Exercise 8

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Exercise 8
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.
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 =
1
∇ × F~ · n̂ = √ (2y − 2z + 2z − 2x + 2x − 2y) = 0
I
Z Z3
⇒ F~ · d~r =
0dS = 0
c
x̂ + ŷ + ẑ
√
3
(1)
(2)
(3)
S
2. (a)
3
Z
2
Z
3
Z
x2 dx = 9
dy(x − y) =
dx
0
x
0
(4)
0
(b)
ZZ
−4
dxdy = −16π
(5)
3. Use Green’s theorem to calculate the area of the region enclosed by the ”deltoid” curve defined
as x = 2 cos t + cos 2t, y = 2 sin t − sin 2t, 0 ≤ t ≤ 2π. Draw a sketch of the curve.
2
1
-1
1
2
3
-1
-2
Using the expressions for x(t), y(t) in the area formula
S=
1
2
I
−ydx + xdy = 2π
(6)
1
p
4. Let S be the surface of the cone z = x2 + y 2 and 0 ≤ z ≤ 1 and let S̃ be the surface of the
lower half of the sphere x2 + y 2 + (z − 1)2 = 1. For the vector field F~ = (0, −2x, 2), calculate
the flux integral through the surfaces S and S̃ in the direction of the outward normal.
~ such that F~ = ∇
~ ×G
~ and apply Stokes’ theorem.
Hint: try to find a vector field G
Using the hint, the flux integral can be converted to a line integral
ZZ
ZZ F~ · n̂dS =
I
~ ×G
~ · n̂dS =
~ · d~r
∇
G
(7)
C
Where C is the unit circle at z = 1. This leads to the (obvious) realization that the fluxes are
the same and can also be calculated by a line integral :
I
ΦS = ΦS̃ =
~ · d~r
G
(8)
c
~ = (−y, x, x2 )
One possible option (there are infinitely many options, why?) is G
I
~ · d~r =
G
I
−ydx + xdy = 2π
(9)
C
The last integral is immediate, since it is just twice that area of the contour C
5. Use the surface integral in Stokes’s theorem to calculate the flux of the rotor of the field F~
across the surface S in the direction of the outward normal n̂.
F~
= (y 2 , z 2 , x)
(10)
S
:
(11)
~r(θ, φ) = (2 sin θ cos φ, 2 sin θ sin φ, 2 cos θ)
0 ≤ θ ≤ π/2 , 0 ≤ φ ≤ 2π
(12)
ZZ
ZZ
(x, y, z)
~
~
(∇ × F ) · n̂dS
(−2z, −1, −2y) ·
sin θdθdφ =
(13)
2
Z 2π Z π/2
=
(−4 cos θ, −1, −4 sin θ sin φ) · (sin θ cos φ, sin θ sin φ, cos θ) sin θdθdφ =
(14)
0
=
0
0
(15)
2