Exercise 9

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Exercise 9
Gauss and Stokes Theorems
1. In the following, use the surface integral in Stokes Theorem to calculate the circulation of the
field F around the curve C in the indicated direction.
(a) F~ = x2 x̂ + 2xŷ + z 2 ẑ
∇ × F~
I
= 2ẑ ⇒ ∇ × F~ · n̂ = 2, dS = dxdy
ZZ
~
2dS = 2(Area of ellipse) = 4π
F · d~r =
c
(1)
(2)
S
C : The ellipse 4x2 + y 2 = 4 in the xy-plane, counter clockwise when viewed from above
(b) F~ = (y 2 + z 2 )x̂ + (x2 + z 2 )ŷ + (x2 + y 2 )ẑ
C : The boundary of the triangle cut from the plane x + y + z = 1 by the first octant,
counterclockwise when viewed from above
∇ × F~
= (2y − 2z)x̂ + (2z − 2x)ŷ + (2x − 2y)ẑ, n̂ =
x̂ + ŷ + ẑ
√
3
(3)
1
∇ × F~ · n̂ = √ (2y − 2z + 2z − 2x + 2x − 2y) = 0
I
Z Z3
⇒ F~ · d~r =
0dS = 0
c
(4)
(5)
S
2. Find the flux of F~ = (3z + 1)ẑ upward across the hemisphere x2 + y 2 + z 2 = a2 , z ≥ 0
(a) With the divergence theorem.
∇ · F~
ZZ
= 3 ⇒ Flux across the hemisphere =
F~ · n̂dS =
S
ZZZ
ZZZ
1 4πa3
~
=
∇ · F dV =
3dV = 3
= 2πa3
2 3
V
V
(6)
(7)
(b) By evaluating the flux integral directly. The flux is the sum of the flux across the
hemisphere and the flux across the base.
ZZ
ZZ
xx̂ + y ŷ + z ẑ a
~
Φhemisphere =
F · n̂dS =
(3z + 1)ẑ ·
dxdy =
(8)
a
z
D
ZZ
ZZ
Z a p
p
=
(3z + 1)dxdy =
(3 a2 − x2 − y 2 + 1)dxdy = πa2 + 2π
2 a2 − r2 rdr (9)
=
D
Φbase
D
= πa2 + 2πa3
ZZ
ZZ
~
=
F (z = 0) · n̂dS = −
dS = −πa2
Φ = Φhemisphere + Φbase = πa2 + 2πa3 − πa2 = 2πa3
1
0
(10)
(11)
(12)
3. Gauss law for gravity.
Let ~g be the gravitational field. The following equation is satisfied by the field
~ · ~g (~r) = −4πGρ(~r)
∇
where G is the universal gravitational constant, and ρ(~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 −4πG) is the source for the gravitational field, and the divergence of the field
is the source density.
For an arbitrary closed surface S and the gravitational field, using the divergence theorem,
we have
ZZZ
ZZZ
ZZ
~ · ~g (~r)dV = −4πG
ρ(~r)dV = −4πGMS
(13)
~g (~r) · n̂(~r) dS =
∇
V
S
V
where MS is the total mass enclosed within the surface S. Since the gravitational force is
radial, we can exploit the spherical symmetry and take S to be a sphere of radius r, so that
n̂ = r̂, since ~g (~r) = g(r)r̂, we get
ZZ
g(r)r̂ · r̂dS = 4πr2 g(r) = −4πGM
(14)
S
From which the gravitational force must be
g(r) = −
GM
r2
(15)
(a) For r > R same as (15), with M =
4πr2 g(r) = −4πG
4π 3
3 R ρ.
For r < R we get
4π 3
r ρ
|3{z }
(16)
M (r)
and therefore
g(r) = −
4πGρ
r
3
(17)
(b) We take (ρc - density at core, ρs -density at surface)
ρ(r) = ρc (1 − αr)
(18)
We are given that
ρs = ρ(R) = ρc (1 − αR)
(19)
The mass of the earth is
Z R
Z
2
ME =
ρ(r)4πr dr = ρc
0
=
R
(1 − αr)4πr2 dr
0
4π
3
3
ρc R 1 − αR
3
4
(20)
(21)
and therefore
4π
ME =
ρs R3
3
1 − 3R
4 α
1 − Rα
!
(22)
2
Plugging in
1 − 34 αR
=
1 − αR
ME
4π
3
3 ρs R
= 1.831
(23)
and
αR = 0.769
(24)
hence
ρc =
ρs
' 13g/cm3
1 − 0.769
(25)
(c) First r < R: The mass within a ball of radius r is
4π
3
r
3
M (r) =
ρc r 1 − (αR)
3
4
R
(26)
and using Gauss’ law for a spherical Gauss surface
g(r)4πr2 = −4πGM (r)
(27)
we get
4π
3
r
g(r < R) = −G ρc r 1 − (αR)
3
4
R
(28)
For r > R it is the same as (3).
(d) 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:
(2πr)hg(r) = −4πG ρ(πr2 )h
| {z }
(29)
M (r)
and hence
g(r < R) = −2πGρr
(30)
For r > R:
(2πr)hg(r) = −4πG ρ(πR2 )h
| {z }
(31)
M (R)
and therefore
g(r > R) = −2πGρ
R2
r
(32)
(e) In Terry Pratchett’s Discworld the world is an infinite, perfectly flat disc, of constant mass
density σ (carrying units of mass/area). Calculate the gravitational field as function of
the distance r away from the disc. What should be σ 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.
3