MATH 1C PRACTICAL HOMEWORK 7 SOLUTIONS
8.1.4
Verify Green’s theorem for the indicated region D and boundary ∂D, and functions P and Q:
D = [−1, 1] × [−1, 1],
P (x, y) = x,
Q(x, y) = y
Solution
RR ∂Q ∂P
∂P
( − ∂y )dA = 0. Now, we parametrize the four sides ofD as c1 (t) = (t, −1), c2 (t) =
Since ∂Q
∂x − ∂y = 0,
D ∂x
(1, t), c3 (t) = (−t, 1), c4 (t) = (−1, −t) where t ∈ [−1, 1]. Then,
R
P4 R
(x, y) · dS
(x, y) · dS =
∂D
R 11 ci
R1
R1
R1
= −1 (t, −1) · (1, 0)dt + −1 (1, t) · (0, 1)dt + −1 (−t, 1) · (−1, 0)dt + −1 (−1, −t) · (0, −1)dt
R1
= 4 −1 tdt = 0
8.1.13
Find the area bounded by one arc of the cycloid x = a(θ − sin θ), y = a(1 − cos θ), where a > 0, and
0 ≤ θ ≤ 2π, and the x axis (use Green’s theorem).
Solution
The region is bounded by the cycloid c1 (t) = (a(t − sin t), a(1 − cos t), t ∈ [0, 2π] and the x-axis c2 (t) = (t, 0)
for t ∈ [0, 2aπ]. Note that because c1 is oriented negatively, we must put a − sign to account for it.
By Green’s theorem we have
ZZ
Z
1
Area(D) =
1dA =
xdy − ydx
2 ∂D
D
Z
Z 2π
−
xdy − ydx = −
(−a + a cos t, a(t − sin t)) · (a − a cos t, a sin t)dt
c1
Z
=−
0
2π
−a2 (1 − cos t)2 + a2 t sin t − a2 sin2 tdt =
0
Z
2π
2a2 − 2a2 cos t + a2 t sin tdt = 3πa2
0
R 2π
R
We can compute the last step using: 1. 0 cos2 tdt = π and; 2. integration by parts for a2 t sin tdt.
Z
Z 2aπ
xdy − ydx =
(0, t) · (1, 0)dt = 0
c2
0
Adding these, we see that the area of the region is 3πa2 .
8.1.15
Evaluate the line integral
Z
(2x3 − y 3 ) dx + (x3 + y 3 ) dy
C
where C is the unit circle, and verify Green’s theorem for this case.
Solution
The unit circle can be parameterized with x = cos(t) and y = sin(t) for t ∈ [0, 2π] Then the integral becomes
Z 2π
3π
[(2 cos(t)3 − sin(t)3 )(− sin(t)) + (cos(t)3 + sin(t)3 )(cos(t))] dt =
2
0
Applying Green’s theorem with P (x, y) = 2x3 − y 3 and Q(x, y) = x3 + y 3 , gives
and the integral becomes
Z Z
3x2 − (−3y 2 ) dx dy
D
1
∂P
∂y
= −3y 2 and
∂Q
∂x
= 3x2
where D is the unit disc. Using polar coordinates, 3x2 + 3y 2 = r2 and dx dy = r dr dθ, and the integral
becomes
Z 2π Z 1
Z 2π
3
3π
2
3r r dr dθ =
dθ =
4
2
0
0
0
Green’s theorem gives the same result, and thus it is verfied for this integral.
8.1.20
Let P (x, y) = −y/(x2 + y 2 ) and Q(x, y) = x/(x2 + y 2 ). Assuming D is the unit disc, investigate why Green’s
theorem fails for this P and Q.
Solution
∂P
We first note that ∂Q
∂x − ∂y = 0. However, parametrizing the boundary of the disc as c(t) = (cos t, sin t), we
R
can compute c P dx + Qdy = 2π. So Green’s theorem fails for this P and Q.
Note that at the origin, (0, 0), P and Q fail to be C 1 and thus, the Green’s theorem fails for this P and
Q. To see that P and Q are not C 1 , we can write P and Q in polar coordinates: P = −r sin θ/r2 = − sin θ/r
and Q = r cos θ/r2 = cos θ/r. Clearly, limr→0 P and limr→0 Q are not defined since the “angle” at which we
approach the origin, i.e. θ, affects the value of P and Q near the origin.
8.2.3.
S = {(x, y, z) : x2 + y 2 + z 2 = 1, z ≥ 0}, ∂S = {(x, y, 0) : x2 + y 2 = 1}, F(x, y, z) = (x, y, z). Verfiy Stokes’
theorem for S, ∂S and F where S is oriented as a graph. (Note that the book says ∂S = {(x, y) : x2 +y 2 = 1}
- which is wrong, since ∂S ⊂ R3 not R2 )
Solution.
∇ ×RFR = 0 .
So,
(∇ × F) · dS = 0.
S
Parametrize ∂S by c(t) = (cos(t), sin(t), 0), t ∈ [0, 2π]. Then, c0 (t) = (− sin(t), cos(t), 0) and F(c(t)) =
F(cos(t), sin(t), 0) = (cos(t), sin(t), 0).
Then F(c(t)) · c0 (t) = 0 .
R
R 2π
So, ∂S F · ds = 0 F(c(t)) · c0 (t) dt = 0.
Since both of the integrals above are 0, the orientation is irrelevant and Stokes’ theorem is verified.
8.2.5
S = {(x, y, z) : z = 1 − x2 − y 2 , z ≥ 0}, ∂S = {(x, y, 0) : x2 + y 2 = 1}, F(x, y, z) = (z, x, 2zx + 2xy).
Verfiy Stokes’ theorem for S, ∂S and F where S is oriented as a graph. (Note that the book says
∂S = {(x, y) : x2 + y 2 = 1} - which is wrong, since ∂S ⊂ R3 not R2 )
Solution.
Since, S is oriented as a graph, we parametrize S by
Ψ(u, v) = (u, v, 1 − u2 − v 2 ), (u, v) ∈ D, where D is the unit disc centered at the origin in R2 .
∂Ψ
∂Ψ
∂u = (1, 0, −2u) and ∂v = (0, 1, −2v).
∂Ψ
∂Ψ
So, ∂u × ∂v = (2u, 2v, 1) .
∇ × F(x, y, z) = (2x, 1 − 2z − 2y, 1).
So, ∇ × F(Ψ(u, v)) = ∇ × F(u, v, 1 − u2 − v 2 ) = (2u, 2u2 + 2v 2 − 2v − 1, 1).
∂Ψ
2
2
3
2
∇ × F(Ψ(u, v)) · ( ∂Ψ
∂u × ∂v ) = 4u + 4u v + 4v − 4v − 2v + 1
2
2
= 4(u + v )v + 4(u2 − v 2 ) − 2v + 1.
RR
RR
∂Ψ
So,
(∇ × F) · dS =
∇ × F(Ψ(u, v)) · ( ∂Ψ
∂u × ∂v ) du dv
S
DR R
2
2
2
=
(4(u + v )v + 4(u − v 2 ) − 2v + 1) du dv.
R 2πDR 1 3
= 0 0 (4r sin(θ) + 4r2 cos(2θ) − 2r sin(θ) + 1)r dr dθ.
R 2π 2
= 0 ( 15
sin(θ) + cos(2θ) + 12 ) dθ = π.
Now the positively oriented boundary of D can be parameterized by c(t) = (cos(t), sin(t)),
t ∈ [0, 2π].
Therefore ∂S is parameterized by p(t) = (cos(t), sin(t), 1 − cos2 (t) − sin2 (t)) = (cos(t), sin(t), 0),
t ∈ [0, 2π].
p0 (t) = (− sin(t), cos(t), 0).
F(p(t)) = F(cos(t), sin(t), 0) = (0, cos(t), 2 cos(t) sin(t)) = (0, cos(t), sin(2t)).
2
F(p(t)) · p0 (t) = cos2 (t) = 12 (1 + cos(2t)).
R
R 2π
R 2π
So, ∂S F · ds = 0 F(c(t)) · c0 (t) dt = 0 21 (1 + cos(2t)) dt = π.
So, Stokes’ theorem is verified.
8.2.8.
Let C be the closed piecewise smooth curve formed by travelling in straight lines
R between the points
(0, 0, 0), (2, 1, 5), (1, 1, 3) and (0, 0, 0), in that order. Use Stokes’ theorem to evaluate C xyz dx+xy dy+x dz.
Solution.
Note that C is the boundary of the triangle with vertices (0, 0, 0), (2, 1, 5) and (1, 1, 3).
So, C = ∂S where S = {u(2, 1, 5) + v(1, 1, 3) : u ≥ 0, v ≥ 0, u + v ≤ 1}
= {(2u + v, u + v, 5u + 3v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 − u}.
(can you see why this is true?)
We can parameterize S by Ψ(u, v) = (2u + v, u + v, 5u + 3v), (u, v) ∈ D, where D = {(u, v) : 0 ≤ u ≤ 1, 0 ≤
v ≤ 1 − u} (which is clearly a simple region i.e. Green’s theorem applies).
Note that if we choose the positive orientation on the boundary of D then the orientation induced on C = ∂S
by Ψ is the orientation given in the problem statement and some elementary algebra shows that Ψ is oneto-one. Therefore, by Stokes’ theorem,
Z
Z
Z
Z Z
xyz dx + xy dy + x dz =
F · ds =
F · ds =
(∇ × F) · dS
C
C
∂S
S
, where F(x, y, z) = (xyz, xy, x).
∂Ψ
∂Ψ
=
(2,
1,
5)
and
=
(1, 1, 3).
∂u
∂v
∂Ψ
So, ∂Ψ
×
=
(−2,
−1,
1)
.
∂u
∂v
∇ × F(x, y, z) = (0, xy − 1, y − xz).
So, ∇ × F(Ψ(u, v)) = ∇ × F(2u + v, u + v, 5u + 3v)
= (0, 2u2 + v 2 + 3uv − 1, u + v − 10u2 − 3v 2 − 11uv).
∂Ψ
∂Ψ
∇ × F(Ψ(u, v)) · ( ∂u × ∂v ) = u + v − 12u2 − 4v 2 − 14uv + 1.
RR
RR
∂Ψ
So,
(∇ × F) · dS =
∇ × F(Ψ(u, v)) · ( ∂Ψ
∂u × ∂v ) du dv
S
DR R
2
2
=
(u + v − 12u − 4v − 14uv + 1) du dv.
R 1D
R 1−u
= 0 0 (u + v − 12u2 − 4v 2 − 14uv + 1) dv du.
R1
2
− 12u2 (1 − u) − 43 (1 − u)3 − 7u(1 − u)2 + 1 − u) du.
= 0 (u(1 − u) + (1−u)
2
R 1 19 3 5 2
= 0 ( 3 u − 2 u − 4u + 16 ) du.
5
1
= 19
12 − 6 − 2 + 6 .
= − 13
12 .
8.2.13.
Let S = S1 ∪ S2 where S1 = {(x, y, z) : x2 + y 2 = 1, 0 ≤ z ≤ 1} and
S2 = {(x, y, z) : x2 + y 2 + (z − 1)2 = 1, z ≥ 1}.
RR
Let F(x, y, z) = (zx + z 2 y + x, z 3 yx + y, z 4 x2 ). Compute
(∇ × F) · dS.
S
Solution.
Clearly, S is a parameterizable surface (for example, S2 can be parameterized by the unit disk in R2 , S1 can
be parametrized by an annulus in R2 whose inner boundary is the boundary of the unit disk and is mapped
to the top boundary circle of the cylinder S1 and these two parameterizations can be “glued togehter” to
give a parametrization of S).
Also, clearly, ∂S = {(x, y, 0) : x2 + y 2 = 1)}.
Parametrize ∂S by c(t) = (cos(t), sin(t), 0), t ∈ [0, 2π]. Then, c0 (t) = (− sin(t), cos(t), 0) and F(c(t)) =
F(cos(t), sin(t), 0) = (cos(t), sin(t), 0).
Then F(c(t)) · c0 (t) = 0 .
RR
R
R 2π
So,
(∇ × F) · dS = ∂S F · ds = 0 F(c(t)) · c0 (t) dt = 0.
S
(Note that the orientation is irrelevant since the integral is 0).
3