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MATH 2263 Name (Print): Summer 2009 Student ID: Exam 4: Solutions July 30, 2009 (corrected August 3, 2009) Signature: Time limit: 60 minutes
This exam contains 9 pages (including this cover page and a scratch page) and 6 problems. Check to make sure you have all 9 pages. Enter all requested information at the top of this page, and put your initials on the top of every page, in case the pages become separated. No calculators or note-sheets are allowed.
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1 15 pts 2 15 pts 3 15 pts 4 10 pts 5 20 pts 6 25 pts Total 100 pts
- (15 pts) Evaluate
C xy (^3) ds, where C is given by the parametric equations
x = 4 sin t, y = 4 cos t, z = 3t, 0 ≤ t ≤
π 2
Ans: We have
ds =
dx dt
dy dt
dz dt
dt
=
(4 cos t)^2 + (−4 sin t)^2 + (3)^2 dt
25 dt = 5 dt.
Therefore the line integral is ∫
C
xy^3 ds =
∫ (^) π/ 2
0
(4 sin t)(4 cos t)^3 (5) dt [u = 4 cos t, du = −4 sin t dt]
4
− 5 u^3 du
u^4
0
4
(4)^4 = 5(64)
- (15 pts) Evaluate (^) ∮
C
(y^2 − tan−^1 x) dx + (3x + sin y) dy,
where C is the boundary of the region in the xy-plane enclosed by the parabola y = x^2 and the line y = 4, positively oriented.
Ans: We note that C is a simple closed curve and is the positively oriented boundary of the region D, where
D =
(x, y) : − 2 ≤ x ≤ 2 , x^2 ≤ y ≤ 4
Therefore Green’s Theorem applies, and we have ∮
C
P dx + Q dy =
D
∂Q
∂x
∂P
∂Q
dA ∮
C
(y^2 − tan−^1 x) dx + (3x + sin y) dy =
D
(3 − 2 y) dA
− 2
x^2
(3 − 2 y) dy dx
− 2
3 y − y^2
∣^4 x 2 dx
− 2
(12 − 16) − (3x^2 − x^4 ) dx
− 2
x^4 − 3 x^2 − 4 dx
x^5 5 − x^3 − 4 x
2
− 2 = 2
- (10 pts) Let F(x, y, z) = x z i + y z j −
z k. Compute the following:
(a)
curl F =
i j k ∂ ∂x
∂ ∂y
∂ x ∂z z
y z −
1 z
= i
y z^2
− j
x z^2
= y z^2 i − x z^2 j.
(b)
div F =
∂x
(x z
∂y
(y z
∂z
z
z
z
z^2
2 z + 1 z^2
- (25 pts) Let C 1 and C 2 be two paths from (0, 0) to (1, −1) in the plane, given by
C 1 : r(t) = t^2 i − t^3 j, 0 ≤ t ≤ 1
C 2 : r(t) = t i − t^2 j, 0 ≤ t ≤ 1.
Compute the following:
(a)
C 1
x^2 y^3 dx − y
x dy:
Ans: For C 1 we have
x = t^2 , dx = 2t dt, y = −t^3. dy = − 3 t^2 dt,
so we can calculate the line integral as ∫
C 1
x^2 y^3 dx − y
x dy =
0
t^4 (−t^9 )2t − (−t^3 )(t)(− 3 t^2 ) dt
0
− 2 t^14 − 3 t^6 dt
t^15 −
t^7
1
0 = −
(b)
C 2
x^2 y^3 dx − y
x dy:
Ans: For C 2 we have
x = t, dx = dt, y = −t^2. dy = − 2 t dt,
so we can calculate the line integral as ∫
C 2
x^2 y^3 dx − y
x dy =
0
t^2 (−t^2 )^3 − (−t^2 )
t(− 2 t) dt
0
−t^2 t^6 − 2 t^3
t dt
0
−t^8 − 2 t^7 /^2 dt
t^9 −
t^9 /^2
1
0 = −
(c) Is F(x, y) = x^2 y^3 i − y
x j a conservative vector field? Explain why or why not.
Ans: No, F is not conservative, because line integrals of conservative vector fields are path-independent. But C 1 and C 2 and two distinct paths from (0, 0) to (1, −1) and the line integral of F is different along the two paths (shown in (a) and (b)), so it is not path-independent. Therefore F cannot be conservative.
