Week 11. Set of problems
The problems marked with an asterix
∗
are optional.
A. The Divergence Theorem
~ in two ways, directly, and using the Di1. Compute the flux integral S F~ · dA
vergence Theorem. In each case, S is closed and oriented outward.
(a) F~ (~r) = ~r and S is the cube enclosing the volume 0 ≤ x ≤ 2, 0 ≤ y ≤ 2,
and 0 ≤ z ≤ 2.
(b) F~ = x2~i + 2y 2~j + 3z 2~k and S is the surface of the box with faces x = 1,
R
x = 2, y = 0, y = 1, z = 0, z = 1.
(c) F~ = y~j and S is a closed vertical cylinder of height 2, with its base a
circle of radius 1 on the xy-plane, centered at the origin.
Answers: (a) 24; (b) 8; (c) 2π.
2. Use the Divergence Theorem to calculate the flux of the vector field through
the surface.
(a) F~ = x2~i + (y − 2xy)~j + 10z~k through a sphere of radius 5 centered at the
origin, oriented outward.
(b) F~ = −z~i + x~k through the sphere of radius a centered at the origin,
oriented outward. Explain your answer geometrically.
(c) F~ = z~i + y~j + x~k out of a sphere of radius 3 centered at the origin.
(d) F~ = xy~i + yz~j + zx~k out of a sphere of radius 1 centered at the origin.
Answer: (a) 5500π/3; (b) 0; the vector field is flowing around the y-axis and
is always tangent to the sphere; (c) 36π; (d) 0.
3.∗ The region W lies between
the spheres x2 +y 2 +z 2 = 4 and x2 +y 2 +z 2 = 9 and
p
2
within the cone z = x + y 2 with z ≥ 0; its bondary is the closed surface, S,
3~
3~
3~
~
is oriented outward.
√ Find the flux of F = x i + y j + z k out of S.
Answer: 633(2 − 2)/5π = 232.98.
4. A cone has its tip at the point (0, 0, 5) and its base the disk D, x2 + y 2 ≤ 1,
in the xy-plane. The surface of the cone is the curved and slanted face, S,
oriented upward, and the flat base, D, oriented downward. The flux of the
constant vector field F~ = a~i + b~j + c~k is given by
Z
~ = 3.32.
F~ · dA
S
~ If so, give the answer. If not, explain what
Is it possible to calculate D F~ ·dA?
additional information you would need to
be able to make
this calculation.
R
R
~
~
~
~ = −3.32.
Answer: By the Divergence Theorem, D F · dA = − S F · dA
R
1
5. Calculate the flux of F~ through the cylinder S given by x2 +y 2 = 2, −3 ≤ z < 3
and its base oriented outward. The cylinder is open at the top. Hint. Close
the cylinder by adding the disk S1 , x2 + y 2 ≤ 2, z = 3, oriented outward and
apply the Divergence Theorem:
Z
~+
F~ · dA
S
Z
~=
F~ · dA
S1
Z
div F~ dV,
W
where W is the solid cylinder.
(a) F~ = z 2~i + x2~j + 5~k
(b) F~ = z~i + x~j + y~k
(c) F~ = x3~i + y 3~j + ~k
Answers: (a) −10π; (b) 0; (c) 34π.
R
~ is negative,
6. Suppose div F~ = x2 + y 2 + 3. Find a surface S such that S F~ · dA
or explain why no such surface exists.
Answer: Any closed surface, S, oriented inward, will work, because div F~ > 0.
7. Let F~ = ~r/k~rk3 . Using the fact that div F~ = 0 for ~r 6= 0, calculate the flux of
F~ out of a box of side a centered at the origin and with edges parallel to the
axes.
Answer: 4π.
B. The curl of the vector field
1. Decide whether the vector fields have a nonzero curl at the origin. The vector
field is shown in the xy-plane, it has no z-component and is independent of z.
Answers: 1. This vector field shows no rotation, and the circulation around
any closed curve appears to be zero, so we suspect a zero curl here.
2. This vector field is definitely swirling, so we suspect a nonzero curl here.
3. The circulation around the boundary of a square in quadrant one is positive,
because the vectors on the top of the square are larger than those at the
bottom. Therefore there is a nonzero curl.
4. This vector field shows no rotation, and the circulation around any closed
curve appears to be zero, so the vector field has zero curl.
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2. Compute the curl of the vector fields
(a) F~ = (x2 − y 2 )~i + 2xy~j
(b) F~ = x2~i + y 3~j + z 4~k
2
(c) F~ = ex~i + cos y~j + ez ~k
(d) F~ = (−x + y)~i + (y + z)~j + (−z + x)~k
(e) F~ (~r) = ~r/|~rk.
Answers: (a) 4y~k; (b) ~0; (c) ~0; (d) −~i − ~j − ~k; (e) ~0.
3. Let F~ = (xy + z 2 )~i + x2~j + (xz − 2)~k.
(a) Find curl F~ at the point (0, −1, 0).
(b) Is this vector field irrotational?
Answer: (a) curl F~ = z~j + x~k; at (0, −1, 0) we have curl F~ = ~0; (b) no.
4. Show that curl (F~ + ~c) = curl F~ for a constant vector field ~c.
5. Use the geometric definition to find the curl of the vector field F~ (~r) = ~r. Check
your answer using the coordinate definition.
Answer. curl F~ = ~0.
6. For any constant vector field ~c, and any vector, F~ , show that
div (F~ × ~c) = ~c · curl F~ .
~ has curl G(0,
~ 0, 0) = 2~i − 3~j + 5~k. Estimate the
7. A smooth vector field G
circulation around a circle of radius 0.01 in each of the following planes:
(a) xy-plane, oriented counterclockwise when viewed form the positive z-axis;
(b) yz-plane, oriented counterclockwise when viewed form the positive x-axis;
(c) xz-plane, oriented counterclockwise when viewed form the positive y-axis.
Answers: (a) 0.0005π, (b) 0.0002π, (c) −0.0003π.
8. Three small circles, C1 , C2 , and C3 , each with radius 0.1 and centered at
the origin are in the xy-, yz-, and xz-planes, respectively. The circles are
oriented counterclockwise when viewed from the positive z-, x-, and y-axes,
respectively. A vector field, F~ , has circulation around C1 of 0.02π, around C2
of 0.5π, around C3 of 3π. Estimate curl F~ at the origin.
Answer: 50~i + 300~j + 2~k.
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C. Stokes’ Theorem
R
1. Can you use Stokes’ Theorem to compute the line integral C (2x~i+2y~j+2z~k·d~r
where C is the straight line from the point (1, 2, 3) to the point (4, 5, 6)? Why
or why not?
Answer: No, because the curve C over which the integral is taken is not a
closed curve, and so it is not the boundary of a surface.
2. At all points in 3-space curl F~ points in the direction of ~i − ~j − ~k. Let C be a
circle in the yz-plane, oriented clockwise when viewed from the positive x-axis.
Is the circulation of F~ around C positive, zero, or negative? Explain.
Answer: By Stokes’ theorem, the circulation of F~ around C is the flux of
curl F~ through the disk S in the yz-plane enclosed by C. By the right hand
rule, a positive normal vector to the disk points in the direction of the negative
x-axis, −~i. Thus,
~ = curl F~ · (−~i) dA = −dA,
curl F~ · dA
so the flux through S is negative. So, the circulation is negative.
3. Use Stokes’ theorem to find the circulation of the vector field around the given
paths.
(a) F~ = (z − 2y)~i + (3x − 4y)~j + (z + 3y)~k and C is the circle x2 + y 2 = 4,
z = 1, oriented counterclockwise when viewed from above.
(b) F~ = (2x − y)~i + (x + 4y)~j and C is the circle of radius 10, centred at
the origin in the xy-plane, oriented clockwise as viewed from the positive
z-axis.
Answers: (a) 20π; (b) −200π.
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