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Calculus III
Coursework 6
Calculus III: Coursework sheet 6 SOLUTIONS
This includes exercises of the final part of chapter 4.
1. Evaluate both sides of Stokes’s Theorem for F = xy j + yz k, for a closed path C in
the x = 0 plane, consisting of four straight path segments as follows: from the origin
to (0, a, 0); from there to (0, a, b); from there to (0, 0, b); and from there back to the
origin; and also for the planar surface bounded by the above path. Show that the two
sides are equal.
Answer: Evaluating curl, we get ∇ × F = z i + y k, and the surface integral should
be taken over the rectangle S in the (y, z) plane: the given path appears anticlockwise
looking from the +x direction, so we take n = +i. So the surface integral is
Z bZ a
Z
(zi + yk) · i dy dz
(∇ × F).dS =
0
0
S
Z bZ a
=
z dy dz = 21 ab2 .
0
0
The line integral along the curve has four straight-line parts, and we need a parametrisation of the straight line for each one, using the usual formula for a finite straight
line
r(t) = r1 + t(r2 − r1 )
0≤t≤1
If our line is parallel to one of x, y, z axes, it is convenient to use t in place of that
one variable, and set its range over the given limits instead of (0,1); this gets the same
result with slightly less arithmetic.
Now taking the four line segments: 1. From (0, 0, 0) to (0, a, 0): we have r = tj,
0 ≤ t ≤ a. Now dr/dt = j so F · dr = 0 .
2. From (0, a, 0) to (0, a, b): let r = aj + tk, 0 ≤ t ≤ b. Here F · dr = at and the
Rb
integral is 0 at dt = ab2 /2.
3. Next, r = (a − t)j + bk, 0 ≤ t ≤ a. Here dr = −j and F · dr = 0.
4. For the last segment, r = (b − t)k, 0 ≤ t ≤ b. Now dr/dt = −k, and F = 0 since
y = 0. So F · dr = 0 again.
The total of these four line integrals is ab2 /2, which agrees with the surface integral
above.
2. For each of the following fields F, evaluate ∇ × F, and either find the general scalar
field V satisfying ∇V = F everywhere, or show that no such V exists:
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Calculus III
Coursework 6
(a) F = z i + x k
(b) F = yz j + zx k
(c) F = y i + (x + z)j + (y + 1)k
(d) F = 2y i + y 2 j + z k .
Answers: For (a) we have ∇×F = 0, so a V will exist. We can either directly observe
that V = xz + const does satisfy ∇V = F; or for a general procedure we can define
Z
V (X, Y, Z) =
F · dr
C
along any path from a fixed point (e.g. the origin) to (X, Y, Z). The simplest such
path is a straight line, which is parametrised by r = (Xt, Y t, Zt) , 0 ≤ t ≤ 1. Hence
dr/dt = (X, Y, Z), and F.dr = [(Zt)X + (Xt)Z]dt = 2XZ t dt. Integrating, we have
V (X, Y, Z) = XZ.
Remember, the use of capital X, Y, Z is just to make clear that the endpoint is fixed,
while x, y, z vary with t along the line integral. Once we’ve evaluated the line integral,
clearly the result only involves X, Y, Z. That answer is valid for any point (X, Y, Z),
so we can replace X, Y, Z by lower-case again, so V (x, y, z) = xz. We can also add an
arbitrary constant to V , which is equivalent to changing our choice of start-point for
the line integral.
(b) We find ∇ × F = −y i − z j; this is non-zero, so no V is possible.
(c) We find ∇ × F = 0; then by a similar line integral method to (a), we find V =
xy + yz + z [+constant]. (Hint: it’s a good idea to check that taking ∇V does recover
the F we started with).
(d) Here ∇ × F = −2k, so no V is possible.
3. The vector field F is given by F = −y i+z j−x k. Verify Stokes’s theorem by evaluating
the following:
R
(a) The surface integral S (∇ × F) · n dS where S is the circular disk x2 + y 2 ≤ a2 ,
z = 0, and n points in the +z direction.
[3]
R
(b) The line integral C F · dr around the circle C bounding S above, taken counterclockwise.
[3]
R
(c) The surface integral S (∇ × F) · dS where S is the surface of the hemisphere
x2 + y 2 + z 2 = a2 , z ≥ 0, and the normal is taken in the outward direction. (Hint:
use a parametrisation of the hemisphere as in Coursework 5 Q3).
[4]
Answers:
(a) We find ∇ × F = −i + j + k.RFor the given surface, n = +k so (∇ × F) · n = +1.
Then the surface integral is just S 1 dS = the area of the circle, i.e. πa2 .
(b) As usual for a line integral, we need a 1-parameter representation of the given
circle. A standard choice is r = a cos t i + a sin t j (which does go counterclockwise seen
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Calculus III
Coursework 6
from +z); this gives dr/dt = −a sin t i + a cos t j, and F(r) = −a sin ti + 0j − a cos tk.
So the line integral is
Z
Z 2π
a2 sin2 t dt
F · dr =
0
which evaluates to πa2 .
(c) As per the hint, we want a 2-parameter equation for the surface of our hemisphere.
The natural one is spherical polar coordinates with r = a fixed, and θ, φ as the two
parameters, so
r = (a sin θ cos φ, a sin θ sin φ, a cos θ) . Since we want the hemisphere with z ≥ 0 i.e.
cos θ > 0, the ranges will be 0 ≤ θ ≤ π/2 (not π), 0 ≤ φ ≤ 2π. (Note that other
hemispheres with x ≥ 0, y ≥ 0 etc can be obtained by changing the limits on φ).
Using methods as in the Hint or Chapter 5.3 of the notes, we get our infinitesimal
surface element dS = a sin θ r dθdφ,
(Note, this can also be written a2 sin θ n dθdφ where n is the unit vector r/a, which is
obviously normal to the hemisphere ).
Now we evaluate the dot product, which is
(∇ × F) · dS = [a2 sin2 θ(sin φ − cos φ) + a2 cos θ sin θ] dθdφ
(remember, this is a dot product of two vectors, so the result must be a scalar ).
Finally we need to integrate the above with limits for the given hemisphere, giving
Z
2
2π
Z
π/2
Z
[sin2 θ(sin φ − cos φ) + cos θ sin θ] dθdφ
(∇ × F) · dS = a
S
0
0
This is easier than it looks; it helps to change the order of integration (dφ as the inner
integral), then the sin φ integrates to [− cos φ]2π
0 = 0, and the cos φ integrates to zero
2
likewise. So both terms with sin θ give zero, without actually integrating the sin2 θ.
For the last term, we use the double-angle result sin θ cos θ = 21 sin 2θ, so we get
Z
π/2
Z
2π
1
sin 2θ dφ dθ
2
0
0
π/2
−1
2
= a (2π)
cos 2θ
4
0
−1
= 2πa2
[−1 − 1]
4
= πa2
Ans = a
2
(note in the first line, the integrand doesn’t contain φ, so the dφ integration just
becomes a multiplier of 2π).
So all these results are equal, as they must be by Stokes’s theorem (which works for
any surface whose boundary is the given curve).
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Calculus III
Coursework 6
Comment: these integrals over (part of) a sphere look somewhat daunting at first,
but once you get used to the 2π’s and π’s and learn to spot things which integrate
to zero or 2π, they are not as hard as they seem. If you’re doing Applied Maths or
Physics courses later on, these sort of integrals will turn up quite often.
Note if we took the downward normal in (a), we’d have got a minus sign; then the
surface integral over the closed surface made by -(a) + (c) is zero, as expected from
div(curl F) = 0 and the Divergence theorem.
4. Show that the following are all zero:
H
(a) r · dr over an arbitrary closed path;
H
(b) dS over an arbitrary closed surface [Hint: Try taking the k component.];
H
(c) r × dS over an arbitrary closed surface;
where in each case the arbitrary curve or surface may be assumed to be sufficiently
smooth for the applicability of any theorems you use.
Answer:
(a) Using Stokes’s theorem,
I
Z
r · dr =
(∇ × r) · dS = 0
since ∇ × r = 0.
(b) This expression is a vector; we can get the k component by dotting it with k:
I
I
Z
k. dS = k.dS = (∇ · k) dV = 0
where we have taken the k inside the integral, then used the divergence theorem.
The same holds for the i and j components, so the original (vector) expression is
zero.
(c) Again consider the k component:
I
I
Z
k · r × dS = (k × r) · dS = ∇ · (k × r) dV ,
using the property of scalar triple product, then the divergence theorem. The
divergence evaluates to zero by product rules (Chapter 3 of notes), since curl r
and curl k are both zero, hence the integral is zero. Finally repeat for the i and
j components as above.
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