Math 235 - Review for Exam 3
Z
(8x + 36xy)ds, where c(t) = (t, t2 , t3 ) on the interval 0 ≤ t ≤ 1.
1. Compute
C
Z 1
Z
Soln:
(8x + 36xy)ds =
C
3
(8t + 36t )
0
p
3
2
2
2
4 3 1
1 + 4t2 + 9t4 dt = (1 + 4t + 9t ) 2 |0 = ((14) 2 − 1).
3
3
2. If a wire with linear density ρ(x, y) lies along a plane curve C, its ‘moment of inertia’ about the x-axis
and y-axis are defined by
Z
Z
Ix :=
y 2 ρ(x, y)ds, Iy :=
x2 ρ(x, y)ds.
C
C
Find the moments of inertia of the right half of the circle x2 + y 2 = 16 if ρ ≡ 1.
Solution: The curve is parameterized by C(t) = (4 cos t, 4 sin t), −π/2 ≤ t ≤ π/2. We compute Ix as
Z π/2
Ix =
2
Z π/2
0
(4 sin t) kC (t)kdt =
Z π/2
2
16 sin t 4 dt = 64
−π/2
−π/2
2
sin t dt = 32π.
−π/2
The other integral is computed in a similar way.
3. If g is continuously differentiable and α, β are constants, compute
Z
αg(x2 + y 2 )dx + βg(x2 + y 2 )dy,
C
where C is the circle of radius 2 (about the origin) rotating counterclockwise.
Solution: Parameterize the circle as C(t) = (2 cos t, 2 sin t), 0 ≤ t ≤ 2π. We then get
Z
2
2
2
2
αg(x + y )dx + βg(x + y )dy
C
Z 2π
=
(αg(4)
0
d
dt
2 cos t + βg(4)
d
dt
2 sin t)dt
Z 2π
=
(−2αg(4) sin t + 2βg(4) cos t)dt
0
=
0
You can also use Green’s theorem to get the same answer.
Z
4. Compute
F · dC, where F (x, y, z) = (3x2 y 2 z, 2x3 yz, x3 y 2 ) and C is a curve from (3, 2, 1) to (1, 2, 3)
C
(hint: there is an easy way to do this problem).
Soln: If you recognize that F = ∇f for f (x, y, z) = x3 y 2 z, then
Z
F · dC = f (1, 2, 3) − f (3, 2, 1) = 12 − 108 = −96.
C
5. Compute the line integral of
−
→
F (x, y) = (2x sin y, x2 cos y − 3y 2 )
along the straight line from (−1, 0) to (5, 1).
→
−
→
−
Solution: Evaluating the line integral directly can be cumbersome. To make things easier, check to see if F is conservative, that is to say F = ∇φ
for some scalar function φ. We need to find (if possible) a function φ such that
→
−
2
2
∇φ = (φx , φy ) = F = (2x sin y, x cos y − 3y ).
Comparing the first components, we get the equation φx = 2x sin y which means that φ = x2 sin y + h(y). Using this and comparing the second
components, we get φy = x2 cos y − 3y 2 = x2 cos y + h0 (y) and so h(y) = −y 3 . Thus φ = x2 sin y − y 3 (plus a constant which is not really
important here). By the ‘fundamental theorem of calculus’ (for line integrals) we get
Z (5,1)
Z (5,1)
→
−
F · dC =
∇φ · dC = φ(5, 1) − φ(−1, 0) = 25 sin 1 − 1.
(−1,0)
(−1,0)
1
2
6. (a) If ZC is a straight line segment from (a1 , b1 ) to (a2 , b2 ), compute (in terms of the a’s and b’s) the
quantity
−ydx + xdy.
C
Z (b) If C is now a positively oriented closed curve enclosing a domain D, what (in words) does the quantity
−ydx + xdy represent.
C
(c) Use (a) and (b) to compute the area of the pentagon with vertices (0, 0), (2, 1), (1, 3), (0, 2) and
(−1, 1).
Solution: (a) The line from (a1 , b1 ) to (a2 , b2 ) is parameterized by
C(t) = (a1 , b1 ) + t(a2 − a1 , b2 − b1 ) = a1 + t(a2 − a1 ), b1 + t(b2 − b1 ) , 0 ≤ t ≤ 1.
The quantity −ydx + xdy can be computed as
n
o
−[b1 + t(b2 − b1 )](a2 − a1 ) + [a1 + t(a2 − a1 )](b2 − b1 ) dt,
which after simplifying equals
n
a1 b2 − b1 a2
o
dt.
Use the above to compute the line integral
Z 1
Z
−ydx + xdy =
C
(a1 b2 − b1 a2 )dt = a1 b2 − b1 a2 .
0
A cautionary note: In the past, students have attempted to use Green’s theorem to compute the above line integral. This is not valid since using
Green’s theorem requires that line integral be taken over a closed loop (with the proper orientation). A line segment does not quality as a closed
loop.
(b) By Green’s theorem (which is valid since we have a closed positively oriented loop)
ZZ
Z
d
d
−ydx + xdy =
x−
(−y) dAx,y ,
dx
dy
C
D
where D is the area enclosed by the curve C. A computation shows this integral is equal to
ZZ
2 · dA = 2 · (area enclosed by C).
D
7. Let F be the vector field
(x, y, z)
,
r6
p
where r = x2 + y 2 + z 2 . Compute the work it takes to move along the straight line from (1, 1, 1) to (2, 2, 2).
F (x, y, z) = −4
Solution: The work is equal to
Z
F · dC,
C
where C is the line from (1, 1, 1) to (2, 2, 2). Evaluating this line integral directly can be complicated. To make things easier, we will use the fact
that F is a conservative vector field, that is F = ∇φ for some scalar function φ. To find φ note that the equation ∇φ = F says that
φx =
−4x
(x2 + y 2 + z 2 )3
and so φ = (x2 + y 2 + z 2 )−2 . Thus, by the ‘fundamental theorem of calculus’ for line integrals,
Z
Z
1
1
−
.
F · dC =
∇φ · dC = φ(2, 2, 2) − φ(1, 1, 1) =
122
32
C
C
8. Let C(t) = (cos t, sin t, t), 0 ≤ t ≤ 2π and F (x, y, z) = (x2 y 3 + y, x3 y 2 + x, z). Calculate the line integral
Z
F · dC.
C
Solution: Calculating the integral directly will be a prohibitive calculation. So try to show that F = ∇φ and then use the fundamental theorem
of calculus. To find φ we note that
2 3
3 2
∇φ = (φx , φy , φz ) = (x y + y, x y + x, z).
Equating the first coordinate entries, we get φx = x2 y 3 + y which means that φ = x3 y 3 /3 + xy + h(y, z). Equating second coordinate entries we
get φy = x3 y 2 + x + hy = x3 y 2 + x which means that hy = 0. Thus, so far, φ = x3 y 3 /3 + xy + h(z). Equating third coordinate entries, we get
φz = h0 (z) = z and so h(z) = z 2 /2 and so φ = x3 y 3 /3 + xy + z 2 /2.
To evaluate the line integral, we get (using the fundamental theorem of calculus for line integrals)
Z
Z
2
F · dC =
∇φ · dC = φ(C(2π)) − φ(C(0)) = φ(1, 0, 2π) − φ(1, 0, 0) = 2π .
C
C
3
9. Let r =
p
er
(x, y, z). Compute
r
x2 + y 2 + z 2 and F (x, y, z) =
Z
F · dC, where C is the straight line from
C
(2, 0, 0) to (0, 0, 1)
q
Solution: The vector field F is conservative with potential function φ(x, y, z) = e
x2 +y 2 +z 2
Z (0,0,1)
, i.e., ∇φ = F . Thus
2
∇φ · dC = φ(0, 1, 0) − φ(2, 0, 0) = e − e .
(2,0,0)
10. Compute
Z
(x2 y 3 + y, x3 y 2 + x, z) · dC,
C
where C is the top portion of the circle of radius 2 in the xy plane, centered at the origin, going counterclockwise.
Solution: From a previous problem, (x2 y 3 + y, x3 y 2 + x, z) = ∇φ, where φ = x3 y 3 /3 + xy + z 2 /2. Thus
Z (−2,0,0)
∇φ · dC = φ(−2, 0, 0) − φ(2, 0, 0) = 0.
(2,0,0)
11. Let F be the vector field
2
2
F (x, y) = (2xex sin y, ex cos y).
Find the work it takes to move along a straight line from (0, 0) to (1, π/2).
Solution: The work is equal to the line integral
Z
F · dC,
C
where C is the line from (0, 0) to (1, π/2). To calculate this line integral, we will find a scalar function φ such that F = ∇φ. Looking at the first
coordinate entries of the equation
x2
∇φ = (φx , φy ) = (2xe
2
sin y, e
x2
cos y)
2
2
2
we see that φx = 2xex sin y which means φ = ex sin y + h(y). Equating the second coordinate entries yields φy = ex cos y + h0 (y) = ex cos y
x2
0
which means that h (y) = 0. Thus we can take φ = e
sin y. Using the fundamental theorem of calculus for line integrals, we get
Z (1,π/2)
Z
F · dC =
C
12. Let F (x, y, z) =
1
r 3 (x, y, z),
∇φ · dC = φ(1, φ/2) − φ(0, 0) = e.
(0,0)
where r =
p
x2 + y 2 + z 2 . Is F conservative? Why?
Solution: To show that F is conservative, we need to show there is a scalar function φ such that ∇φ = F . To this end,
∇φ = (φx , φy , φz ) = (
x
(x2 + y 2 + z 2 )3/2
,
y
(x2 + y 2 + z 2 )3/2
,
z
(x2 + y 2 + z 2 )3/2
).
Equating the first coordinate entries in the above equation gives us φx = x(x2 + y 2 + z 2 )−3/2 which says φ = −(x2 + y 2 + z 2 )−1/2 . You can
check this potential function φ works for the other entries. Thus F is conservative with potential function φ = −(x2 + y 2 + z 2 )−1/2 .
13. Compute
Z
F · dC
where F (x, y, z) = (yz, xz, xy) and C(t) = (2 cos t, 3 sin t, 4), 0 ≤ t ≤ 2π.
4
Solution: One can show that F = ∇φ for some potential function φ. Indeed
∇φ = (φx , φy , φz ) = (yz, xz, xy)
and so equating the first components, we get φx = yz. This means that φ = xyz + h(y, z). Proceeding as in the previous problems, one can check
that φ can be taken to be φ = xzy. Thus, by the fundamental theorem of calculus for line integrals,
Z
∇φ · dC = φ(C(2π)) − φ(C(0)) = 0
C
since C(2π) = C(0).
14. Let C be any path joining any point on the sphere x2 + y 2Z + z 2 = a2 to any point on the sphere
→
−
→
→
→
v , where −
v = (x, y, z), what is
F · dC?
x2 + y 2 + z 2 = b2 . If F = 5k−
v k3 −
C
Solution: Notice that
2
2
2 3/2
F (x, y, z) = (5x(x + y + z )
2
2
2 3/2
, 5y(x + y + z )
2
2
2 3/2
, 5z(x + y + z )
)
and a computation shows that if φ(x, y, z) = (x2 + y 2 + z 2 )5/2 , then ∇φ = F . Thus if A = (a1 , a2 , a3 ) lies on the sphere of radius a (i.e.,
2
2
2
2
2
2
2
a2
1 + a2 + a3 = a ) and B = (b1 , b2 , b3 ) lies on the sphere of radius b (i.e., b1 + b2 + b3 = b ), then by the fundamental theorem of calculus for
line integrals, the line integral is
5
5
φ(B) − φ(A) = b − a .
15. Parameterize the following surfaces:
p
(a) z = x2 + y 2 , 1 ≤ z ≤ 4.
√
√
(b) x2 + y 2 + z 2 = 1, −1/ 2 ≤ z ≤ 1/ 2.
(c) x2 + z 2 = 4, 0 ≤ y ≤ 2.
Solution:
p
(a) X(u, v) = (u, v, u2 + v 2 ), 1 ≤ u2 + v 2 ≤ 16.
(b) X(θ, φ) = (sin φ cos θ, sin φ sin θ, cos φ), 0 ≤ θ ≤ 2π, π/4 ≤ φ ≤ 3π/4.
(c) X(θ, t) = (2 cos θ, t, 2 sin θ), 0 ≤ θ ≤ 2π, 0 ≤ t ≤ 2.
16. Compute the surface area of the solid bounded by
p
x2 + y 2 < z < 2.
Solution: The surface S is a cone parameterized by X : D → S
X(r, θ) = (r cos θ, r sin θ, r), D = {(r, θ) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
The surface are is
ZZ
1 · dS.
S
Here
dS = kXr × Xθ kdAr,θ = k(−r cos θ, −r sin θ, r)kdAr,θ =
√
2 r dAr,θ .
The surface integral becomes
ZZ
1 · dS =
S
ZZ √
Z 2π Z 2 √
√
2 r dAr,θ =
2 r dr dθ = 4π 2.
0
D
0
17. Compute the surface area of the surface bounded by z = x2 + y 2 , z < 4.
Solution: The surface S is parameterized by the function X : D → S given by
2
2
X(u, v) = (u, v, u + v ),
2
D = {(u, v) : u + v
2
< 4}.
Surface area is given by the surface integral of the constant function 1, that is
ZZ
1 · dS
S
which can be computed as
ZZ
1 · kXu × Xv kdAu,v .
D
A computation shows that
kXu × Xv k = k(−2u, −2v, 1)k =
p
1 + 4u2 + 4v 2
5
and so the surface area is then
ZZ p
1 + 4u2 + 4v 2 dAu,v
D
which, after switching to polar coordinates, becomes
Z 2π Z 2 p
√
π
(17 17 − 1)
r 1 + 4r 2 dr dθ =
6
0
0
18. Consider the surface parameterized by
X(u, v) = (u cos v, u sin v, u2 ), D = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π}
(a) Draw this surface (b) Compute its surface area.
Solution: (a) If x = u cos v, y = u sin v, z = u2 , notice that z = x2 + y 2 which is a paraboloid opening upwards with height 1.
(b) The surface are is
ZZ
ZZ
1 · dS =
kXu × Xv kdAu,v .
S
D
A computation yields
2
2
kXu × Xv k = k(−2u cos v, −2u sin v, u)k = u
Thus
Z 2π Z 1
ZZ
kXu × Xv kdAu,v =
u
0
D
0
p
1 + 4u2 .
p
π √
1 + 4u2 du dv =
(5 5 − 1).
6
19. If z = f (x, y), where (x, y) ∈ D. Write down a double integral, involving the function f , which represents
the surface area of the graph of f (x, y).
Solution: Parameterize the surface by X(u, v) = (u, v, f (u, v)), (u, v) ∈ D. The surface area is then
ZZ
kXu × Xu kdAu,v .
D
Note that
Xu × Xv = (1, 0, fu ) × (0, 1, fv ) = (−fu , −fv , 1)
and so
q
2 + f 2.
1 + fu
v
kXu × Xv k =
Put this altogether to get the surface area is then
ZZ q
2 + f 2 dA
1 + fu
u,v .
v
D
20. Evaluate the surface integral
ZZ
z 2 dS
S
where S is the portion of the cone z =
p
x2
y2
+
for which 1 ≤ x2 + y 2 ≤ 4.
Solution: First parameterize the surface S by X : D → S,
X(r, θ) = (r cos θ, r sin θ, r), D = {(r, θ) : 1 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
Note that
dS = kXr × Xθ kdAr,θ = k(−r cos θ, −r sin θ, r)kdAr,θ =
√
2 r dAr,θ .
The surface integral becomes
ZZ
2
ZZ
z dS =
S
r
D
2√
2rdArθ =
Z 2π Z 2 √
15 √
3
2π
2r drdθ =
2
0
1
ZZ
21. Evaluate
(x + y + z)dS across the rectangle with vertices (1, 1, 1), (2, 3, 4), (−1, 2, 1), and (0, 4, 4).
S
i
Soln: Parametrize the surface by (1, 1, 1) + u(1, 2, 3) + v(−2, 1, 0), 0 ≤ u ≤ 1, 0 ≤ v ≤ 1. In this case, the length of Tu × Tv is 1
−2
ZZ
√ Z 1Z 1
√
11 √
70.
which is 70. Thus,
(x + y + z)dS = 70
(3 + 6u − v)dudv =
2
S
0
0
j
2
1
k
3
0
,
6
ZZ
22. Evaluate
(x2 + y 2 )dS, where S is the surface of the cone z 2 = 3(x2 + y 2 ) bounded by z = 0 and
S
z = 3.
Solution: This cone is parameterized by
√
√
X(u, t) = (u cos t, u sin t, 3u), 0 ≤ t ≤ 2π, 0 ≤ u ≤ 3/ 3.
√
√
Note that Xu × Xt = (− 3u cos t, − 3u sin t, u) and so kXu × Xt k = 2u. Thus
ZZ
Z 2π Z 3/√3
2
2
3
(x + y )dS =
2u dudt = 9π.
0
S
0
23. Compute the following surface integral
ZZ
xdS,
S
where S is part of the plane x + y + z = 1 in the first octant
{(x, y, z) : x ≥ 0, y ≥ 0, z ≥ 0}.
Solution: The surface S can be parameterized by X : D → S by
X(u, v) = (u, v, 1 − u − v), D = {(u, v) : 0 ≤ v ≤ 1 − u, 0 ≤ u ≤ 1}.
√
3dAu,v and so
√
ZZ
ZZ √
√ Z 1 Z 1−u
3
xdS =
u 3 dv du = 3
u dv du =
.
6
0
0
A computation shows that dS = kXu × Xv kdAu,v = k(1, 1, 1)kdAu,v =
S
D
→ −
−
→
24. Let the velocity field of a fluid be described by F = x i +y j (measured in meters per second). Compute
how many cubic meters of fluid per second are crossing the surface of z = 4 − x2 − y 2 , z ≥ 0, in the direction
of increasing z.
Z 2π Z 2
ZZ
(x, y, 0) · (2x, 2y, 1)dxdy =
Soln:
S
2
Z 2π
2r rdrdθ =
0
0
8dθ = 16π.
0
25. Use Green’s Theorem to show that the area contained by an ellipse
x2
a2
+
y2
b2
= 1 is πab.
Soln: Parametrize the ellipse by x = a cos (θ) and y = b sin (θ). Green’s Theorem states that
ZZ
Z
Z 2π
1
1
xdy − ydx =
(a cos (θ)b cos (θ) − (b sin (θ))(−a sin (θ)))dθ = πab.
Area =
dxdy =
2 C
2 0
D
26. Compute
Z
(x, y) · dC,
C
where C is the triangle with vertices (1, 0), (0, 1), (−1, 0) in the clockwise direction.
Solution: By Green’s theorem (were D denotes the interior of the above triangle),
Z
ZZ
d
d
y−
x)dA = 0.
(x, y) · dC =
(
dx
dy
C
D
27. Evaluate the line integral
Z
F · dC
where F (x, y) = (−3y 3 + x3 , 3x3 − y 3 ) and C is the circle of raduis 1 centered at the origin going counterclockwise.
7
Solution: By Green’s theorem
Z
3
3
3
d
ZZ
3
(−3y + x , 3x − y ) · (dx, dy) =
C
x2 +y 2 <1
dx
3
3
(3x − y ) −
d
dy
3
3
(−3y + x )
dA.
Converting to polar coordinates yields
Z 2π Z 1
9π
2
.
r r dr dθ =
2
0
0
28. Compute
Z
F · dC
where F (x, y) = (yexy + ex , xexy + ey ) and C(t) is the path along the boundary of R = [0, 1] × [0, 1] going
counterclockwise.
Solution: By Green’s theorem,
Z
xy
(ye
x
+ e , xe
xy
ZZ
y
+ e ) · (dx, dy) =
C
R
d
dx
xy
(xe
y
+e )−
d
xy
dy
(ye
x
+e )
dA = 0.
29. Suppose that fxx + fyy = 0. Compute
Z 2π
fx (cos t, sin t) cos t + fy (cos t, sin t) sin t dt.
0
Hint: Try to write the above as a line integral.
Solution: Notice that the above integral can be written as
Z 2π
(fx (cos t, sin t)
0
d
dt
sin t − fy (cos t, sin t)
d
dt
cos t)dt.
Letting C(t) = (cos t, sin t), 0 ≤ t ≤ 2π, we note that the above is equal to
Z
fx dy − fy dx,
x2 +y 2 =1
where the integration is counterclockwise. By Green’s theorem, this integral is equal to
ZZ
(fxx + fyy )dA
x2 +y 2 ≤1
which is equal to zero, by the hypothesis of the problem.
Z Z
30. Compute the flux integral
−
→
−
→
F · n dS, where F = (3x, xz, z 2 ) and S is the surface of the solid
S
bounded by z = 4 − x2 − y 2 and the xy-plane.
Solution: You can compute the surface integral directly but it might take a while since the surface is in two pieces (the top part of the upsidedown paraboloid and the xy-plane). However, using the divergence theorem (Gauss’ theorem), the computation is much easier. By the divergence
theorem we get
ZZ
ZZZ
→
−
→
−
F · n dS =
∇ · F dV,
Ω
S
where Ω is the solid enclosed by S (which in this case is a upside-down paraboloid which intersects the xy plane in a circle of radius 2 and peaks
→
−
along the z-axis at z = 4). A computation shows that ∇ · F = 3 + 2z and so
ZZZ
ZZZ
→
−
∇ · F dV =
(3 + 2z)dV,
Ω
Ω
which after switching to cylindrical coordinates becomes
Z 2π Z 2 Z 4−r2
(3 + 2z)r dz dr dθ =
0
0
0
136π
3
.
8
31. Evaluate the surface integral
Z
F~ · ~ndS,
bd(W )
where F = (1, 1, z(x2 + y 2 )) and W is the solid cylinder x2 + y 2 ≤ 1, 0 ≤ z ≤ 1.
Solution: By Gauss’ theorem,
Z
~ ·~
F
ndS =
bd(W )
ZZZ
ZZZ
∇ · F dV =
W
2
2
(x + y )dV
W
which, after switching to polar coordinates, becomes
Z 2π Z 1 Z 1
π
3
r dz dr dθ =
.
2
0
0
0
32.
Z Z ZIf n is the outward unit normal for a smooth closed surface S which encloses a solid Ω, what is
div n dV ?
Ω
Solution: In this problem you need to work the divergence theorem backwards. Note, by the divergence theorem, that
ZZ
ZZZ
n · ndS =
∇ · ndV,
Ω
S
where S is the surface which bounds Ω. However n · n = knk2 = 1, since n is the outward unit normal (and hence has norm equal to 1). Thus
ZZZ
∇ · ndV = surface area of S.
Ω
ZZZ
33. If v is a tangent vector to a smooth closed surface S which bounds a solid Ω, what is
div v dV ?
Ω
Solution: Again, use the divergence theorem backwards to get
ZZZ
ZZ
∇ · vdV =
v · ndS = 0
Ω
S
since v is tangent to S and n is the normal to S.
ZZ
34. Evaluate
− −
→
−
→
F ·→
n dS, where F (x, y, z) = (x, y, z) and S is the boundary of the solid region enclosed
S
by z = 1 − x2 − y 2 and z = 0.
→
−
Solution: Use Gauss’ (divergence) theorem to evaluate the integral. Note that ∇ · F = 3 and so
ZZ
ZZZ
ZZZ
→
−
→
−
F · ndS =
∇ · F dV =
3dV,
0≤z≤1−x2 −y 2
S
0≤z≤1−x2 −y 2
which, after switching to cylindrical coordinates, becomes
Z 2π Z 1 Z 1−r2
3 r dz dr dθ =
0
−
35. Let →
v = (x, y, z) and
ZZ
(a) Compute
−
→
→
→
F =−
v /k−
v k3 .
− −
→
F ·→
n dS directly.
x2 +y 2 +z 2 =1
ZZZ
(b) Compute
x2 +y 2 +z 2 ≤1
(c) Comment.
(∇ · F )dV directly.
0
0
3π
2
.
9
→
−
−
−
−
−
−
Solution: (a) Since we are dealing with the unit sphere, then →
n =→
v /k→
vk=→
v . Also notice that F = →
v (on the sphere). Thus
ZZ
ZZ
ZZ
→
− →
2
−
→
−
F · n dS =
k v k dS =
1dS
x2 +y 2 +z 2 =1
x2 +y 2 +z 2
x2 +y 2 +z 2 =1
which is equal to the surface area of the unit sphere, i.e., 4π.
ZZZ
(b) A routine computation shows that ∇ · F = 0 and so
(∇ · F )dV = 0.
x2 +y 2 +z 2 ≤1
(c) Thought it seems like something is wring with Gauss’ theorem (which says (a) and (b) should be the same), nothing is wrong here since
→
−
the vector field F is not differentiable at the origin. Thus Gauss’ theorem cannot be applied.
36. Let S be the unit sphere centered at the origin (0, 0, 0) and
→
−
F (x, y, z) = (sin(xyz), sin(x2 y 2 z 2 ), 1).
Compute
→
−
∇ × F · v dS,
ZZ
S
where
1
v=p
x2
+ y2 + z2
(x, y, z).
Solution: The actual vector field is not important here. The important thing to notice is that since we are dealing with the unit sphere, the
vector v is the outward unit normal to S (check this!). By Gauss’ theorem (since we are working with a closed surface)
ZZ
ZZZ
→
−
→
−
∇ × F · v dS =
∇ · (∇ × F )dV = 0
x2 +y 2 +z 2 <1
S
→
−
→
−
since ∇ · (∇ × F ) = 0 for any vector field F .
37. A function g(x, y, z) is said to be harmonic if
gxx + gyy + gzz = 0.
If g is harmonic and S is the boundary surface of a region R, with outer normal n, compute
Z Z
∇g · n dS.
S
Solution: By Gauss’ theorem,
ZZ
ZZZ
∇g · n dS =
S
ZZZ
∇ · (∇g)dV =
R
(gxx + gyy + gzz )dV = 0
R
since we are assuming g is harmonic.
38. Verify the Divergence Theorem for the region bounded by the cone z =
and F (x, y, z) = (x, 3xz, y).
p
x2 + y 2 and the plane z = 2
p
Solution: Doing this directly: There are two parts to the surface; S1 : the plane z = 2, and S2 : the cone z =
x2 + y 2 . For the S1 , the
parameterization is X(u, v) = (u, v, 2), u2 + v 2 ≤ 4. The outward unit normal is clearly n = (0, 0, 1) and dS = kXu × Xv kdA = dA. Thus
ZZ
ZZ
ZZ
F · ndS =
(u, 6u, v) · (0, 0, 1)dA =
vdA.
S1
u2 +v 2 ≤4
u2 +v 2 ≤4
Switching to polar coordinates gives
ZZ
F · ndS = 0.
S1
For the surface S2 , the parameterization is X(u, v) = (u, v,
p
u2 + v 2 ), u2 + v 2 ≤ 4. Note that
u
−v
, p
, 1).
Xu × Xv = (− p
u2 + v 2
u2 + v 2
Unfortunately, this is neither unit not outward (it points up - not down!). Notice that kXu × Xv k =
1
u
v
n = √ (p
, p
, −1).
2
u2 + v 2
u2 + v 2
√
2 and so the outward unit normal is
10
A computation shows that
ZZ
u2
(p
+ 3uv − v)dA,
u2 + v 2
ZZ
F · ndS =
u2 +v 2 ≤4
S2
which, after switching to polar coordinates, is equal to π8/3.
By using Gauss’ theorem, we get
ZZ
ZZZ
ZZZ
Z 2π Z 2 Z 2
8
F · ndS =
∇ · F dV =
1dV =
rdzdrdθ = π.
3
0
0
r
S1 ∪S2
Ω
Ω
→
−
→
−
39. Is there a vector field G such that ∇ × G = (xy 2 , yz 2 , zx2 )?
Solution: No. Note that ∇ · (∇ × G) 6= 0. We know from elementary facts that the the ‘divergence of the curl’ is equal to zero.
−
→
40. For the vector field F = (3z, 4x, 2y) and the surface z = 4 − x2 − y 2 , z ≥ 0, evaluate
ZZ
→
−
(∇ × F ) · n dS
S
in two ways: (a) directly and (b) using Stokes’ theorem.
Solution: The surface S is parameterized by X : D → S2
2
2
2
X(u, v) = (u, v, 4 − u − v ), D = {(u, v) : u + v
2
≤ 4}.
A computation shows that
p
kXu × Xv k = k(2u, 2v, 1)k = 1 + 4u2 + 4v 2
and moreover the outward unit normal n is given by
Xu × Xv
n=
Thus
→
−
(∇ × F ) · ndS =
Z
(2u, 2v, 1)
.
= p
1 + 4u2 + 4v 2
kXu × Xv k
S
Z Z
(2, 3, 4) · (2u, 2v, 1)dAu,v ,
D
which, after switching to polar coordinates, is equal to 16π.
Stokes’ theorem says
ZZ
→
−
(∇ × F ) · ndS =
Z
F · dC,
C
S
where here C(t) = (2 cos t, 2 sin t, 0), 0 ≤ t ≤ 2π is the circle which bounds the surface S. The line integral can be computed as
Z
Z 2π
Z 2π
Z 2π
0
2
F · dC =
F (C(t)) · C (t)dt =
(0, 8 cos t, 4 sin t) · (−2 sin t, 2 cos t, 0)dt =
16 cos tdt = 16π,
C
0
0
0
which, thankfully, matches the answer we got before.
41. Let S be a region in the plane which is bounded by a smooth positively oriented curve C. Imagining S
as a surface and letting F (x, y, z) = (P (x, y), Q(x, y), 0), use Stokes’ theorem to derive Green’s theorem.
Solution: Stokes’ theorem says that
ZZ
Z
(∇ × F ) · ndS =
F · dC.
C
S
In this special case where S is a surface consisting of a region in the plane, n = (0, 0, 1) and dS = dA (regular area measure in the plane). A
computation shows that
∇ × F = ∇ × (P, Q, 0) = Qx − Py .
Putting this altogether we get
Z
ZZ
F · dC =
C
ZZ
(∇ × F ) · ndS =
S
(Qx − Py )dA
S
which is Green’s theorem.
ZZ
42. Evaluate
→ →
−
−
→
∇× F ·−
n dS, where F (x, y, z) = (x2 eyz , y 2 exz , z 2 exy ) and S is the hemisphere x2 + y 2 +
S
z 2 = 4, z > 0.
Solution: Without the use of Stokes’ theorem, the direct calculation of his integral would be prohibitive. By Stokes’ theorem,
ZZ
Z
→
−
→
− −
F · dC,
∇× F ·→
n dS =
C
S
where C(t) = (2 cos t, 2 sin t, 0) is the boundary (with the proper orientation) of the hemisphere. The line integral can be computed directly as
Z
Z 2π
Z 2π
→
−
→
−
0
2
2
F · dC =
F (C(t)) · C (t)dt =
(4 cos t, 4 sin t, 0) · (−2 sin t, 2 cos t, 0)dt,
C
0
which after a computation (using substitution) equals zero.
0
11
43. Consider the surface S, x2 + y 2 + 3z 2 = 1 and any differentiable vector field F . Compute the following
integral
ZZ
∇ × F · n dS,
S
where n is a outward unit normal to S in two distinct ways.
Solution: The first way is to use Gauss’ theorem.
ZZ
ZZZ
∇ × F · n dS =
S
∇ · (∇ × F )dV = 0
R
since ∇ · (∇ × F ) = 0 (the divergence of the curl is zero).
For the second way, cut the surface in half (say along the xy plane) into two pieces S+ (the top half) and S− (the bottom half) and let C
denote the curve in the plane which lies at their intersection. Use Stokes’ theorem on S+ to get that
ZZ
Z
∇ × F · n dS =
F · dC,
C
S+
where C circulates counterclockwise (important!). Use Stokes’ theorem on S− to get that
ZZ
Z
∇ × F · n dS =
F · dC,
C
S−
where C circulates clockwise (important!). Notice that the last line integral is the negative of the other (since they circulate in opposite directions)
and so their sum is zero.
44. Verify Stokes’ theorem for F~ = (z 2 , x2 , y 2 ) and the surface is the part of the plane x + y + z = 1 lying
in the first octant.
Solution: Doing this integral directly: First parameterize the triangle as
X(u, v) = (u, v, 1 − u − v), 0 ≤ v ≤ 1 − u, 0 ≤ u ≤ 1.
From the equation of the plane x + y + z = 1 we know that the vector (1, 1, 1) is normal to this triangle and is points out. The outward unit
√
normal is thus n = √1 (1, 1, 1). The dS = kXu × Xv k = 3dA and so
3
Z 1 Z 1−u
ZZ
(∇ × F ) · ndS =
0
S
0
√
1
(2v, 2(1 − u − v), 2u) · √ (1, 1, 1) 3dvdu = 1.
3
Using Stokes’ theorem: We need to compute
Z
F · dC,
C1 ∪C2 ∪C3
where C1 , C2 , C3 are the sides of the triangle. They are parameterized by
C1 (t) = (1 − t, t, 0),
Note that
C2 (t) = (0, 1 − t, t),
Z
C3 (t) = (t, 0, 1 − t),
Z
2
F · dC =
C1 ∪C2 ∪C3
2
0 ≤ t ≤ 1.
2
z dx + x dy + y dz
C1 ∪C2 ∪C3
and it is easy to check that
Z
2
2
Z 1
2
2
(1 − t) dt =
z dx + x dy + y dz =
Ci
0
1
3
and so the total line integral is equal to 1 (as it should be).
→
−
2
2
45.
Z Z Let F = (x + y, yz, x − z ) and S be the triangle defined by 2x + y + 2z = 2, x, y, z ≥ 0. Compute
(∇ × F ) · n dS.
S
Solution: By Stokes’ theorem, the above integral is equal to
Z
Z
2
2
F · dC =
(x + y)dx + yzdy + (x − z )dz,
C
C
where C is the triangle with vertices (1, 0, 0), (0, 2, 0), (0, 0, 1) - traversed in that order. The three boundary curves are
C1 (t) = (1 − t, 2t, 0),
C2 (t) = (0, 2 − 2t, t),
One can show that
Z
2
C3 (t) = (t, 0, 1 − t),
Z 1
2
(x + y)dx + yzdy + (x − z )dz =
C1
Z
2
((1 − t) + 2t)(−1)dt = −
0
2
2
Z 1
(x + y)dx + yzdy + (x − z )dz =
C2
Z
0 ≤ t ≤ 1.
4
3
2
((2 − 2t)t(−2) − t )dt = −1
0
2
2
Z 1
(x + y)dx + yzdy + (x − z )dz =
C3
The integral is the sum of the integrals above which is −7/6.
2
2
(t + t − (1 − t) (−1))dt =
0
7
6
12
→
−
→
−
→
−
46. If F (x, y, z) = A(x, y, z) i + B(x, y, z) j + C(x, y, z) k such that there is some point P such that
(∇ × F )(P ) 6= (0, 0, 0).
Does there exist a scalar function φ such that F = ∇φ?
Solution: No! If F = ∇φ, then ∇ × (∇φ) = (0, 0, 0).
47. State Stokes’ Theorem, and explain why both integrals in Stokes’ Theorem will be 0 if the function
F = ∇f for some f (there is a different reason for the two integrals).
ZZ
Z
Soln: Stokes’ Theorem states that under suitable conditions on the function and the surface,
(curlF ) · dS =
F · dS. If F = ∇f for some
S
∂S
ZZ
Z
f , then curlF = ∇ × ∇f = 0, so
(curlF ) · dS = 0. In the line integral, if F = ∇f for some f , then
F · dS = f (c(b)) − f (c(a)). Since ∂S
S
∂S
Z
is a simple closed curve, c(b) = c(a) and hence
F · dS = 0.
∂S