SOLUTION — MATH 52 SECOND MIDTERM
SPRING 2010
1.(10) Let C be the spaceZ curve given by x(t) = (t, t, t2 ), 0 ≤ t ≤ 1.
a. Evaluate the integral
x ds;
C
Z
b. for F = i + z j − y k, evaluate he integral
F · ds;
C
Z
c. evaluate the integral
x dx + z dy.
C
√
Solution. a. x0 = (1, 1, 2t), k x0 k= 2 + 4t2 ;
Z
Z 1 p
3 1
3
1
1 3
x ds =
t 2 + 4t2 dt =
(2 + 4t2 ) 2 =
(6 2 − 2 2 ).
12
12
0
C
0
Z
Z 1
Z 1
2
b.
F · ds =
(1, t2 , −t) · (1, 1, 2t) dt =
(1 − t2 ) dt = .
3
C
0
0
Z
Z 1
5
c.
x dx + z dy =
(t + t2 ) dt = .
6
C
0
Z
Comment. Distinguish
Z
f ds and
F · ds. This is testing the ability to under
the notion of line integrals.
2.(10) Let F be the plane vector field F = (x + y) i + 2xy j; let D be the region
bounded by y = 1 and y + x2 = 2, and let C be its boundary.Z
(F · T) ds;
a. Find by evaluating a line integral directly the circulation
b. find by Green’s theorem the circulation of F along C.
C
Solution. a. The curves consists of two parts: C1 the top part, from right to
left; the lower line segment, from left to right. (See figure 1 at the end.) Their
parameterizations:
C1 : x1 = (t, 2 − t2 ), t : 1 → −1;
C : x = (t, 1), t : −1 → 1.
Z2 2
Z −1
2
(F · T) ds =
(t + 2 − t2 ) + 2t · (2 − t2 )(−2t) dt = ;
5
ZC1
Z1 1
(F · T) ds =
(t + 1) dt = 2;
C2Z
−1
Z
Z
12
So
(F · T) ds =
(F · T) ds +
(F · T) ds =
.
5
C
C1
C2
1
2
SPRING 2010
Z
C
12
.
5
Z
(F · T) ds =
b. Using Green’s theorem,
1) dydx =
ZZ
1
Z
2−x2
(2y − 1) dA =
(2y −
−1
D
1
Comment. Testing the circulation, and applying the Green’s theorem. Circulation
means the whole loop.
2
2
3.(10) Consider the vector
Z field F = x i + y j and the region D = {1 ≤ x + y ≤ 2}.
(F · n) ds of the vector field F across the boundary of D.
Find the outward flux
∂D
Z
ZZ
(F·n) ds =
(Mx +Ny ) dA =
Solution. Using the Green’s theorem,
∂D
D
√ 2
√
1) dA = 2(π( 2) − π) = 2π. Here D is the annulus of radius 2 and
Z 1.
ZZ
(1+
D
An alternative is to calculate the flux directly, using the formula
(F · n) ds =
∂D
Z
√
√
−N dx + M dy. In this case, ∂D = C2 − C1 , where C2 : ( 2 cos t, 2 sin t) and
∂D
C1 : (cos t sin t), both from 0 → 2π.
Comment. The understanding of flux. Applying the formula.
4.(10) Indicate and justify your conclusion whether the following vector fields are
conservative in their naturally defined regions; in case it is conservative, find one
of its potential functions.
a. F1 = (6xy + y cos x)i + (3x2 + sin x)j;
b. F2 = (yz + x2 )i + (xz + y 2 )j + (−xy + z 2 )k.
Solution. a. Computing: Ny = 6x + cos x; Mx = 6x + cos x. They are equal.
Since the field is defined on the region R2 , which is simply connected, so F1 is
conservative.
For potential f , first solve fx = 6xy + y cos x, get f (x, y) = 3x2 y + y sin x + g(y).
Using fy = 3x2 + sin x, and fy = 3x2 + sin x + g 0 (y), we get g 0 (y) = 0. We pick
g = 0. thus f = 3x2 + sin x.
b. Since Py = −x and Nz = x, they are different. Thus F2 is not conservative.
Comment. Criterion using curl.
5.(10) Let F be the vector field F =
−y + x
x+y
i+ 2
j.
x2 + y 2
x + y2
a. Show thatI ∇ × F = 0;
b. evaluate
(F · T) ds for C th circle x2 + y 2 = 1, clockwise;
C
c. determine whether F is a conservative vector field in its naturally defined region
R2 \ {0}; justify your answer.
SOLUTION — MATH 52 SECOND MIDTERM
Solution. a. Direct computation, Nx =
that Nx = My .
1
x2 +y 2
−
(x+y)2x
(x2 +y 2 )2 .
3
Computing My shows
b. Using x = (cos t, sin t), t : 2π → 0, (since clockwise). Evaluating (using
I
Z 0
(sin2 t−sin t cos t+cos2 t+sin t cos t) dt = −2π.
sin2 t+cos2 t = 1) (F·T) ds =
2π
C
c It is not conservative since a necessarily and sufficient condition for F to be
conservative is that the circulation of F along all closed arcs are zero. b shows for
one circle the circulation is non-zero; so not conservative.
Comment. Another must have problem. The vector field has ∇ × F = 0; the
region is not simply connected, with one hole; conservative or not depend on the
circulation of one loop around the hole.
6. (10) Let F = (x + z) i + y j + (x + z) k.
a. Show that F is a conservative vector
field;
√
2
b. let C be the arc x(t) = (sin πt, t2 + 1, et −1 ) for t ∈ [0, 1]; find the work done
by F along the arc C.
Solution. a. We calcularte Nx = 0 = My , Py = 0 = Nz , Mz = 1 = Px . And the
region is simply connected. So conservative.
R
b. We find f = 12 (x2 + y 2 + z 2 ) + xz. Then C F · ds = f (q) − f (p), where
√
q = x(1) = (0, 2, 1) and p = x(0) = (0, 1, 1e ). So the work is 1 − 2e12 .
Comment. Work is the difference of the values of the potential for a conservative
field.
7. (10) Let D be a region with piecewise smooth boundary
∂D.
I
a. Use Green’s theorem to confirm that Area(D) =
−y dx;
∂D
b. Let D be the region bounded by x(t) = (2 sin t, 3 cos t), t ∈ [0, 2π]. Use Green’s
theorem to find the area of D.
I
ZZ
ZZ
Solution. a.
−y dx =
(Nx − My ) dA =
(−(−1)) dA = Area(D).
∂D
D
D
b.
of the curve gives it a clockwise orientation. So area is
I The parameterization
Z 2π
−y dx = −
−3 cos t(2 cos t) dt = 6π.
∂D
0
Comment. This is a proof; and an easy application, though with a trick embedded
in that the parameterization is given with clockwise orientation.
8. (6) Which of the following regions are simply connected:
a. D = {x2 + y 2 < 1} \ [0, 1): the unit disk {x2 + y 2 < 1} taking out the line
segment [0, 1);
b. D = [0, 1] × [0, 1] \ {0}: the square [0, 1] × [0, 1] taking out the origin;
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SPRING 2010
c. R = {x2 + y 2 + z 2 < 1} \ {x2 + y 2 + z 2 ≤ 12 }: the unit ball {x2 + y 2 + z 2 < 1}
taking out a sub-ball {x2 + y 2 + z 2 ≤ 12 }.
Solution. All three are simply connected.
Comment. The understanding of simply connected region. Note that having a
hole in two dimensional region makes it non-simply connected; having a hole in
three dimensional region does not affect the simply-connectedness.
9. (8) Let D be the region {1 ≤ x2 + y 2 ≤ 5} in the xy-plane; let F be a smooth
vector field on D such that ∇ × F = 0 over D.
a. Can one conclude that F is conservative over D?I
b. If for C the curve x = (cos t, sin t, 0), t ∈ [0, 2π],
F · ds = 0; can one conclude
C
that F is conservative?
I
c. If for all simple closed arcs C in D,
F · ds = 0; can one conclude that F is
C
conservative?
I
d. If for all closed arcs C in D,
F · ds = 0; can one conclude that F is conservative?
Briefly justify your answer.
C
Solution. a. Not sufficient: Example see problem 5.
b. It is conservative. Since the region has one hole, we only need to test one curve
surround this hole. The reason is that for this particular region, any closed arc can
be deformed continuously within the region to a multiple of this curve. So if the
line integral along this curve is zero, the line integral
R along any closed curve is zero.
Here we have used that in case ∇ × F = 0, then C F · ds remain unchanged when
we deform the closed arc C. This is sufficient for the field to be conservative.
c. Conservative. See reason b; or Theorem 3.2 on page 391.
c. Conservative. Same reason as c.
Comment. Understanding of the criterions of conservative vector field.
10. (6) Let D be the unit disk {x2 + y 2 < 1} and F a smooth vector field F on the
unit disk D satisfying ∇ × F = 0.
I
a. Use Green’s theorem to show that for the arc C1 shown below,
F · ds = 0;
IC1
b. Use Green’s theorem to show that for the arc C2 shown below,
F · ds = 0.
C2
(Using the notion of conservative vector field is not allowed.)
Solution. a. SinceI C1 is the opposite
of the boundary of D1 , as shown, then by
ZZ
Green’s theorem,
F · ds =
∇ × F dA = 0.
C1
D1
SOLUTION — MATH 52 SECOND MIDTERM
5
I
b. The curve C2 = ∂D2 −∂D3 +∂D4 , as indicated. So by Green’s therem
F·ds =
C2
I
ZZ
ZZ
ZZ
F · ds =
∇ × F dA −
∇ × F dA +
∇ × F dA = 0. ∂D2 −∂D3 +∂D4
D2
D3
D4
Comment. Apply Green’s theorem in the abstract form; curves can be recombined
to apply the Green’s theorem.