14. v = e x yz (xi + y 2 j + zk). The field lines satisfy dx dy dz = 2 = , x y z so they are given by z = C 1 x, ln |x| = ln |C 2 | - (1/y) (or, equivalently, x = C 2 e-1/y ).
28. The surface area element for a conical surface , z =h 1- x 2 + y2 a ,
having base radius a and height h, was determined in the solution to Exercise 23 to be a2 + h2 d x dy. dS = a The mass of , which has areal density , was also determined in that exe
13. v = x zi + yzj + xk. The field lines satisfy dx dy dz = = , xz yz x or, equivalently, d x/x = dy/y and d x = z dz. Thus the field lines have equations y = C 1 x, 2x = z 2 + C 2 , and are therefore parabolas.
27.
is the spherical shell, x 2 + y 2 + z 2 = a 2 , with areal density . Its mass is 4 a 2 . Its moment of inertia about the z-axis is
2
=
0
= 2 a 4 = 2 a
4
The radius of gyration is D =
I =
(x 2 + y 2 ) d S
d
0 1 0
a 2 sin2 a 2 sin d sin (1 - cos2 ) d
12. v =
xi + yj . (1 + z 2 )(x 2 + y 2 )
The streamlines satisfy dz = 0 and
dy dx = . Thus z = C 1 and y = C 2 x. The streamlines x y are horizontal half-lines emanating from the z-axis.
26.
is the cylindrical surface x 2 + y 2 = a 2 , 0 z h, with areal density . Its mass is m = 2 ah . Since all surface elements are at distance a from the z-axis, the radius of gyration of the cylindrical surface about the z-axis is D = a. Therefore the mo
11. v(x, y, z) = yi - xj + k. dx dy The streamlines satisfy =- = dz. Thus y x 2 x d x + y dy = 0, so x 2 + y 2 = C 1 . Therefore, dz 1 = = dx y 1
2 C1 - x 2
.
This implies that z = sin-1
x + C 2 . The streamlines are the spirals in which the surfaces C1 2
25. The surface element d S = a d dz at the point with cylindrical coordinates (a, , z) attracts mass m at point (0, 0, b) with a force whose vertical component (see the figure) is dF = km d S km a(b - z) d dz cos = 2 D D3 km a(b - z) d dz = . 3/2 a 2 + (
10. v(x, y, z) = xi + yj - xk. dx dy dz The streamlines satisfy = = - . Thus x y x z + x = C 1 , y = C 2 x. The streamlines are straight half-lines emanating from the z-axis and perpendicular to the vector i + k.
24. By symmetry, the force of attraction of the hemisphere shown in the figure on the mass m at the origin is vertical. The vertical component of the force exerted by area element d S = a 2 sin d d at the position with spherical coordinates (a, , ) is dF
9. v(x, y, z) = yi - yj - yk. The streamlines satisfy d x = -dy = -dz. Thus y + x = C 1 , z + x = C 2 . The streamlines are straight lines parallel to i - j - k.
23. The cone z = h 1 -
x 2 + y2 a
has normal z z i- j+k x y h a xi + yj x 2 + y2 + k,
n=- =- so its surface area element is dS = The mass of the conical shell is m=
h2 + 1 d x dy = a2
a2 + h2 d x dy. a
a2 + h2 ( a 2 ) = a a 2 + h 2 . dS = a x 2 +y 2 a 2
8. F = cos yi - cos xj. The field lines satisfy dx dy =- , that is, cos y cos x cos x d x + cos y dy = 0. Thus they are the curves sin x + sin y = C.
y
x
Fig. 1-8
22. Use spherical coordinates. The area of the eighth-sphere
a2 1 2 sq. units. A = (4 a ) = 8 2 The moment about z = 0 is Mz=0 = =
0
z dS
/2 /2
d
0 /2 0
a cos a 2 sin d a3 sin 2 d = . 2 4
a3 = 2 Thus z =
a Mz=0 = . By symmetry, x = y = z, so the centroi
29. By Exercise 27, the moment of inertia of a spherical shell of radius a about its diameter 2 is I = ma 2 . Following the argument given in Example 4(b) of Section 5.7, the kinetic 3 energy of the sphere, rolling with speed v down a plane inclined at an
1. F = xi + zj. The surface of the tetrahedron has four faces: ^ ^ On 1 , x = 0, N = -i, F N = 0. ^ ^ On 2 , y = 0, N = -j, F N = -z, d S = d x dz. ^ = -k, F N = 0. ^ On 3 , z = 0, N i + 2j + 3k x + 2z d x dy 14 ^ ^ ,FN = , dS = = On 4 , x + 2y + 3z = 6,
M
ETU
Department of Mathematics
MATH 260
SECOND MIDTERM EXAM
Date
Time
:22.11.2010
Duration
: 10A min.
:17.10
Name
Student ID
Section
Signature
3 Quetions on 4 Pages
100 Points
SHOW YOUR WORK
Q.1. (30 pts) a) (5 pts.) Show that the foliowing functions
x,I
5. The part of z = a - x 2 - y 2 lying above z = b < a lies inside the vertical cylinder x 2 + y 2 = a - b. For z = a - x 2 - y 2 , the upward vector surface element is
2xi + 2yj + k ^ N dS = d x dy. 1
Thus the flux of F = xi + yj + zk upward through
1
On
1:
1 ^ N= 2
xi + yj
x 2 + y2
+ k , dS =
2 d x dy. Thus
1
^ F N dS xy
x 2 +y 2 1
=
x 2 + y2
2
+1-
1
x 2 + y2
= 0 + 12 - 2 =- = . 3 3 On
2:
d
0 0
r 2 dr
^ ^ N = -k and z = 0, so F N = 0. Thus, the total flux of F out of the cone is /3.
z
1
^ N
1
z=1-
3. F = xi + yj + zk. ^ The box has six faces. F N = 0 on the three faces x = 0, y = 0, and z = 0. On the face ^ ^ x = a, we have N = i, so F N = a. Thus the flux of F out of that face is a (area of the face) = abc. By symmetry, the flux of F out of the fa