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Solutions to Problem Set 2, Physics 370, Spring 2014
1
TOTAL POINTS POSSIBLE: 65 points.
1. Conceptual Question: The fundamental theorem of Calculus states
“the integral of a derivative over an interval is given by the value of the
function at the end points (boundaries). In other words, if F (x) is the
d
derivative of f (x) such that F (x) = dx
f (x), then the integral of F (x)
is determined by the values of f (x) at the endpoints such that
Z b
Z b
df (x)
F (x)dx =
dx = f (b) − f (a).
(1)
x=a
x=a dx
This is essentially a one-dimensional statement of the theorem of Calculus. Extending this idea of the relationship between F (x) and f (x)
into more dimensions (and to scalars from vectors), you can derive
other versions of the “fundamental theorem.”
(a) Explain how the fundamental theorem of gradients (equation 1.55)
Z
~b
~a
~ · d~l = T (~b) − T (~a).
∇T
(2)
is the equivalent statement to “fundamental theorem of Calculus.”
I am not asking for a proof, I am asking you to explain in clear
English how this fundamental theorem of gradients is conceptually
equivalent to the fundamental theorem of Calculus.
(b) Explain how the divergence theorem a.k.a. the fundamental theorem of divergences (equation 1.56)
Z I
~ · ~v dτ = ~v · d~a.
∇
(3)
V
S
is the equivalent statement to “fundamental theorem of Calculus.”
Again, a clear English explanation of how this is equivalent is
needed for full credit.
(c) Explain how Stokes’ theorem a.k.a. the fundamental theorem of
curls (equation 1.57)
Z I
~ × ~v · d~a =
∇
~v · d~l.
(4)
S
P
Solutions to Problem Set 2, Physics 370, Spring 2014
2
is the equivalent statement to “fundamental theorem of Calculus.”
As with the last two parts, a clear English explanation of how this
is equivalent is needed for full credit.
I will note, this solution is me somewhat “winging it,” I won’t claim
these solutions are the only possible way to state this clearly (or that
they are perfectly clear, for that matter).
~ · d~l is just how much the
(a) (5 points possible) Recall that ∇T
function T changes along infinitesimal change in position d~l. As
such, this integral is simply stating that if I sum up how much
the function changes in infinitesimal steps along the path, it is
equal to the difference in the value of T at the beginning and end
~ is a three-dimensional analog to the
of the path. In essence, ∇T
derivative of the function f in the fundamental theorem of calculus
~ along a path takes on a value determine
and the integral of ∇T
by the value of the function T at the endpoints of the path (just
as in the fundamental theorem of calculus, where the value of the
integral of the derivative of f is determined by the value of f at
the endpoints of the integration interval).
~ · ~v represents how
(b) (5 points possible) Recall the divergence ∇
much the vector field ~v “spreads out” from a point. If the vector
field is spreading out from the interior of a surface, than the vector
field ~v must pass through that surface. Now consider that ~v · d~a
represents how much of the vector ~v is in the direction perpendicH
ular to the infinitesimal surface element d~a. Therefore, S ~v · d~a
~ · ~v is the
represents the total “flow” of ~v through the surface. ∇
vector derivative of the field ~v , so summing the value this derivative summed over a volume should be determined by the value of
~v at the bounds of the volume, which is a surface. In this way, it
is analogous to the fundamental theorem of calculus.
~ × ~v
(c) (5 points possible) Recall that the curl of a vector field ∇
represents how the vector field “twists.”
The integral of the curl
R
~
over the entire surface, S ∇ × ~v · d~a should represents the total
amount of twisting in the vector field over this surface. This should
be equal to the integral of the projection of ~v parallel to the line
Solutions to Problem Set 2, Physics 370, Spring 2014
3
H
bounding the surface, which can be written ~v ·d~l. In essence, the
surface integral of a vector form of the derivative of a vector field
~v is determined by the values of the vector field ~v at the bounds
of that surface, which is a path around the surface. In this sense,
Stoke’s theorem is also analogous to the fundamental theorem of
calculus.
2. Griffiths Problem
1.35 (reworded): Corollary 1 of Stoke’s theorem
R
~
which states “ (∇ × ~v) · d~a depends only on the boundary line, not on
the particular surface used.” Check Corollary 1 by using the same
function and boundary as in Example 1.11 but integrating over the
sides of the cube in Fig. 1.35 (below). The back of the cube is open.
A proper check shows that the surface integral over these five sides of
the cubeHis equal to the surface integral over the sixth side and the line
integral ~v · d~l, which were shown to both be equal to 43 in Example
1.11. (HINT: This problem makes much more sense if you first review
Example 1.11 on pages 35-36 of Griffiths.)
(10 points possible) As I stated, the real goal of this problem is to
see the corollary works by showing the complete surface integral
Z
Z
~
(∇ × ~v ) · d~a =
[4z 2 − 2x]ˆ
x + 2zˆ
z · d~a
(5)
S
S
~ × ~v expression from Example 1.10 where it was
(where I got the ∇
already worked out) is equal to 43 for the case where we consider the
Solutions to Problem Set 2, Physics 370, Spring 2014
4
five sides of the cube indicated in Fig. 1.35, bounded by the same
boundary line as the sixth side of the cube, to be a single surface S.
Using the same numbering scheme as indicated in the figure, I compute
this surface integral piecemeal, taking each side one at a time:
(i) For this side of the unit cube, x = 1, y = 0 → 1, and z = 0 → 1.
Furthermore, the infinitesimal area d~a = dydzˆ
x. Putting this
surface in equation 5, we get:
Z 1 Z 1
Z
~ × ~v ) · d~a =
(∇
[4z 2 − 2x]ˆ
x + 2zˆ
z · dydzˆ
x (6a)
z=0
1
=
Z
y=0
dy
y=0
Z 1
Z
1
z=0
(4z 2 − 2)dz
(4z 2 − 2)dz
1
4 3
= z − 2z
3
z=0
2
4
= −2 =−
3
3
=
(6b)
(6c)
z=0
(6d)
(6e)
where in equation 6b I used the fact that there is no y dependence
in the integrand to separate the y and z integrations and the fact
that x = 1 to eliminate x.
(ii) For this side, z = 0, x = 0 → 1, y = 0 → 1, and d~a = −dxdyˆ
z,
which means equation 5 for this side becomes:
Z
Z 1 Z 1
~
(∇ × ~v ) · d~a = −
[4z 2 − 2x]ˆ
x + 2zˆ
z · dxdyˆ
z (7a)
x=0
1
=−
=−
Z
x=0
Z 1
x=0
y=0
dx
dx
Z
1
y=0
Z 1
dy(2z)
(7b)
dy(0) = 0
(7c)
y=0
where in the last step there I used the fact that z = 0 for this
surface to finish the integral.
(iii) For this side, y = 1, x = 0 → 1, z = 0 → 1, and d~a = dxdz yˆ, so
Solutions to Problem Set 2, Physics 370, Spring 2014
the surface integral is
Z
Z
~ × ~v ) · d~a =
(∇
1
x=0
Z
5
1
[4z 2 − 2x]ˆ
x + 2zˆ
z · dxdz yˆ
z=0
=0
(8a)
(8b)
~ × ~v term, the intewhere since there is no y component in the ∇
grand is zero.
(iv) For this side, y = 0, x = 0 → 1, z = 0 → 1, and d~a = −dxdz yˆ.
~ × ~v
As for surface (iii), since there is no y component to the ∇
term
Z
Z 1 Z 1
~ × ~v ) · d~a = −
(∇
[4z 2 − 2x]ˆ
x + 2zˆ
z · dxdz yˆ (9a)
x=0
z=0
= 0.
(9b)
(v) For the fifth side, z = 1, x = 0 → 1, y = 0 → 1, and d~a = dxdyˆ
z.
So the surface integral is similar to that of surface (ii), except that
z = 1, therefore
Z
Z 1 Z 1
~
(∇ × ~v) · d~a =
[4z 2 − 2x]ˆ
x + 2zˆ
z · dxdyˆ
z (10a)
=
Z
x=0
1
y=0
dx
x=0
= 2.
Z
1
dy(2)
(10b)
y=0
(10c)
Summing together these surface integrals of these five surfaces
into one complete surface integral (bounded by the path around
the sixth side), I have:
Z
~ × ~v ) · d~a = − 2 + 0 + 0 + 0 + 2 = 4 .
(11)
(∇
3
3
S
Corollary 1 holds, the surface integral is determined by the line integral around the boundary, not by the specific surface you choose
to integrate over.
Solutions to Problem Set 2, Physics 370, Spring 2014
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ˆ φˆ in terms of
3. Griffiths Problem 1.38: Express the unit vectors rˆ, θ,
xˆ, yˆ, zˆ (that is, derive Eqn. 1.64). Check your answers several ways (
does rˆ · rˆ = 1, θˆ · φˆ = 0, rˆ × θˆ = φˆ ...). Also work out the inverse
ˆ φ.
ˆ
formulas, giving xˆ, yˆ, zˆ in terms of rˆ, θ,
NOTE: This is probably going to be computationally the longest problem of this problem set, not because the math is difficult, but it is rather
tedious. Sorry.
(10 points possible) The easiest approach to determine the spherical unit vectors is to first use the relationship between spherical and
Cartesian (x, y, z) coordinates from equation 1.62
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
(12a)
(12b)
(12c)
to note that the position vector can be written
~a = r sin θ cos φˆ
x + r sin θ sin φˆ
y + r cos θˆ
z.
(13)
(NOTE: I have chosen to call the position vector ~a instead of ~r to avoid
confusion with the coordinate with the same name, r.) If we take an
infinitesimal step, we can represent the increase in each coordinate as:
d~a =
∂~a
∂~a
~ = ∂~a dφ.
dr; d~θ =
dθ; dφ
∂r
∂θ
∂φ
(14)
And if we divide these infinitesimal vectors by their lengths, we have
the coordinate unit vectors:
∂~a
∂~a
∂~a
∂~a
∂~a
∂~a
dφ
dr
d~a
∂φ
∂r
∂r
ˆ = ∂θ dθ = ∂θ ; φˆ = ∂φ
=
;
θ
=
.
= ∂~
a
∂~a
∂~a
∂~a
∂~a
dr
|
dθ|
|
dφ|
|
| ∂ra dr|
| ∂~
|
|
|
|
∂r
∂θ
∂θ
∂φ
∂φ
(15)
So all we have to do is apply equation 15 by taking the appropriate
derivatives of the ~a defined by equation 13. We should note that as
is typical, the length of a vector is the square root of its components
squared. As such, here begin the derivatives:
rˆ =
Solutions to Problem Set 2, Physics 370, Spring 2014
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For the r direction:
∂~a
= sin θ cos φˆ
x + sin θ sin φˆ
y + cos θˆ
z
∂r q
∂~a = sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ
∂r p
= sin2 θ + cos2 θ = 1
(16a)
(16b)
(16c)
For the θ direction:
∂~a
= r cos θ cos φˆ
x + r cos θ sin φˆ
y − r sin θˆ
z
∂θ q
∂~a = r 2 cos2 θ cos2 φ + r 2 cos2 θ sin2 φ + r 2 sin2 θ = r
∂θ (17a)
(17b)
For the φ direction:
∂~a
= −r sin θ sin φˆ
x + r sin θ cos φˆ
y
∂φ
q
∂~a = r 2 sin2 θ sin2 φ + r 2 sin2 θ cos2 φ = r sin θ
∂φ (18a)
(18b)
where in all the above cases I used the identity cos2 x + sin2 x = 1. Now
we solve for the unit vectors:
rˆ =
θˆ =
φˆ =
∂~a
∂r
∂~a
| ∂r |
∂~a
∂θ
∂~a
| ∂θ |
∂~a
∂φ
∂~a
| ∂φ
|
= sin θ cos φˆ
x + sin θ sin φˆ
y + cos θˆ
z
(19a)
= cos θ cos φˆ
x + cos θ sin φˆ
y − sin θˆ
z
(19b)
= − sin φˆ
x + cos φˆ
y
(19c)
At this stage, we can perform the recommended checks to see if every-
Solutions to Problem Set 2, Physics 370, Spring 2014
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thing is behaving as expected:
√
rˆ · rˆ = sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ = 1
√
θˆ · φˆ = − cos θ cos φ sin φ + cos θ cos φ sin φ = 0
xˆ
yˆ
zˆ rˆ × θˆ = sin θ cos φ sin θ sin φ cos θ cos θ cos φ cos θ sin φ − sin θ
= (− sin2 θ sin φ − cos2 θ sin φ)ˆ
x
2
2
+ (cos θ cos φ + sin θ cos φ)ˆ
y
+ (sin θ cos φ cos θ sin φ − sin θ sin φ cos θ cos φ)ˆ
z
= − sin φˆ
x + cos φˆ
y
√
= φˆ
(20a)
(20b)
(20c)
(20d)
(20e)
(20f)
As a final task, we are asked to solve for the inverse relations. To do
this, we’ll resort to algebra (unless you want to use this same derivatives
technique in reverse). First I’ll try to solve for xˆ by combining equations
19a and 19b since I see a way to eliminate the zˆ terms there.
sin θˆ
r = sin2 θ cos φˆ
x + sin2 θ sin φˆ
y + sin θ cos θˆ
z
cos θθˆ = cos2 θ cos φˆ
x + cos2 θ sin φˆ
y − cos θ sin θˆ
z
sin θˆ
r + cos θθˆ = cos φˆ
x + sin φˆ
y
(21a)
(21b)
(21c)
Notice that these expressions, multiplied by a sine or cosine, can be
combined with equation 19c to eliminate yˆ and solve for xˆ
ˆ = cos2 φˆ
cos φ(sin θˆ
r + cos θθ)
x + cos φ sin φˆ
y
2
sin φφˆ = − sin φˆ
x + sin φ cos φˆ
y
ˆ − sin φφˆ
xˆ = cos φ(sin θˆ
r + cos θθ)
(22a)
(22b)
(22c)
and yˆ
ˆ = sin φ cos φˆ
sin φ(sin θˆ
r + cos θθ)
x + sin2 φˆ
y
2
cos φφˆ = − cos φ sin φˆ
x + cos φˆ
y
ˆ + cos φφˆ
yˆ = sin φ(sin θˆ
r + cos θθ)
(23a)
(23b)
(23c)
Solutions to Problem Set 2, Physics 370, Spring 2014
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And finally, returning to equations 19a and 19b, I can eliminate terms
leaving just zˆ by
cos θˆ
r = cos θ sin θ cos φˆ
x + cos θ sin θ sin φˆ
y + cos2 θˆ
z
sin θθˆ = sin θ cos θ cos φˆ
x + sin θ cos θ sin φˆ
y − sin2 θˆ
z
zˆ = cos θˆ
r − sin θθˆ
(24a)
(24b)
(24c)
4. Griffiths Problem 1.39:
(a) Check the divergence theorem for the function ~v1 = r 2 rˆ, using as
your volume the sphere of radius R, centered at the origin.
(b) Do the same for ~v2 =
at Problem 1.16.)
1
rˆ.
r2
(If the answer surprises you, look back
(5 points possible) The divergence theorem states (from equation
1.56) that:
I
Z
~ · ~v dτ =
~v · d~a
(25)
∇
V olume
Surf ace
Using equation 1.71 which tells us that in spherical coordinates
~ · ~v = 1 ∂ r 2 vr + 1 ∂ (sin θvθ ) + 1 ∂vφ
∇
r 2 ∂r
r sin θ ∂θ
r sin θ ∂φ
(26)
therefore:
(a) Starting with
~ · ~v1 = ∇
~ r 2 rˆ
∇
1 ∂ 2 2
= 2
r r
r ∂r
1
= 2 4r 3
r
= 4r
(27a)
(27b)
(27c)
(27d)
10
Solutions to Problem Set 2, Physics 370, Spring 2014
and thus the volume integral in the divergence theorem becomes
Z
Z
~
∇ · ~v1 dτ =
4r(r 2 sin θdrdθdφ)
(28a)
V olume
V olume
Z R
Z π
Z 2π
3
=4
r dr
sin θdθ
dφ
(28b)
r=0
θ=0
φ=0
1
= 4( R4 )(2)(2π)
4
= 4πR4
(28c)
(28d)
For this spherical volume, the surface area vector is always radial,
with a value of d~a = r 2 sin θdθdφˆ
r, therefore the surface integral
of a sphere with r = R becomes:
I
I
~v1 · d~a =
R2 rˆ · R2 sin θdθdφˆ
r
(29a)
Surf ace
Surf ace
Z π
4
=R
sin θdθ
θ=0
Z
2π
dφ
(29b)
φ=0
= R4 (2)(2π)
= 4πR4
(29c)
(29d)
which we expect assuming the divergence theorem holds.
(b) (5 points possible) Trying this for the function ~v2 = r12 rˆ we find
that the divergence is:
1
~ · ~v1 = ∇
~
∇
rˆ
(30a)
r2
1 ∂
2 1
r 2
(30b)
= 2
r ∂r
r
=0
(30c)
R
~
So the volume integral V olume ∇ · ~v2 dτ = 0. Now let’s check
the surface integral in the divergence theorem (when r = R and
Solutions to Problem Set 2, Physics 370, Spring 2014
d~a = R2 sin θdθdφ):
I
I
~v2 · d~a =
1
rˆ · R2 sin θdθdφˆ
r
2
R
Surf ace
Z π
Z 2π
=
sin θdθ
dφ
Surf ace
θ=0
11
(31a)
(31b)
φ=0
= 4π
(31c)
Oh dear, should we be shocked that the divergence theorem isn’t
holding? Examining the solution to 1.16, we are reminded that
the divergence is zero everywhere except the origin, where the
value
vector
field ~v is undefined. Therefore the calculation
R of the
~ · ~v2 dτ is flawed. The correct answer, taking into
of V olume ∇
account the ~v (0, 0, 0) = ∞ condition, is 4π as indcated by the
divergence theorem (which recall is also the “fundamental theorem
for divergences”).
5. Griffiths Problem 1.44: Evaluate the following integrals:
R2
(a) −2 (2x + 3)δ(3x)dx
R2
(b) 0 (x3 + 3x + 2)δ(1 − x)dx
R1
(c) −1 9x2 δ(3x + 1)dx
Ra
(d) −∞ δ(x − b)dx
Given that the Dirac delta function is defined as:
0, =if x 6= a
δ(x − a) =
∞, =if x = a
(from equation 1.90) and therefore
Z ∞
f (x)δ(x − a)dx = f (a)
(32)
(33)
−∞
(from equation 1.92). Now, the key is as long as the range of integration
includes the Dirac delta function (in this case, long as x = a is in the
range of integration), then it takes on the value f (a). Let’s apply these
rules to the integrals we’ve been asked about:
Solutions to Problem Set 2, Physics 370, Spring 2014
12
(a) (2 points possible) As in all cases where I am multiplying the
variable in the Dirac delta function, I will need to properly deal
with the constant. So in this case, citing equation 1.94 which
1
δ(x), I can rewrite this integral as:
states δ(kx) = |k|
Z 2
Z 2
1
(2x + 3)δ(3x)dx =
(2x + 3) δ(x)dx
(34a)
3
−2
−2
1
(34b)
= (0 + 3)
3
=1
(34c)
(b) (3 points possible) The integral range 0 to 2 covers x = 1,
therefore
Z 2
(x3 + 3x + 2)δ(1 − x)dx = (13 + 3 + 2) = 6
(35)
0
(c) (3 points possible) As with part (a), we can rewrite the equation
as
Z 1
Z 1
1
2
2
dx
(36a)
9x δ (3x + 1) dx =
9x δ 3 x +
3
−1
−1
Z 1
1
21
=
9x δ x +
dx
(36b)
3
3
−1
(36c)
and so it looks like the delta function should be zero everywhere
except at x = − 31 (which is in the integration range), therefore:
Z 1
Z 1
1
1
2
(37a)
9x δ(3x + 1)dx =
9x2 δ(x + )dx
3
3
−1
−1
2
1
=3 −
(37b)
3
1
(37c)
=
3
Ra
(d) (2 points possible) To evaluate −∞ δ(x − b)dx all we need to
consider is that δ(x−b) = 0 everywhere except at x = b. Therefore
the integral equals 1 if the range of the integral includes b, that is
if a > b, and the integral equals 0 if a < b.
Solutions to Problem Set 2, Physics 370, Spring 2014
13
6. Griffiths Problem 1.53: Check the divergence theorem for the funcˆ using as your volume one
tion ~v = r 2 cos θˆ
r + r 2 cos φθˆ − r 2 cos θ sin φφ.
octant of the sphere of radius R (Figure 1.48, shown below). Make sure
you include the entire surface. [Answer: π4 R4 ]
(10 points possible) The divergence theorem (equation 1.56) states:
Z
I
~
∇ · ~v dτ =
~v · d~a.
(38)
V olume
Surf ace
To verify this theorem for the function ~v = r 2 cos θˆ
r + r 2 cos φθˆ −
ˆ we will first compute the left hand side, then the right
r 2 cos θ sin φφ,
hand side.
Since ~v is given in spherical coordinates, we will use equation (1.71), the
expression for divergence in spherical coordinates, and apply it here:
~ · ~v = 1 ∂ r 2 [r 2 cos θ]
∇
2
r ∂r
1 ∂
1 ∂
+
sin θ[r 2 cos φ] +
[−r 2 cos θ sin φ] (39a)
r sin θ ∂θ
r sin θ ∂φ
1
1
(r 2 cos θ cos φ) +
(−r 2 cos θ cos φ)
= 4r cos θ +
r sin θ
r sin θ
(39b)
= 4r cos θ
(39c)
Having determined the divergence, we can now work on the left hand
side of equation 1.56. Since we are only working on the first octant, θ
14
Solutions to Problem Set 2, Physics 370, Spring 2014
goes from 0 to π2 and φ goes from 0 to π2 , therefore the integral takes
the form:
Z
Z R Z π/2 Z π/2
~
∇ · ~v dτ =
4r cos θr 2 sin θdφdθdr
(40a)
V olume
r=0
θ=0
R
φ=0
Z π/2
π
3
r dr
cos θ sin θdθ
= ·4
2
r=0
θ=0
Z
π
1 4 1 π/2
= ·4· R ·
cos(2θ)dθ
2
4
2 θ=0
Z
πR4 1 π/2
=
cos(u)du
·
4
2 θ=0
πR4
π/2
=
· [− sin(u)]θ=0
8
πR4
π/2
· [sin(2θ)]θ=0
=−
8
π 4
= R
4
Z
(40b)
(40c)
(40d)
(40e)
(40f)
(40g)
where I used the half-angle identity sin(2θ) = 2 sin θ cos θ and then usubstitution of u = 2θ to manhandle the integral in θ. For the left hand
side, we have a surface integral that covers a surface with four parts:
the bottom, left, back, and curved portions. We will need to integrate
over all four of these surfaces and then sum the integrals. Several of
you only integrated over the curved surface, “got the right
answer,” and stopped. You need to integrate over the entire
closed surface, not just the curved side!
• Curved: For the curved surface, d~a = (rdθ)(r sin θdφ)ˆ
r (using
equations 1.66 and 1.67) where r = R. Thus:
Z
Surf ace
~v · d~a =
Z
π/2
θ=0
=R
4
Z
Z
π 4
R
4
(R2 cos θ)(Rdθ)(R sin θdφ)
φ=0
π/2
θ=0
=
π/2
cos θ sin θdθ
Z
(41a)
π/2
dφ
(41b)
φ=0
(41c)
Solutions to Problem Set 2, Physics 370, Spring 2014
15
• Bottom: For the bottom surface, d~a = (r sin θdφ)(dr)θˆ (using
equations 1.65 and 1.67) where θ = π2 . Therefore:
Z
Surf ace
R
π/2
π
(r 2 cos φ)(r sin )dφdr
2
r=0 φ=0
Z R
Z π/2
=
r 3 dr
(cos φ)dφ
Z
~v · d~a =
Z
r=0
(42a)
(42b)
φ=0
1
= R4
4
(42c)
• Left: For the left surface, d~a = −(rdθ)(dr)φˆ (using equations 1.65
and 1.66) where φ = 0. Therefore:
Z
Surf ace
~v · d~a = −
Z
R
r=0
Z
π/2
(r 2 cos θ sin(0))(r)dθdr
(43a)
θ=0
=0
(43b)
• Back: For the back surface, d~a = (rdθ)(dr)φˆ (using equations 1.65
and 1.66) where φ = π2 . Thus:
Z
Surf ace
~v · d~a =
Z
R
r=0
=−
Z
Z
π/2
(−r 2 cos θ sin(π/2))(r)dθdr
(44a)
θ=0
R
3
r dr
r=0
1
= − R4
4
Z
π/2
cos θdθ
(44b)
θ=0
(44c)
So summing up results of these four surface integrals we find the surface
integral over the entire closed surface is:
I
π
(45)
~v · d~a = R4
4
Surf ace
which is the same result as we saw for the volume integral of the divergence in equation 40g