James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

section 16.7 Surface Integrals 1133
6. yy S
xyz dS,
S is the cone with parametric equations x − u cos v,
y − u sin v, z − u, 0 < u < 1, 0 < v < y2
7. yy S
y dS, S is the helicoid with vector equation
rsu, vd − ku cos v, u sin v, vl, 0 < u < 1, 0 < v <
8. yy S
sx 2 1 y 2 d dS,
S is the surface with vector equation
rsu, vd − k2uv, u 2 2 v 2 , u 2 1 v 2 l, u 2 1 v 2 < 1
9. yy S
x 2 yz dS,
S is the part of the plane z − 1 1 2x 1 3y that lies above
the rectangle f0, 3g 3 f0, 2g
10. yy S
xz dS,
S is the part of the plane 2x 1 2y 1 z − 4 that lies in the
first octant
11. yy S
x dS,
S is the triangular region with vertices s1, 0, 0d, s0, 22, 0d,
and s0, 0, 4d
12. yy S
y dS,
S is the surface z − 2 3 sx 3y2 1 y 3y2 d, 0 < x < 1, 0 < y < 1
13. yy S
z 2 dS,
S is the part of the paraboloid x − y 2 1 z 2 given by
0 < x < 1
14. yy S
y 2 z 2 dS,
S is the part of the cone y − sx 2 1 z 2 given by 0 < y < 5
15. yy S
x dS,
S is the surface y − x 2 1 4z, 0 < x < 1, 0 < z < 1
16. yy S
y 2 dS,
S is the part of the sphere x 2 1 y 2 1 z 2 − 1 that lies above
the cone z − sx 2 1 y 2
17. yy S
sx 2 z 1 y 2 zd dS,
S is the hemisphere x 2 1 y 2 1 z 2 − 4, z > 0
18. yy S
sx 1 y 1 zd dS,
S is the part of the half-cylinder x 2 1 z 2 − 1, z > 0, that
lies between the planes y − 0 and y − 2
19. yy S
xz dS,
S is the boundary of the region enclosed by the cylinder
y 2 1 z 2 − 9 and the planes x − 0 and x 1 y − 5
20. yy S
sx 2 1 y 2 1 z 2 d dS,
S is the part of the cylinder x 2 1 y 2 − 9 between the planes
z − 0 and z − 2, together with its top and bottom disks
21–32 Evaluate the surface integral yy S
F dS for the given
vector field F and the oriented surface S. In other words, find
the flux of F across S. For closed surfaces, use the positive
(outward) orientation.
21. Fsx, y, zd − ze xy i 2 3ze xy j 1 xy k,
S is the parallelogram of Exercise 5 with upward orientation
CAS
CAS
CAS
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22. Fsx, y, zd − z i 1 y j 1 x k,
S is the helicoid of Exercise 7 with upward orientation
23. Fsx, y, zd − xy i 1 yz j 1 zx k, S is the part of the
para boloid z − 4 2 x 2 2 y 2 that lies above the square
0 < x < 1, 0 < y < 1, and has upward orientation
24. Fsx, y, zd − 2x i 2 y j 1 z 3 k, S is the part of the cone
z − sx 2 1 y 2 between the planes z − 1 and z − 3 with
downward orientation
25. Fsx, y, zd − x i 1 y j 1 z 2 k, S is the sphere with radius 1
and center the origin
26. Fsx, y, zd − y i 2 x j 1 2z k, S is the hemisphere
x 2 1 y 2 1 z 2 − 4, z > 0, oriented downward
27. Fsx, y, zd − y j 2 z k,
S consists of the paraboloid y − x 2 1 z 2 , 0 < y < 1,
and the disk x 2 1 z 2 < 1, y − 1
28. Fsx, y, zd − yz i 1 zx j 1 xy k,
S is the surface z − x sin y, 0 < x < 2, 0 < y < , with
upward orientation
29. Fsx, y, zd − x i 1 2y j 1 3z k,
S is the cube with vertices s61, 61, 61d
30. Fsx, y, zd − x i 1 y j 1 5 k, S is the boundary of the
region enclosed by the cylinder x 2 1 z 2 − 1 and the planes
y − 0 and x 1 y − 2
31. Fsx, y, zd − x 2 i 1 y 2 j 1 z 2 k, S is the boundary of the
solid half-cylinder 0 < z < s1 2 y 2 , 0 < x < 2
32. Fsx, y, zd − y i 1 sz 2 yd j 1 x k,
S is the surface of the tetrahedron with vertices s0, 0, 0d,
s1, 0, 0d, s0, 1, 0d, and s0, 0, 1d
33. Evaluate yy S
sx 2 1 y 2 1 z 2 d dS correct to four decimal
places, where S is the surface z − xe y , 0 < x < 1,
0 < y < 1.
34. Find the exact value of yy S
xyz dS, where S is the surface
z − x 2 y 2 , 0 < x < 1, 0 < y < 2.
35. Find the value of yy S
x 2 y 2 z 2 dS correct to four decimal
places, where S is the part of the paraboloid
z − 3 2 2x 2 2 y 2 that lies above the xy-plane.
36. Find the flux of
Fsx, y, zd − sinsxyzd i 1 x 2 y j 1 z 2 e xy5 k
across the part of the cylinder 4y 2 1 z 2 − 4 that lies above
the xy-plane and between the planes x − 22 and x − 2
with upward orientation. Illustrate by using a computer
algebra system to draw the cylinder and the vector field on
the same screen.
37. Find a formula for yy S
F dS similar to Formula 10 for
the case where S is given by y − hsx, zd and n is the unit
normal that points toward the left.
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