Date: 10/06/2016
Instructor: Jay Taylor
Math 223 - Vector Calculus (Fall 2016)
Exam 3-A
Solutions
1. (a)
−→
P Q = (3 − 2)i + (4 − (−1))j + (−1 − 3)k = i + 5j − 4k
r (t) = (2 + t)i + (−1 + 5t)j + (3 − 4t)k
(b)
2(2 + t) + 3(−1 + 5t) + 4(3 − 4t) = 6 ⇒ 4 − 3 + 12 + 2t + 15t − 16t = 6
⇒ t = −7
The line meets the point at (−5, −36, 31).
2. The arrangement is
dependent
independent
independent
dependent
3.
r 0 (t) = − sin(t)i + cos(t)j + k
F (r (t)) = (3 cos(t) − sin(t))i + (cos(t) + 3 sin(t))j + 7k
F (r (t)) · r 0 (t) = − sin(t)(3 cos(t) − sin(t)) + cos(t)(cos(t) + 3 sin(t)) + 7
= cos2 (t) + sin2 (t) + 7
= 8.
Therefore we have
Z
Z
F · dr =
C
4π
8 dt = 32π.
0
4. (a) The region of integration is a triangle
y
4
3
2
1
0
1
x
2
(b) We need to know the bounds on θ and r for the polar integration. Considering the
above triangle we have
r
4
θ
2
The maximal value for θ is thus arctan(4/2) = arctan(2) and the minimal value is
0. As we change θ the bottom line remains constant, so we have cos(θ) = 2/r or
r = cos(θ)/2. Therefore,
Z
0
2
Z
0
2x
x2
dy dx =
x2 + y2
Z
arctan(2)
Z
arctan(2)
Z
0
Z
2/ cos(θ)
0
2/ cos(θ)
r cos2 (θ) dr dθ
=
0
Z
0
arctan(2)
=
0
Z
=
r 2 cos2 (θ)
r dr dθ
r2
r2
cos2 (θ)
2
arctan(2)
2 dθ
0
= 2 arctan(2).
2/ cos(θ)
dθ
0
5. (a) In cylindrical coordinates we have the equations are
)
r 2 + (z − 2)2 = 2
z =r
2
⇒ r 2 + (r 2 − 2)2 = 2
⇒ r 2 + r 4 − 4r 2 + 4 = 2
⇒ r 4 − 3r 2 + 2 = 0
⇒ (r 2 − 1)(r 2 − 2) = 0.
Therefore either r 2 = 1 or r 2 = 2. In cartesian coordinates we thus have the
paraboloid and the sphere meet at
x2 + y2 = 1
and
x 2 + y 2 = 2.
(b) In cylindrical coordinates we have the volume is given by
Z
2π
0
√
Z
1
2
Z
r2
√
2+ 2−r 2
r dzdr dθ.
6. We have
x = ρ sin φ cos θ
y = ρ sin φ sin θ
dV = ρ2 sin φ dρdφdθ.
x = ρ cos φ
7. The only curve that satisfies the assumptions of Green’s Theorem is
8. We can’t apply the curl test because F is not defined at (0, 0). Therefore, we need to
find a potential function for F . In other words, a function f (x, y ) such that F = grad f .
We must have
∂f
3x 2 + 2y
= F1 (x, y ) = 3
.
∂x
x + 2xy + y 3
Integrating with respect to x we thus have
Z
f (x, y ) =
3x 2 + 2y
dx = ln(|x 3 + 2xy + y 3 |) + g(y )
x 3 + 2xy + y 3
for some function g(y ). Differentiating this expression with respect to y we have
2x + 3y 2
∂f
= 3
+ g 0 (y ).
∂y
x + 2xy + y 3
As we require
∂f
3y 2 + 2x
= F2 (x, y ) = 3
∂y
x + 2xy + y 3
we can assume g(y ) = 0 and so f (x, y ) = ln(|x 3 + 2xy + y 3 |) is a potential function
for F , which means F is a gradient vector field.