ax + 3y
(6)
2
(7)
(8)
) a = 6, b = 3
.2
f (x, y)
=
ˆ
fy0 (x, y)
=
2x3 + 3xy 2 + c0 (y) = Fy = 2x3 + 3xy 2 + 2 ) c(y) = 2y + const
Fx dx + c(y) =
x
f (x, y)
=
3
ˆ
(6x2 y + y 3 + 1)dx + c(y) = 2x3 y + xy 3 + x + c(y) (9)
3
2x y + xy + x + 2y + const
,( e⇡ , 0)
,
ˆ
C
.
:
F~ · d~r = f ( e⇡ , 0)
=
2x
v
=
J
=
x+y 1
ux uy
vx vy
dudv
=
V
=
=
=
y)2 + (x + y
.z = (2x
u
f (1, 0) =
1)2
e
⇡
(1, 0)
C .3
z=4
(13)
(14)
=
2
1
1
1
(15)
=3
1
3dxdy ) dxdy = dudv
3
¨
(4 (2x y)2 (x + y
(2x y)2 +(x+y 1)2 <4
¨
1
(4 u2 v 2 ) dudv =
3
u2 +v 2 <4
ˆ 2⇡ ˆ 2
ˆ 2⇡ 
2 2
1 4
2 1
(4 r ) rdrd✓ =
r
r
3
3
12
0
0
0
(16)
1)2 )dxdy =
2
d✓ =
0
˜
R
R
C
(17)
(18)
ˆ
0
.F~ =
.x2 + y 2 + z 2 = 1
(11)
(12)
1
y
.x2 + y 2 + z 2 = 1
(10)
2⇡
4
8⇡
d✓ =
3
3
(19)
2xz x̂ + y 2 ẑ
~ ⇥ F~
.r
.1
~ ⇥ F~ · n̂dS = 0
r
¸
F~ · d~r = 0
.2
C
.3
.1
x̂
@x
2xz
~ ⇥ F~ =
r
ŷ
@y
0
ẑ
@z
y2
= 2yx̂
(20)
2xĵ
.2
n̂
=
~ ⇥ F~ ) · n̂
(r
=
(21)
(x, y, z)
(2y, 2x, 0) · (x, y, z) = 0 )
¨
R
~ ⇥ F~ · n̂dS = 0
r
˜
¸
~ ⇥ F~ · n̂dS =
R
r
F~ · d~r ,
R
C
˜
¸
~ ⇥ F~ · n̂dS =
. Rr
F~ · d~r = 0
C
C
.f (x, y) = x + 4y +
(22)
.3
,
2
xy
.1
2
=0
x2 y
2
=0
xy 2
fx
=
1
fy
=
4
2
)
xy 2
=
)x
=
2
1
2
4y
)
1
y= , x=2
2
x y
=
(23)
(24)
(25)
(26)
(27)
(28)
(29)
.(x, y) = (2, 1/2)
.2
fxx
=
A
=
4
4
2
, fyy = 3 , fxy = fyx = 2 2
x3 y
xy
x y
fxx (2, 1/2) = 1, C = fyy (2, 1/2) = 16, B = fxy (2, 1/2) = 2
) AC
2
B = 12 > 0, A > 0 ) (2, 1/2)IS MINIMUM
(1.02)2 · (1
p
(30)
(31)
(32)
3.98)
f (x, y) = x2 (1
p
y)
(33)
.x = 1.02, y = 3.98
f (x, y)
⇡
f (1, 4)
=
fx0 (1, 4)
=
f (x0 , y0 ) + fx0 (x0 , y0 )(x x0 ) + fy0 (x0 , y0 )(y
p
12 (1
4) = 1 2 = 1
p
2x(1
y)|(1,4) = 2
2
fy0 (1, 4)
=
f (1.02, 3.98)
⇡
(x0 , y0 ) = (1, 4)
x
p |(1,4) =
2 y
1
2(1.02
1
4
1)
y0 )
(34)
(35)
(36)
(37)
1
(3.98
4
4) =
1.035
(38)
Gauss law for gravity.
Let ~g be the gravitational field. The following equation is satisfied by the field
~ · ~g (~r) =
r
4⇡G⇢(~r)
where G is the universal gravitational constant, and ⇢(~r) is the mass density at each point of space,
carrying units of mass/volume. Using the terminology we learned in class, the mass (multiplied by
4⇡G) is the source for the gravitational field, and the divergence of the field is the source density.
1. For an arbitrary closed surface S show that the following holds for the gravitational field:
¨
~g (~r) · n̂(~r) dS = 4⇡GMS
S
where MS is the total mass enclosed within the surface S.
~ g (~r) = 0, and it goes to zero at infinity. Using previous
2. Gravity is a conservative force, i.e. r⇥~
results (question 1), prove Newtonian gravity: show that for a point mass M located at the
origin the following holds
r̂
~g (~r) = GM 2
r
Hint: assume from symmetry considerations that the gravitational field is radial, i.e. that
~g (~r) = g(r)r̂, and choose the surface S wisely.
3. Repeat the previous question for a spherical mass of constant mass density ⇢: what is the
gravitational field as function of the distance from the center of the mass? (use the result
from question 1).
4. The Earth is hardly point-like; however, let us assume it is approximately a sphere. Its
density is not constant either - however we can assume it is constant within each layer; we
shall further assume that it changes linearly with the distance from the center of the Earth,
where the density is maximal. The density at the crust is approximately 3 g/cm3 . What is
the density at the core? (the mass of the earth is 5.97 · 1024 Kg and its radius is 6378 Km).
5. Using the previous results (questions 1 and 4), calculate the gravitational field along a straight
line starting at the center of the earth and ending in an airplane some 10km above China.
Draw a graph showing the magnitude of the gravitational field |~g (rr̂)| as function of r, the
distance from the center of the earth. Indicate your units on your x-axis and y-axis.
6. USS Enterprise encounters an infinite cylindrical shape of constant radius R and a constant
mass density ⇢ (within the region of space 1 < z < 1, r  R), pulling the starship towards
it! Captain Kirk’s orders: “Mr. Spock, quickly calculate the gravitational field as function of
the distance from the center of the shape! Both for r < R and r > R” (assume you are Mr.
Spock).
7. In Terry Pratchett’s Discworld the world is an infinite, perfectly flat disc, of constant mass
density (carrying units of mass/area). Calculate the gravitational field as function of the
distance r away from the disc. What should be so that g = 9.8 m/s2 ? (Hint: use a surface
that is a cylinder of radius R extending both above and below the disc, r  z  r, where
the disc is at z = 0) . You can neglect the elephants and the turtle.
Solutions to Ex 8
˜
˝
~ ·~g dV =
1. S ~g (~r) · n̂(~r) dS = V r
within S.
4⇡G
˝
4⇡GMS where MS is the mass trapped
⇢(~r) =
V
2. Take S to be a sphere of radius r, so that n̂ = r̂. Since ~g (~r) = g(r)r̂ we get
¨
g(r)r̂ · r̂ = 4⇡r2 g(r) = 4⇡GM
(39)
S
and therefore
g(r) =
3. For r > R same as (2), with M =
4⇡ 3
3 R ⇢.
1
r2
GM
(40)
For r < R we get
4⇡r2 g(r) =
4⇡G
4⇡ 3
r ⇢
|3{z }
(41)
M (r)
and therefore
4⇡G⇢
r
3
g(r) =
(42)
4. We take (⇢c - density at core, ⇢s -density at surface)
⇢(r) = ⇢c (1
(43)
↵r)
We are given that
⇢s = ⇢(R) = ⇢c (1
↵R)
(44)
The mass of the earth is
ME =
ˆ
R
2
⇢(r)4⇡r dr
=
0
ˆ
R
↵r)4⇡r2 dr
✓
◆
4⇡
3
3
⇢c R 1
↵R
3
4
⇢c
(1
(45)
0
=
and therefore
4⇡
ME =
⇢s R3
3
Plugging in
1
1
3
4 ↵R
↵R
=
1
1
ME
4⇡
3
3 ⇢s R
!
(47)
= 1.831
(48)
3R
4 ↵
R↵
and
hence
⇢c =
1
(46)
↵R = 0.769
(49)
⇢s
' 13g/cm3
0.769
(50)
5. First r < R: The mass within a ball of radius r is

4⇡
3
r
3
M (r) =
⇢c r 1
(↵R)
3
4
R
(51)
and using Gauss’ law for a spherical Gauss surface
g(r)4⇡r2
we get
g(r < R) =
=
4⇡GM (r)

4⇡
G ⇢c r 1
3
3
r
(↵R)
4
R
(52)
(53)
For r > R it is the same as (3).
6. We take a cylindrical Gauss surface of height h and radius r. Notice that the two flat surfaces
at the top and bottom do not contribute as the normal is orthogonal to the field. For r < R:
(2⇡r)hg(r) =
4⇡G ⇢(⇡r2 )h
| {z }
(54)
M (r)
and hence
g(r < R) =
For r > R:
(2⇡r)hg(r) =
2⇡G⇢r
4⇡G ⇢(⇡R2 )h
| {z }
(55)
(56)
M (R)
and therefore
g(r > R) =
2⇡G⇢
R2
r
(57)
#14
,
2015
19
x = f (t), y = g(t), ti  t  tf
.(f (tf ), g(tf )) (f (ti ), g(ti ))
x = t2 , y = t + 1,
1<t<1
,
.
,
x = t2 = (y
1)2
.
.
.
a
.
1
P
P
.
t
P
x = at + a cos ✓, y = a + a sin ✓
t + ✓ = 3⇡/2
x
=
y
=
t
✓
3⇡
t) = at a sin t
2
3⇡
a + a sin(
t) = a + a sin t
2
at + a cos(
.#7
S
:
I
C
F~ (~r) · d~r
=
ZZ
S
~ ⇥ F~ ) · n̂dS
(r
,S
C
.
,
,S
.
n̂
C
,
I
Fx dx + Fy dy =
C
ZZ ✓
D
@Fy
@x
,
@Fx
@y
◆
dxdy
3
~v (~r) = (ex sin y
3y + x2 , ex cos y
x2 + y 2 = 2x
.
1, 0)
R
C
C
,2014
~v (~r) · d~r
,
f (~r0 +
~r)
=
f (~r0 ) +
⇣
⌘2
⇣
⌘n
~ (~r0 ) + 1
~ f (~r0 ) + ... 1
~
~r · rf
~r · r
~r · r
f (~r0 ) + ...
2!
n!
2014
p
2
.f (x, y) = sin(x) + sin(y)2
.(x = 320 , y = 320 )
,
y 00 (t) + p(t)y 0 (t) + q(t)y(t)
=
g(t)
y(t)
=
c1 y1 (t) + c2 y2 (t)
Y (t)
=
u1 (t)
=
u2 (t)
=
W (y1 , y2 )
=
u1 (t)y1 (t) + u2 (t)y2 (t)
Z
y2 (t)g(t)
dt
W (y1 , y2 )
Z
y1 (t)g(t)
dt
W (y1 , y2 )
y1 y20 y10 y2
2014
ẍ(t) + 4ẋ(t) + 4x(t)
=
et
x(0)
=
0
ẋ(0)
=
1