ff (z + z²y) ds, where S is the part of the cylinder y² + ² = 1 that lies between the planes x = 0 and x = 3 in the first octant. 12 CC - Evaluate
ff (z + z²y) ds, where S is the part of the cylinder y² + ² = 1 that lies between the planes x = 0 and x = 3 in the first octant. 12 CC - Evaluate
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Can you please solve problem #3 and show all of your work in pictures do not type it, please! make sure you do the right work, thank you for your help.
![[[ (a²z+ y²2) dS, where S is the top half of the sphere
S
x² + y² + z² = 4. 16
Problem 2. Evaluate
][ (z+x²y) ds, where S is the part of the cylinder
S
y² + ² = 1 that lies between the planes x = 0 and x = 3 in the
first octant. 12
Problem 3. Evaluate
Problem 4. Evaluate
that lies in the first octant. 4
Problem 5. Evaluate
• [[ 20
S
that lies between the planes z = 1 and 2 = 3. 364√/2π/3
J xz dS, where S is the part of the plane 2x + 2y +z = 4
Problem 6. Evaluate
]] x ds, where S is the triangular region with vertices
S
(1,0,0), (0, -2, 0), and (0,0,4). √21/3
Problem 7. Evaluate
]] xz dS, where S is the boundary of the region enclosed
by the cylinder y² + ² = 9 and the planes x = 0 and x + y = 5. 0
!! dS, where S is the paraboloid y = 9-x² - 2²
S
together with its cap when y = 0. (7/6)[37/37 1] +9T
Problem 8. Evaluate.
Problem 9. Evaluate.
x²² ds, where S is the part of the cone z² = x² + y²
//*
F. ds, where F(x, y, z) = xyî + 4x²ĵ + yzk, and S is
S
the surface with parameterization F(u, v) = ui + vj + ueºk, where
0 ≤u≤ 2 and 0 ≤v≤1. 16(1-e)
Problem 10. Evaluate
· // ₁ F. ds, where F(x, y, z) = (x, y, z²) and S is the
S
part of the unit sphere where x > 0, y > 0, and z < 0. 57/24
Problem 11. Evaluate
[] F. ds, where F(x, y, z) = (y, z-y, x) and S is the surface
S
of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (0,0,1). -1/6
Problem 12. One important law of electrostatics is Gauss' Law, which says that the
net charge enclosed by a closed surface S is Q = €0 E ds, where
•!! B
€0 is the constant permittivity of free space. Use Gauss' Law to find
the charge enclosed by the cube with vertices (±1, ±1, ±1) if the electric
field is E = xi+yj + zk. 24€0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0655393b-8df3-4633-b13c-e0d6983d2306%2Fcb62cebf-ca2c-41d7-ba59-117ae4b52f95%2Feapmk83h_processed.png&w=3840&q=75)
Transcribed Image Text:[[ (a²z+ y²2) dS, where S is the top half of the sphere
S
x² + y² + z² = 4. 16
Problem 2. Evaluate
][ (z+x²y) ds, where S is the part of the cylinder
S
y² + ² = 1 that lies between the planes x = 0 and x = 3 in the
first octant. 12
Problem 3. Evaluate
Problem 4. Evaluate
that lies in the first octant. 4
Problem 5. Evaluate
• [[ 20
S
that lies between the planes z = 1 and 2 = 3. 364√/2π/3
J xz dS, where S is the part of the plane 2x + 2y +z = 4
Problem 6. Evaluate
]] x ds, where S is the triangular region with vertices
S
(1,0,0), (0, -2, 0), and (0,0,4). √21/3
Problem 7. Evaluate
]] xz dS, where S is the boundary of the region enclosed
by the cylinder y² + ² = 9 and the planes x = 0 and x + y = 5. 0
!! dS, where S is the paraboloid y = 9-x² - 2²
S
together with its cap when y = 0. (7/6)[37/37 1] +9T
Problem 8. Evaluate.
Problem 9. Evaluate.
x²² ds, where S is the part of the cone z² = x² + y²
//*
F. ds, where F(x, y, z) = xyî + 4x²ĵ + yzk, and S is
S
the surface with parameterization F(u, v) = ui + vj + ueºk, where
0 ≤u≤ 2 and 0 ≤v≤1. 16(1-e)
Problem 10. Evaluate
· // ₁ F. ds, where F(x, y, z) = (x, y, z²) and S is the
S
part of the unit sphere where x > 0, y > 0, and z < 0. 57/24
Problem 11. Evaluate
[] F. ds, where F(x, y, z) = (y, z-y, x) and S is the surface
S
of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), (0,0,1). -1/6
Problem 12. One important law of electrostatics is Gauss' Law, which says that the
net charge enclosed by a closed surface S is Q = €0 E ds, where
•!! B
€0 is the constant permittivity of free space. Use Gauss' Law to find
the charge enclosed by the cube with vertices (±1, ±1, ±1) if the electric
field is E = xi+yj + zk. 24€0
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