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MTH 255 Exam II Solutions
- (16 pts) In this problem the curve C is the quarter of the circle of radius 2, centered
at the origin, which lies in the first quadrant, oriented from (2, 0) to (0, 2).
(a) Compute
C
f ds where f (x, y) =
xy
x
2
2
The first thing which is needed is a parametrization of the curve. While it is
possible to use a representation as a graph, the most convenient is ~r(t) = cos t~ı +
sin t~. In terms of the parametrization,
C
f ds =
π/ 2
0
f (x(t), y(t))|~r
′ (t)|dt. An
easy computation gives |~r
′ (t)| = 2, so the integral becomes
∫ π
2
0
2 cos t2 sin t
4 cos
2 t + 4 sin
2
t
2 dt =
∫ π
2
0
2 sin t cos tdt = sin
2
t
π
2
0
(b) Compute
C
F · d~r for
F = y~ı − x ~.
This is a line integral given in terms of the parametrization by
F (~r(t)) · ~r
′
(t)dt
which evaluates to
(2 sin t~ı − 2 cos t~) · (−2 sin t~ı + 2 cos t~)dt =
∫ π
2
0
− 4 dt = − 2 π
- (16 pts) In this problem C is the half of the circle x
2
2
= 9 in the right half
plane (x ≥ 0) together with the segment on the y-axis from (0, 3) to (0, −3) oriented
counterclockwise, and
F = −x
2
y~ı + xy
2
~.
(a) Compute
C
F · d~r.
This curve is a closed curve, bounding the half disk of radius three in the right
half plane. Thus Green’s theorem is applicable. Using Green’s theorem
C
F · d~r =
Half disk
∂
∂x
(xy
2 ) −
∂
∂y
(−x
2 y)
dA
y
2
2 dA =
∫ π
2
−
π
2
3
0
r
2 rdrdθ
= π
1
4
r
4
|
3
0
81 π
4
(b) Compute
C
F ·
N ds, where
N is the outer unit normal.
Using the variation of Green’s theorem for this flux integral (integral of the normal
component of
F ) gives
C
F ·
N ds =
div
F dA =
0 dA = 0
- (9 pts) Compute
C
F · d~r where
F = grad ((x + 2y + 3z)e
xz
) and C is the piecewise
smooth curve made of the straight line segment from (0, 1 , 1) to (1, 2 , 1) followed by
the straight line segment from (1, 2 , 1) to (2, 1 , 0).
Since this is the line integral of a gradient, the fundamental theorem for line integrals
says that it is equal the difference of the function evaluated at the terminal point less
the value at the initial point. Substituting in the function, the difference is 4 − 5 = − 1
- (9 pts) Let
F = (3y
2
− 3 x
2
- y)~ı + (6xy + x + z) ~ + y
k
(a) Compute curl
F =
(b) Compute div
F = 0
(c) Is
F conservative on R
3 ? Justify. Yes, since R
3 is simply connected, and the curl
of
F is zero.
- (a) (4 pts) Let u = 2x + y and v = −x + 4y. Compute the Jacobian
∂(x, y)
∂(u, v)
The easiest method is to compute
∂(u, v)
∂(x, y)
which is
and then use that
∂(x, y)
∂(u, v)
is the reciprocal, which is
1
9
(b) (6 pts) Compute
D
sin(2x + y)dA where D is the region bounded by the lines
−x + 4y = 4, −x + 4y = 12, 2x + y = 0 and 2x + y = π.
Using the change of variables from above
D
sin(2x + y)dA =
12
4
π
0
sin(u)
∂(x,y)
∂(u,v)
du dv
12
4
π
0
sin(u)
1
9
du dv = 8 ∗ 2 ∗
1
9
16
9
