Solution to Set 5, Friday May 7th
Section 5.6: 15, 17. Section 5.7: 3, 5, 7, 16.
(1) (Section 5.5, Problem 2) Find a parametrization of the suface
x2 + y 2 = 9 between z = −1 and z = 2.
Solution: x = 3 cos t, y = 3 sin t and z = s with 0 ≤ t ≤ 2π
and −1 ≤ z ≤ 2.
(2) (Section 5.5, Problem 4) Find a parametrization of hyperboloid
of one sheet x2 + y 2 − z 2√= 1 between z = 0 and
√ z = 1.
2
Solution: z = s, x = 1 + s cos t and y = 1 + s2 sin t for
0 ≤ s ≤ 1 and 0 ≤ t ≤ 2π.
(3) (Section 5.5, Problem 6) Find parametrization of the plane x +
6y + z = 12 for in the first octant. We can choose x = t, y = s
and z = 12 − t − 6s with 0 ≤ t ≤ 12 − 6s, 0 ≤ s ≤ 2.
Alternately, we take obvious three points on the plane (12, 0, 0),
(0, 2, 0) and (0, 0, 12). Vectors a = (−12, 2, 0) and b = (0, −2, 12)
are in the plane of the the plane. So, parametric equation of
the plane
(x, y, z) = (12, 0, 0)+t(−12, 2, 0)+s(0, −2, 12) , with 0 ≤ t ≤ 1, 0 ≤ s ≤ 1
(4) (Section 5.5, Problem 20) Set up the double integral that represents the area of the parametrized surface
x = es+t , 2s − t, s + t , 0 ≤ s ≤ 1, 0 ≤ t ≤ ln 2
Solution: We have fs = (es+t , 2, 1) and ft = (es+t , −1, 1). We
s+t
note
√ that fs × ft = (3, 0, −3e ). So, we have kfs × ft k =
3 1 + e2s+2t . Therefore, area of the given surface is given by
Z ln 2 Z 1 √
σ=3
1 + e2s+2t dsdt
0
0
(5) (Section 5.5, Problem 24) Find the area of the surface M, where
M is the portion of the plane z = 8x + 5y + 2 that lies above
the region in the x − y-pane bounded by the parabola x = y 2
and the line x = 4.
Solution: Since surface M is given by z = 8x+5y+2 = h(x, y),
we have area of the surface
Z Z
Z Z q
√ Z Z
√
2
2
1 + hx + hy dA =
1 + 64 + 25dA = 90
dA
σ(M) =
R
R
R
Now R is the region in the x − y plane with left curve x = y 2
and right curve x = 4. Note that the curves intersect where
4 = y 2 , i.e. y = ±2. So, y ranges from -2 to 2in R. So, from
1
2
above
σ(M) =
√
90
Z
2
−2
Z
4
dx =
√
90
y2
√
= 2 90
Z
2
Z
2
−2
√
y3
(4 − y )dy = 2 90 4y −
3
2
0
4 − y 2 dy
2
√
= 32 10
0
RR p
x2 + y 2dσ,
(6) (Section 5.6, Problem 2) Evaluate the surface integral
M
where M is the cylindrical surface parametrized by x = (2 cos s, 2 sin s, t)
for 0 ≤ s ≤ 2π, −1 ≤ t ≤ 1.
Solution: We note that fs = (−2 sin s, 2 cos s, 0) and ft =
(0, 0, 1). So, fs ×ft = (2 cos s, 2 sin s, 0) and therefore, kfs ×ft k =
2. Therefore,
Z Z p
Z 1 Z 2π p
Z 1 Z 2π
2
2
2
2
x + y dσ =
4 cos s + 4 sin skfs ×ft kdsdt = 4
dsdt = 16π
M
0
−1
−1
RR
0
p
(7) (Section 5.6, Problem 4) Evaluate
z 2 x2 + y 2 dσ, where
M
M is parametrized by x = (θ cos θ, θ sin θ, w), 0 ≤ θ ≤ 6π,
−2 ≤ w ≤ 2.
Solution: Here the parameters (s, t) have been replaced by
(θ, w). We note that fθ = (cos θ − θ sin θ, sin θ + θ cos θ, 0), fw =
(0, 0, 1). So,
So,
fθ × fw = (sin θ + θ cos θ, − cos θ + θ sin θ, 0)
p
sin2 θ + 2θ sin θ cos θ + θ2 cos2 θ + cos2 θ − 2θ sin θ cos θ + θ2 sin2 θ
√
= 1 + θ2
√
p
We also have on M, x2 + y 2 = θ2 cos2 θ + θ2 sin2 θ = |θ| = θ
(since θ ≥ 0). So,
Z 2 Z 6π
Z Z
p
√
2
2
2
w 2 θ 1 + θ2 dθdw
z x + y dσ =
kfθ ×fw k =
=
1
3
M
2 −2
0
3 w=2
1
16 2 3/2
2 3/2
(1
+
36π
)
−
1
w
(1
+
36π
)
−
1
=
θ=0
w=−2
9
9
−2
RR
(8) (Section 5.6, Problem 6) Calculate
(y + z)dσ, where M is
M
the part
of the plane 2x + y + z = 3 that
lies above the region
R = (x, y)|0 ≤ x ≤ 31 , 0 ≤ y ≤ 1 − 3x in the x − y-plane.
Solution: Note in the representation given the surface is conveniently decribed by z = 3 − 2x − y = h(x, y). We have
Z
w 2 (1 + θ2 )
3/2 θ=6π
dw =
3
√
√
1 + h2x + h2y = 1 + 4 + 1 = 6. So, treating R as a y-simple
region in the x − y plane,
Z Z
Z Z
Z 1/3 Z 1−3x
q
√
(y+z)dσ =
(y+h(x, y)) 1 + h2x + h2y dA =
(3−2x) 6dydx
p
M
R
√ Z
= 6
1/3
0
0
√ Z
y=1−3x
dx = 6
[(3 − 2x)y]y=0
0
1/3
(3 − 2x)(1 − 3x)dx
0
1/3
√
√
11
11 2
2
3
= 6 1+
−
3 + 6x − 11x dx = 6 3x + 2x − x
2
27 18
0
0
RR
(9) (Section 5.6, Problem 13) Calculate
F · ndσ, where F =
M
(x, −z, −y) and M is the surface of revolution parametrized
and oriented by x = (s, (s3 − s) cos t, (s3 − s) sin t) ≡ f(s, t),
−1 ≤ s ≤ 1, 0 ≤ t ≤ 2π.
Solution: Note fs = (1, (3s2 − 1) cos t, (3s2 − 1) sin t) and ft =
(0, −(s3 − s) sin t, (s3 − s) cos t). We find
fs × ft = (3s2 − 1)(s3 − s), −(s3 − s) cos t, −(s3 − s) sin t
√ Z
= 6
1/3
2
So,
Z Z
=
Z
0
F · ndσ =
M
2π Z 1
Z
0
−1
=
2π
0
Z
1
−1
F · (fs × ft ) dsdt
(x, −z, −y)· (3s2 − 1)(s3 − s), −(s3 − s) cos t, −(s3 − s) sin t dsdt
−1
1 Z 2π
Z
Z
Z
1
−1
1
−1
s(3s2 − 1)(s3 − s) + 2(s3 − s)2 sin t cos t dtds =
t=2π
ts(3s2 − 1)(s3 − s) + (s3 − s)2 sin2 t t=0 ds
2
3
2πs(3s − 1)(s − s)ds = 4π
7
3
s=1
Z
0
1
s(3s2 − 1)(s3 − s)ds
16
1 2 1
=−
= 2π − + −
π
7 5 3
105
s=0
RR
(10) (Section 5.6, Problem 15) Calculate
F · ndσ, where F =
M
(−x, −y, −z) and M is the sphere parametrized and oriented
by (x, y, z) = (sin φ cos θ, sin φ sin θ, cos φ), where 0 ≤ θ ≤ 2π,
0 ≤ φ ≤ π.
Solution: fφ = (cos φ cos θ, cos φ sin θ, − sin φ), fθ = (− sin φ sin θ, sin φ cos θ, 0).
We have
fθ × fφ = − sin2 φ cos θ, − sin2 φ sin θ, − cos φ sin φ
s
2
s
= 2π s(s − s) − + s5 −
7
5
3
3
2
4
From the statement about orientation, this is to be taken as the
direction of n.
Therefore,
Z Z
Z
F · ndσ
M
2π Z π
(cos θ sin φ)(sin2 φ cos θ) + (sin φ sin θ)(sin2 φ sin θ) + cos φ(cos φ sin φ) dφdθ
0
0
Z 2π Z π
Z π
3
2
=
sin φ + sin φ cos φ dφdθ = 2π
sin φdφ = −2π [cos φ]π0 = 4π
=
0
0
0
RR
(11) (Section 5.6, Problem 17) Calculate
F · ndσ, where F =
M
2
3
(y , x , z) and M is the boundary of the cylindrical solid bounded
by x2 + y 2 = 4, z = 2 and z = 0. Use the outward normal.
Solution: Note that the closed boundary of the solid is composed of three separate pieces: the bottom flat cap
M1 : (x, y, z) : z = 0, x2 + y 2 ≤ 4 with outward normal n = (0, 0, −1)
top cap
M2 : (x, y, z) : z = 0, x2 + y 2 ≤ 4 with outward normal n = (0, 0, 1)
and the cylindrical boundary
M3 = {(x, y, z) : (x, y, z) = (2 cos θ, 2 sin θ, z), 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 2}
We note that on M3 , fθ = (−2 sin θ, 2 cos θ, 0) and fz = (0, 0, 1)
and so fθ × fz = (2 cos θ, 2 sin θ, 0), which is indeed directed
outwards from the cylinder.
RR
On M1 F · n = (y 2 , z 3 , z) · (0, 0, −1) = −z = 0. So,
F·
M1
2 3
ndσ = 0. On M2 , F · n = (y , z , z) · (0, 0, 1) = z = 2. So,
Z Z
Z Z
F · ndσ = 2
dA = 2(π22 ) = 8π
x2 +y 2 ≤4
M2
On M3 ,
Z Z
M3
F · ndσ =
=
Z 2Z
0
0
Z 2Z
0
2π 2π
0
F · (fθ × fz ) dθdz
(2 sin θ)2 (2 cos θ) + (z 3 )(2 sin θ) dθdz
Z 2n
θ=2π o
=
8 sin3 θ − 2z 2 cos θ θ=0 dz = 0
0
So, we have total contribution
RR
M
F · ndσ = 0 + 8π + 0 = 8π.
5
(12) (Section 5.7, Problem 3) Evaluate the integral
RR
R
√dxdy
x2 +y 2
by
converting into polar coordinates where R is the annular region
4 ≤ x2 + y 2 ≤ 9.
Solution: In polar coordinates (r, θ), region R is described by
= r as
2 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. We also have Jacobian ∂(x,y)
∂(r,θ)
worked out in class. So, we have
Z Z
R
dxdy
p
=
x2 + y 2
Z
2π
0
Z
2
3
rdrdθ
r
Z 2π Z
=
0
3
2
dr dθ = (3 − 2)(2π) = 2π
RR
(13) (Section 5.7, Problem 5) Evaluate the double integral
arctan xy dA
R
by converting into polar coordinates, where R is the first qudrant
between the circles 41 = x2 + y 2 and x2 + y 2 = 1 and the lines
y = √x3 and y = x.
y=x
y=x/sqrt(3)
R
r=1/2
r=1
√
Figure 1. Region R between y = x, y = x/ 3, x2 +
y 2 = 41 and x2 + y 2 = 1
Solution: Note integrand arctan xy = θ in polar coordinates.
Since y = x in polar coordinates implies tan θ = 1, we have
θ = π4 since we are in the first quadrant. Again y = √x3 implies
tan θ = √13 , so θ = π6 since we are in the first quadrant. Therefore, the description of given region R in polar coordinates is
R=
π
π
1
(r, θ) : ≤ r ≤ 1, ≤ θ ≤
2
6
4
6
So, recalling from class Jacobian
∂(x,y)
∂(r,θ)
= r, we have
Z π/4 Z 1
y
arctan dA =
θrdrdθ
1
x
R
π/6
2
Z π/4
Z π/4 2 r=1
15π 2
3
3 2 θ=π/4
r θ
dθ =
θ θ=π/6 =
θdθ =
=
2 r=1/2
8 π/6
16
(16)(144)
π/6
Z Z
(14) (Section 5.7, Problem√7) Using polar coordinates evaluate the
R 2 R 4−y2
2
2
double integral −2 √ 2 e−x −y dxdy.
4−y
−
y=2
2
x=−sqrt(4−y
)
x=sqrt(4−y2 )
R
y=−2
p
Figure
2.
Region
R
between
x
=
−
4 − y 2 and x =
p
4 − y 2 for −2 ≤ y ≤ 2. Clearly a circle of radius 2
centered at (0, 0)
Solution: From the order of integration, it is clear that the
above is the area integral
treated as an x-simple
p over region R p
2
region between x = − 4 − y amd x = 4 − y 2 for −2 ≤ x ≤
2. This is clearly the interior of the circle x2 + y 2 = 4. So, polar
representation of domain R is {(r, θ), 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.
So, it follows that
Z
2
−2
Z √4−y2
−
√
−x2 −y 2
e
dxdy =
4−y 2
=
Z
0
2π
Z
0
2
−r 2
e
Z Z
rdrdθ =
Z
2
e−r dA
R
2π
0
1 2
− e−r
2
r=2
r=0
dθ = π 1 − e−4
R R x2 sin(xy)
(15) (Section 5.7, Problem 16) Evaluate the double integral
dxdy
R
y
by making an appropriate change of variables where R is bounded
2
x2 = πy, y 2 = x2 and y 2 = x. We choose s = xy and
by x2 = πy
2
7
t=
y2
.
x
Then the domain R in the s − t variable becomes
π
1
(s, t) : ≤ s ≤ π, ≤ t ≤ 1
2
2
Also, note that s2 t = x3 and st2 = y 3 , so x = s2/3 t1/3 , y =
s1/3 t2/3 and we obtain
∂(x, y)
4 1
1
J=
= xs yt − ys xt = − =
∂(s, t)
9 9
3
2
Therefore, noting that the integrand xy sin(xy) = s sin(st), we
obtain
Z Z
Z Z
x2 sin(xy)
1 π 1
[s sin(st)] dtds
dxdy =
y
3 π/2 21
R
Z
Z π 1
1
s
1 π
t=1
ds
[− cos(st)]t=1/2 ds =
− cos s + cos
=
3 π/2
3
3
2
π/2
√
s=π
2
1
2
s
2
2 1
= − sin s + sin
= + − √ =1−
3
3
2 s= π
3 3 3 2
3
2
(16) (Section 5.8, Problem 3) By converting into cylindrical coordinates, evaluate
Z 0 Z √1−x2 Z 3−x2 −y2
zdzdydx
√
−1
− 1−x2
−1
z
2 2
z=3−x −y
S
R
y
x
Figure 3. Region S between z = −1 and z = 3−x2 −y 2
whose projection on the x−y plane is the unit semi-circle
centered at the origin for which −1 ≤ x ≤ 0
8
Z
0
−1
Solution: We note from the order of the integration that the
solid S that corresponds to the volume integral above is a zsimple with an upper surface at z = 3 − x2 − y 2 = h2 (x, y),
which is a paraboloid facing down and a lower-surface z =
−1 = h1 (x, y), which is a plane parallel to the x − y plane.
The projection of the solid region S in the x − y plane, call
it R is√ being treated as a y simple√ region with lower curve
y = − 1 − x2 and upper-curve y = 1 − x2 , each being on the
circle x2 + y 2 = 1. However, R is not the full circle since from
the outermost limit, −1 ≤ x ≤ 0. Hence R is only the semicircle
{(x, y) : x2 + y 2 ≤ 1, −1 ≤ x ≤0}. We note that in polar coor
dinates the description of R is (r, θ) : 0 ≤ r ≤ 1, π2 ≤ θ ≤ 32 π .
So, S is the region between the two surfaces z = h1 (x, y) and
z = h2 (x, y) whose projection on the x − y plane is the semicircle R. It is useful to sketch S as given above. Converting
into cylindrical coordinates and doing the w (or z) integration
first, we have
Z Z Z
Z √1−x2 Z 3−x2 −y2
zdzdydx =
zdV
√
− 1−x2
=
3π/2
Z
π/2
=
Z
0
3π/2
π/2
S
−1
1
Z
Z
1
Z
0
3−r 2
wrdwdrdθ =
−1
Z
3π/2
π/2
Z
1
0
1 2
w r
2
w=3−r2
drdθ
w=−1
r=1
1
1
π
r2
2 2
2 3
− (3 − r ) −
r(3 − r ) − r dr =
2
2
6
2 r=0
8 27 1
4
π
− +
= π
−
=
2
6
6
2
3
(17) (Section 5.8, Problem 5) Using cylindrical coordinates calculate
Z 1/√2 Z √1−x2 Z √x2 +y2
(x2 + y 2 )dzdydx
0
Z
0
x
x2 +y 2 −1
Solution: Looking at the limits, we have
√ 2 2
√
√
Z Z Z
1−x2 Z
1/ 2 Z
x +y
2
2
(x + y )dzdydx =
(x2 + y 2)dV
x
x2 +y 2 −1
S
where S is being treated as a z-simple region in 3-D
p going from
2
2
the paraboloid z = x + y − 1 to the cone z = x2 + y 2. So,
we may write
Z Z Z
Z Z Z r
2
2
2
(x + y )dV =
r dz dA
S
R
r 2 −1
9
The projected region R in the x−y plane is obtained by looking
at the two outer integrals. Since y integration is done first, the
region R is being described by
n
o
√
R = (x, y) : x ≤ y ≤ 1 − x2 , 0 ≤ x ≤ 2−1/2
√
Note that the upper and lower curves intersect when x = 1 − x2 ,
i.e. x = 1/sqrt2. We plot this as shown below:
y
y=x
−1/2
R
x
−1/2
(2 , 2 )
x
Figure
4. Region R between curves y = x and y =
√
2
1 − x for 0 ≤ x ≤ 2−1/2
Since the part of the straightline y = √
x for x > 0 in the x − y
π
plane corresponds to θ = 4 , while y = 1 − x2 corresponds to
x2 + y 1 = 1 = r 2 , i.e. r = 1, we have the description of region
R in the (r, θ) domain as
n
πo
π
R = (r, θ) : 0 ≤ r ≤ 1, ≤ θ ≤
4
2
Therefore, since Jacobian J = r, it follows that
Z Z Z
Z π/2 Z 1 Z r
2
2
(x + y )dV =
r 2 (r)dwdrdθ
S
=
Z
π/2 Z 1 Z
π/4
0
π/4
r
r 2 −1
0
r 2 −1
π/2
1
π
r w w=r2−1 drdθ =
r (r−r +1)drdθ =
4
π/4
0
17
π 1 1 1
=
− +
π
=
4 5 6 4
240
3
w=r
Z
Z
3
2
Z
(18) (Section 5.8, Problem 8) Using spherical coordinates, evaluate
Z Z Z
x
p
dxdydz,
2
x + y2 + z2
S
0
1
r 4 − r 5 + r 3 dr
10
where S is the
octant bounded by the planes
√ solid in the firstp
y = x, y = 3x, the cone z = x2 + y 2, the plane z = 0 and
the spheres x2 + y 2 + z 2 = 2 and x2 + y 2 + z 2 = 8.
Soluton: Note that
of the spheres in spherical
√
√ the description
coordinates is ρ = 2 and ρ = 8. Since tan θ = xy , the description of the part of the plane in the first octant y = x is θ = π4 and
√
the plane y = 3x is θ = π3 , Further, since z = ρ cos φ, it follows
p
that z = 0 plane corresponds to φ = π2 . Again, z = x2 + y 2
p
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ cos2 θ = ρ sin φ.
implies ρpcos φ =
So, z = x2 + y 2 implies φ = π4 . So, the region S in spherical
coordinate description is
n
√ π
√
π π
πo
S = (ρ, θ, φ) : 2 ≤ ρ ≤ 8, ≤ θ ≤ , ≤ φ ≤
4
3 4
2
We recall the absolute value of the Jacobian |J| between the
transformation from cartesian to spherical coordinates is ρ2 sin φ
So,
Z Z Z
x
dxdydz =
p
Z
π/2
Z
π/3
Z
√
8
√
π/4
2
ρ=√8
ρ sin φ cos θ 2
(ρ sin φ)dρdθdφ
ρ
x2 + y 2 + z 2
π/4
Z π/2 Z π/3 1 3
ρ cos θ sin2 φ √ dθdφ
=
3
π/4
π/4
ρ= 2
Z
3/2
3/2 Z π/2 Z π/3
θ=π/3
(83/2 − 23/2 ) π/2 (8 − 2 )
2
cos θ sin φdθdφ =
sin θ sin2 φ θ=π/4 dφ
=
3
3
π/4
π/4
π/4
!Z
√
π/2
3
1
(83/2 − 23/2 )
−√
sin2 φdφ
=
3
2
2
π/4
S
R
R
Since sin2 φ = 21 (1 − cos 2φ)dφ = φ2 − 41 sin(2φ) , we obtain
answer
!
√
3/2
3/2
(8 − 2 )
π 1
3
1
−√
+
3
2
8 4
2
(19) (Section 5.8, Problem 11) Using spherical coordinates evaluate
Z 3Z
0
√
0
9−x2
Z √9−x2 −y2
0
xydzdydx
11
Solution: Looking at the order of integration and limits we
have
#
Z 3 Z √9−x2 Z √9−x2 −y2
Z Z "Z √9−x2 −z 2
xydzdydx =
xydz dA
0
0
0
R
0
where R is the projected region in the x−y plane between which
is given by the description
o
n
√
R = (x, y) : 0 ≤ y ≤ 9 − x2 , 0 ≤ x ≤ 3
This is clearly the first quadrant in the interior of the circle of
radius 3 centered at the
So, the triple integral represents
R R origin.
R
the volume integral
xydV
, where S is described as the
S
3-D region whose projection
p on the x − y plane is R and lies
between z = 0 and z = 9 − x2 − y 2, i.e. in the first octant
of the sphere of radius 3, centered at the origin. Therefore,
recalling that Jacobian |J| = ρ2 sin φ, we obtain
Z Z Z
Z π/2 Z π/2 Z 3
2 2
xydV =
ρ sin φ cos θ sin θ (ρ2 sin φ)dρdθdφ
S
0
0
0
ρ=3
Z
Z
Z π/2 Z π/2 243 π/2 π/2 3
1 5 3
dθdφ =
sin φ cos θ sin θdθdφ
ρ sin φ cos θ sin θ
=
5
5 0
0
0
0
ρ=0
θ=π/2
Z
Z
243 π/2 3
243 π/2 1 3
2
dφ =
sin φ sin θ
sin φdφ
=
5 0
2
10 0
θ=0
φ=π/2
Z
−243 π/2 81
1
−243
3
2
=
=
cos φ − cos φ
1 − cos φ d[cos φ] =
10 0
10
3
5
φ=0
(20) (Section 5.8, Problem
appropriate changes of
R R 13)
R By making
2
variables, calculate
(x
+
z)
dxdydz,
where S is the solid
S
bounded by the planes x + y + 3z = 0, x + y + 3z = 1, y + z = 0,
y + z = 1, x + z = −1 and x + z = 1.
Solution: We choose s = x+y +3z, t = y +z, u = x+z. So the
image of region S under the transformation (x, y, z) → (s, t, u)
is the region
S ∗ ≡ {(s, t, u) : 0 ≤ s ≤ 1, 0 ≤ t ≤ 1, −1 ≤ u ≤ 1}
Now note


1 1 3
∂(s, t, u)
= det0 1 1 = −1
∂(x, y, z)
1 0 1
12
So, needed jacobian J = ∂(x,y,z)
= 1/(−1) = −1 and |J| = 1.
∂(s,t,u)
So,
Z Z Z
Z Z Z
Z 1 Z 1Z 1
2
2
(x + z) dV =
u (1)dsdtdu =
u2 dsdtdu
S
S∗
1
1
1
−1
1
0
0
3 1
2
u
=
3 u=−1 3
−1 0
−1 0
(21) (Section 6.1, Problem 2) Determine if F(x, y) = 2xy 3 + 1, 3x2 y 2 − y12
is conservative in D = {(x, y)|y > 2}.
Solution: Note F1 (x, y) = 2xy 3 + 1, F2 (x, y) = 3x2 y 2 − y12 . We
also note
∂F2
∂F1
= 6xy 2 =
∂y
∂x
Further the function F is continuously differentiable in D since
domain D does not contain y = 0 where the function is not continous/differentiable. Therefore, from theorem given in class, F
is conservative.
(22) (Section 6.1, Problem
4) Determine if F(x, y, z)(F1 , F2 , F3 ) =
1
x ln y, z, x − y is conservative in the domain D = {(x, y, z)|y > 0}.
Solution: We note that


i
j
k
∂y
∂z 
∇ × F =  ∂x
x ln y z x − y1
∂
1
∂
1
∂
∂
∂
∂
=
x−
−
x−
, (z) −
(z), (x ln y) −
(x ln y)
∂y
y
∂z
∂z
∂x
y ∂x
∂y
x
1
6= 0
− 1, −1, −
=
y2
y
=
Z
Z
u2 s
1
s=0
dtdu =
Z
Z
u2 dtdu =
Hence F is not conservative.
R
(23) (Section 6.1, Problem 16) Calculate C y 5/3 dx+ 53 xy 2/3 dy, where
C is the parabolic path y = 2x2 from (0, 0) to (2, 8).
Solution: First, check if F ≡ (F1 , F2 ) = y 5/3 , 35 xy 2/3 is
2
1
= 53 y 2/3 = ∂F
. So, F is
conservative or not. We note ∂F
∂y
∂x
conservative and therefore from Theorem in class, there exists
scalar potential f so that ∇f = F. Now we calculate f . Since
∇f = (fx , fy ) = (F1 , F2 ),
5
fx = y 5/3 , and fy = xy 2/3
3
13
Partially integrating the first relation, we have f (x, y) = xy 5/3 +
g(y). Substituting into second equation, we have
5 2/3
5
xy + g ′ (y) = xy 2/3
3
3
′
Therefore, g (y) = 0 and so g(y) = C. So, f (x, y) = xy 5/3 + C.
So, with F defined above, we note
(0.1)
Z
Z
Z
5 2/3
5/3
F·dx = (∇f )·dx = f (2, 8)−f (0, 0) = 2(8)5/3 = 64
y dx+ xy dy =
3
C
C
C
The answer does not depend on the path since F was checked
to be conservative (path-independent).
H
(24) (Section 6.1, Problem 18) Calculate C ydx−xdy, where C is the
circle centered at the origin traversed counter-clockwise once.
F(x, y) = (y, −x) ≡ (F1 , F2 ) noting that
HSolution. We define
H
2
1
ydx−xdy
=
F·dx.
We first note that ∂F
= −1 6= ∂F
; so
∂x1
∂x
C
C
F is not conservative and we do not have a short cut to get to
the answer. We have to use the basic line integral definition and
start first with parametrization of path C: x = (cost, sin t) for
0 ≤ t ≤ 2π in order to traverse the unit circle counterclockwise
once. So,
I
Z 2π
Z 2π
ydx−xdy =
[sin t(− sin tdt) − cos t(cos tdt)] =
[−1]dt = −2π
C
0
0
R
(25) (Section 6.1, Problem 20) Determine C yex dy +ex dy +dz where
C is the line segment from (ln 3, 2, −4) to (0, 1, 2).
Solution Let’s check first, if the F = (yex , ex , 1) is conservative.
We note that


i
j k
∇ × F ==  ∂x ∂y ∂z 
yex ex 1
∂(1) ∂(ex ∂(yex ) ∂(1) ∂(ex ) ∂(yex
=
= (0, 0, 0)
−
,
−
,
−
∂y
∂z
∂z
∂x
∂x
∂y
So, F is conservative and the answer is not dependent on the
path C but only on the end points. We also have some scalar
potential f so that F = (yex , ex , 1) = ∇f = (fx , fy , fz ). So,
fx = yex , fy = ex , fz = 1
So, partially integrating first expression, f (x, y, z) = yex +
g(y, z). Substituting into second expression fy = ex + gy (y, z) =
ex . So, gy (y, z) = 0 implying g(y, z) = h(z) and we have so
14
far f (x, y, z) = yex + h(z). Subsituting into third expression,
fz = h′ (z) = 1. So, h(z) = z + C. So, we have scalar potential
f (x, y, z) = yex + z + C
Z
C
Therefore, using theorems in class
Z
Z
x
x
ye dy + e dy + dz =
F · dx = (∇f ) · dx
C
C
= f (0, 1, 2) − f (ln 3, 2, −4) = e0 + 2 − 2eln 3 − (−4) = 3 − 6 + 4 = 1
H
(26) (Section 6.1, Problem 22). Calculate C 2xyzdx+x2 zdy +x2 ydz
where C is the closed contour along the intersection of the
sphere x2 + y 2 + z 2 = 1 and the plane z = 21 .
Solution: First check if F = (2xyz, x2 z, x2 y) is conservative,
because
if this is true, than according to the theorem in class,
H
F · dx = 0 and we don’t have to do any line integral calcuC
lations. We calculate


i
j
k
∂y ∂z 
∇ × F =  ∂x
2xyz x2 z x2 y
∂(x2 y) ∂(x2 z) ∂(2xyz) ∂(x2 y) ∂(x2 z) ∂(2xyz)
=
−
,
−
,
−
∂y
∂z
∂z
∂x
∂x
∂y
2
2
= (x − x , 2xy − 2xy, 2xz − 2xz) = 0
So, ∇ × F = 0. Therefore, F is conservative and therefore from
theorem in class the closed path integral
I
I
2
2
2xyzdx + x zdy + x ydz =
F · dx = 0
C
C