MATH 317 MIDTERM 2, NOVEMBER 2011: SOLUTIONS
1. Evaluate
the line integrals. (Finish the calculations. Your answer should be a number.)
Z
t3
y ds, if C is the curve in R3 parametrized by r(t) = h , 2t, t2 i, 0 ≤ t ≤ 1.
(a)
3
C
p
√
Solution. We have r0 (t) = ht2 , 2, 2ti, so that |r0 (t)| = t4 + 4 + 4t2 = (t2 + 2)2 = t2 + 2, ds = (t2 + 2)dt,
Z
1
Z
Z
(2t3 + 4t)dt =
2t(t + 2)dt =
y ds =
0
0
C
1
Z
2
1 1
t4
5
+ 2t2 = + 2 − 0 = .
2
2
2
0
(x + y)dx + (y − x)dy, if C is the arc of the circle x2 + y 2 = 4 that lies in the quadrant x ≥ 0, y ≥ 0,
(b)
C
oriented counterclockwise.
Solution. Here x = 2 cos t, y = 2 sin t, 0 ≤ t ≤ π/2, so that dx = −2 sin t dt, dy = 2 cos t dt.
Z
π/2
(2 cos t + 2 sin t)(−2 sin t)dt + (2 sin t − 2 cos t)(2 cos t)dt
0
Z
=
π/2
(−4 cos t sin t − 4 sin2 t + 4 sin t cos t − 4 cos2 t)dt
0
Z
−π/2
(−4)dt = −2π.
=
0
Z
(c)
F · dr, where F = xy k and C is the line segment in R3 from (1, 0, −1) to (0, 2, 2).
C
Solution. The line segment can be parametrized by r(t) = (1 − t)h1, 0, −1i + th0, 2, 2i = h1 − t, 2t, 3t − 1i,
0 ≤ t ≤ 1. Then
Z
Z
Z 1
Z 1
1
2
2
3
F · dr =
xy dz =
(1 − t)(2t)3dt =
(6t − 6t )dt = 3t − 2t = 3 − 2 = 1.
C
C
0
0
0
y
x
i− p
j in R2 .
2. Sketch the vector field F = − p
x2 + y 2
x2 + y 2
Solution. The vectors all have length 1 and point towards the origin.
3. For each statement below, decide whether it is true or false. (Read and think carefully. For this question
only, no explanation or justification is needed, and no credit will be given for an incorrect answer.)
R
(a) Let F be a vector field that is continuous on an open connected region D in R3 . If C F·dr is independent
of path in D (i.e. depends only on the initial and terminal points of C), then F is conservative. – TRUE.
(b) If F = P i + Qj is a vector field on an open connected region D in R2 , P and Q have continuous first
∂P
∂Q
order derivatives on D, and if
=
on D, then there is a function f (x, y) such that F = ∇f on D.
∂y
∂x
– FALSE. (It would be true if we assumed that the region is simply connected.)
(c) The region {(x, y) ∈ R2 : 1 ≤ x2 + y 2 ≤ 2} is simply connected. – FALSE.
4. (a) The vector field F = cos y i − (x sin y + sin z) j + (sin z − y cos z) k on R3 is conservative. Find a
function f (x, y, z) such that F = ∇f .
Solution.
fx = cos y, f (x, y, z) = x cos y + g(y, z),
fy = −x sin y + gy = −x sin y − sin z, gy = − sin z, g(y, z) = −y sin z + h(z),
f (x, y, z) = x cos y − y sin z + h(z),
fz = −y cos z + h0 (z) = −y cos z + sin z, h0 (z) = sin z, h(z) = − cos z,
f (x, y, z) = x cos y − y sin z − cos z.
(b) Let F = ∇f , where f (x, y, z) = cos(x + Zy − 2z). (Note: this is not the field from part (a).) Find a
F · dr = 1.
smooth curve C that is not closed such that
C
R
Solution. C F·dr = f (x2 , y2 , z2 )−f (x1 , y1 , z1 ), where (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are the initial and terminal
points of C. It suffices for example to find two points such that f (x2 , y2 , z2 ) = 1 and f (x1 , y1 , z1 ) = 0,
then any smooth curve from the first to the second point will do. For example, f (0, 0, 0) = cos 0 = 1 and
f (π/2, 0, 0) = 0, so C can be the line segment from (π/2.0, 0) to (0, 0, 0).
5. Use Green’s theorem to evaluate the following line integrals.
Z
(a)
(y 2 − xy)dx + 2xy dy, where C is the closed curve in R2 consisting of the line segment from (0, 0)
C
to (2, 0), the arc of the parabola y = 4 − x2 from (2, 0) to (0, 4), and the line segment from (0, 4) to (0, 0).
R
RR
RR
Solution. C (y 2 − xy)dx + 2xy dy = D (2y − 2y + x)dA = D xdA, where D is the region in the first
quadrant bounded by the parabola and the positive x- and y-axes. This is equal to
Z
ZZ
2
Z
x dA =
Z
(b)
Z
x dydx =
0
D
4−x2
0
2
(4x − x3 )dx = 2x2 −
0
x4 2
16
− 0 = 4.
=8−
4 0
4
2
F · dr, where F = ex i + e2x+y j and C is the boundary of the rectangle in R2 with vertices (0, 0),
C
(0, 1), (2, 1) and (2, 0), oriented clockwise.
R
RR
Solution. C F · dr = − D (2e2x+y − 0)dA, where D is the rectangle as above. (The minus sign is because
C is negatively oriented.) So
Z
Z 2Z 1
Z 2
Z 1
2x+y
2x
F · dr = −
2e
dydx = −
2e dx
ey dy = − e2x |20 ey |10 = −(e4 − 1)(e − 1).
C
0
0
0
0