Jim Lambers
MAT 280
Summer Semester 2012-13
Practice Final Exam Solution
1. Evaluate the line integral
Z
where C is given by r(t) =
√
C
4
2
ht , t , t3 i,
xy dx + ey dy + xz dz,
0 ≤ t ≤ 1.
Solution From r0 (t) = h4t3 , 2t, 3t2 i, we obtain
Z
Z 1p
dx
dy
dz
√
y
xy dx + e dy + xz dz =
x(t)y(t)
+ ey(t)
+ x(t)z(z) dt
dt
dt
dt
C
0
Z 1
2
=
t3 (4t3 ) + et (2t) + t7 (3t2 ) dt
0
Z 1
2
=
4t6 + 2tet + 3t9 dt
0
1
3t10 4t7
t2
+e +
=
7
10 0
3
4
+e+
− 1.
=
7
10
2. Show that the vector field
F(x, y, z) = hey , xey + ez , yez i
R
is conservative, and use this fact to evaluate C F· dr where C is the line segment from (0, 2, 0)
to (4, 0, 3).
Solution A vector field F = hP, Q, Ri is conservative if
Ry = Qz ,
Pz = Rx ,
Qx = Py .
From
Ry = ez = Qz ,
Pz = 0 = Rx ,
Qx = ey = Py ,
we find that F is in fact conservative. To evaluate the line integral efficiently, we need to find
a function f such that ∇f = F. To that end, we obtain
Z
f (x, y, z) = P dx = xey + g(y, z).
The requirement that fy = Q yields the equation
gy (y, z) = Q(x, y, z) − (xey )y = ez .
Solving the equation for gy yields
g(y, z) = yez + h(z).
The requirement that fz = R yields the equation
h0 (z) = R(x, y, z) − (xey + yez )z = 0.
It follows that h(z) = K where K is an arbitrary constant, and therefore
f (x, y, z) = xey + yez + K.
From the Fundamental Theorem of Line Integrals, we obtain
Z
Z
∇f · dr
F · dr =
C
C
= f (4, 0, 3) − f (0, 2, 0)
= (4e0 + 0e3 ) − (0e2 + 2e0 )
= 2.
3. Use Green’s Theorem to evaluate
Z
x2 y dx − xy 2 dy,
C
where C is the circle x2 + y 2 = 4 with counterclockwise orientation.
Solution Let D = { (x, y) | x2 + y 2 ≤ 4 } be the interior of C. Then, by Green’s Theorem,
Z
Z Z
Z Z
2
2
2
2
x y dx − xy dy =
(−xy )x − (x y)y dA =
−y 2 − x2 dA.
C
D
D
Converting to polar coordinates, we obtain
Z
Z 2π Z 2
2
2
x y dx − xy dy =
(−r2 )r dr dθ
C
0
0
Z 2π
Z 2
=
dθ
−r3 dr
0
0
2
= 2π −
4 0
= −8π.
r4
4. If f (x, y, z) and g(x, y, z) are twice differentiable functions, show that
∇2 (f g) = f ∇2 g + g∇2 f + 2∇f · ∇g,
where ∇2 f = ∇ · (∇f ).
Solution We have
∇2 (f g) = ∇ · (∇(f g))
= ∇ · h(f g)x , (f g)y , (f g)z i
= ∇ · hfx g + f gx , fy g + f gy , fz g + f gz i
= (fx g + f gx )x + (fy g + f gy )y + (fz g + f gz )z
= fxx g + 2fx gx + f gxx + fyy g + 2fy gy + f gyy + fzz g + 2fz gz + f gzz
= f (gxx + gyy + gzz ) + g(fxx + fyy + fzz ) + 2(fx gx + fy gy + fz gz )
= f ∇ · (∇g) + g∇ · (∇g) + 2∇f · ∇g
= f ∇2 g + g∇2 f + 2∇f · ∇g.
5. Evaluate the surface integral
Z Z
(x2 z + y 2 z) dS,
S
where S is the part of the plane z = 4 + x + y that lies inside the cylinder x2 + y 2 = 4.
Solution We use the parametric equations
x = u cos v,
y = u sin v,
0 ≤ u ≤ 2,
z = 4 + u cos v + u sin v,
0 ≤ v ≤ 2π,
for which
ru = hcos v, sin v, cos v + sin vi,
rv = h−u sin v, u cos v, −u sin v + u cos vi,
√
ru × rv = h−u, −u, ui, kru × rv k = 3|u|.
This yields
Z Z
Z
2
2
(x z + y z) dS =
√
[(u cos v)2 + (u sin v)2 ](4 + u cos v + u sin v) 3u du dv
0
0
√ Z 2π Z 2 3
=
3
4u + u4 cos v + u4 sin v du dv
S
2π
Z
2
0
√ Z
=
3
0
2π
0
√ Z
=
3
2
u5
u + (cos v + sin v) dv
5
0
4
2π
32
(cos v + sin v) dv
5
0
√
32
= 16 3(2π) +
(sin v − cos v)|2π
0
5
√
= 32 3π.
16 +
6. Evaluate the surface integral
Z Z
F · dS,
S
where
F(x, y, z) = hxz, −2y, 3xi
and S is the sphere x2 + y 2 + z 2 = 4 with outward orientation.
Solution We use spherical coordinates
x = 2 sin φ cos θ,
y = 2 sin φ sin θ,
z = 2 cos φ,
0 ≤ θ ≤ 2π,
0 ≤ φ ≤ π,
which yields
rφ = h2 cos φ cos θ, 2 cos φ sin θ, −2 sin φi,
rθ = h−2 sin φ sin θ, 2 sin φ cos θ, 0i,
rφ × rθ = 4 sin φhsin φ cos θ, sin φ sin θ, cos φi,
which has outward orientation since 4 sin φ ≥ 0 for 0 ≤ φ ≤ π. We then have
Z 2π Z π
Z Z
2 sin φh2 cos θ cos φ, −2 sin θ, 3 cos θi · 4 sin φhsin φ cos θ, sin φ sin θ, cos φi dφ dθ
F · dS =
0
0
S
Z 2π Z π
= 8
sin2 φ(2 cos2 θ sin φ cos φ − 2 sin2 θ sin φ + 3 cos θ cos φ) dφ dθ
0
0
Z 2π
Z π
Z 2π
Z π
= 16
cos2 θ dθ
sin3 φ cos φ dφ − 16
sin2 θ dθ
sin3 φ dφ +
0
0
0
0
Z π
Z 2π
sin2 φ cos φ dφ
cos θ dθ
24
0
0
Z
Z 2π
1 + cos 2θ dθ u3 du − (u = sin φ, du = cos φ dφ)
= 8
0
Z 2π
Z π
Z π
8
1 − cos 2θ dθ
(1 − cos2 φ) sin φ dφ + 24 sin θ|2π
sin2 φ cos φ dφ
0
0
0
0
π
Z 1
sin4 φ = 16π
+ 16π
(1 − v 2 ) dv (v = cos φ, dv = − sin φ dφ)
4 0
−1
π
cos3 φ = 16π cos φ −
3
0
2
= 16π −2 +
3
64π
.
= −
3
R
7. Use Stokes’ Theorem to evaluate C F· dr, where F(x, y, z) = hxy, yz, xzi and C is the triangle
with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1), oriented counterclockwise as viewed from above.
Hint: to obtain an equation for the surface enclosed by C, compute the equation of a plane
containing the vertices.
Solution Using the given vertices, it can be determined that S is contained within the plane
x + y + z = 1. We therefore describe S using the parametric equations
x = u,
y = v,
z = 1 − u − v,
0 ≤ u ≤ 1,
0 ≤ v ≤ 1 − u,
which yields
ru = h1, 0, −1i,
rv = h0, 1, −1i,
and the normal vector
ru × rv = h1, 1, 1i,
which is consistent with the counterclockwise orientation of C (that is, when traversing C
such that this normal vector, which points upward, is visible, then the region S is on the left.
Applying Stokes’ Theorem yields
Z
Z 1 Z 1−u
F · dr =
curl F(u, v) · (ru × rv ) dv du.
C
0
0
From
curl F = hRy − Qz , Pz − Rx , Qx − Py i
= h(xz)y − (yz)z , (xy)z − (xz)x , (yz)x − (xy)y i
= h−y, −z, −xi.
We then have
curl F · (ru × rv ) = h−y, −z, −xi · h1, 1, 1i = −(x + y + z),
and thus
curl F(u, v) · (ru × rv ) = −(u + v + 1 − u − v) = −1.
We conclude that
Z
1 Z 1−u
Z
F · dr = −
C
1 dv du = −A(D),
0
0
where D is the triangle { (u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u }. This triangle has base and height
1, which yields
Z
1
F · dr = − .
2
C
RR
8. Use the Divergence Theorem to evaluate the surface integral
S F · dS, where F(x, y, z) =
hx3 , y 3 , z 3 i and S is the surface of the solid E bounded by the cylinder x2 + y 2 = 1 and the
planes z = 0 and z = 2.
Solution Using cylindrical coordinates,
Z Z
Z
F · dS =
S
Z
=
Z
=
we obtain
Z Z
div F dV
Z ZE
[(x3 )x + (y 3 )y + (z 3 )z ] dV
Z ZE
3(x2 + y 2 + z 2 ) dV
E
Z 2π Z 1 Z 2
= 3
(r2 + z 2 )r dz dr dθ
0
0
0
2
z 3 3
dr dθ
= 3
r z+r
3 0
0
0
Z 2π
Z 1
8
= 3
dθ
2r3 + r dr
3
0
0
1
4
2
4r r
= 6π
+
2
3 Z
2π
Z
1
0
= 11π.
9. (Bonus) Let
f (x, y, z) = x2 y 3 z 4 ,
g(x, y, z) = x4 + y 3 + z 2 ,
and let
ω = f dx + g dy,
η = f dx dz + g dy dz
be a 1-form and a 2-form, respectively. Compute ω ∧ η and dω.
Solution We have
ω ∧ η = (f dx + g dy)(f dx dz + g dy dz)
= f 2 dx dx dz + f g dy dx dz + f g dx dy dz + g 2 dy dy dz
= 0 − f g dx dy dz + f g dx dy dz + 0
= 0,
and
dω = df dx + dg dy
= (fx dx + fy dy + fz dz) dx + (gx dx + gy dy + gz dz) dy
= fx dx dx + fy dy dx + fz dz dx + gx dx dy + gy dy dy + gz dz dy
= (gx − fy ) dx dy + fz dz dx − gz dy dz
= (4x3 − 3x2 y 2 z 4 ) dx dy + 4x2 y 3 z 3 dz dx − 2z dy dz.