How to slove this question For Problems 4 9, calculate the flux of the vector field through the given surface.4. F = zk through a square of side length 5 in the plane z = 2. The square is centered on the z-axis, has sidesparallel to the axes, and is oriented in the positive z-direction.5. F(x2 – 2) i through a square of side length 3 in the yz-plane, oriented in the negative r-direction.6. F = z(x2 + y²) k through the disk of radius 4 in the plane z = 2, centered on the z-axis, oriented upward.7. F = (e-x²-y')k through the disk of radius 2 in the ry-plane, oriented upward.8. F = 7+2xi+ 4y k, and S is the part of the surface z = 2x2 + y + 3 above the square 0 < x < 1, 0 < y < 1,oriented upward.9. F = xỉ+ (x + 2y – 8)3+ 2z k, and S is the part of the surface z = 4 – x? – yin the first octant (see figure to the right), oriented upward.

Flux flux is something that is constantly changing and passes through any cross-sectional area.

Answered: 9. F = xỉ+ (x + 2y – 8)j+ 2z k, and S…

9. F = xỉ+ (x + 2y – 8)j+ 2z k, and S is the part of the surface z = 4 – x2 . in the first octant (see figure to the right), oriented upward.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The theory of special relativity explains the connection between space and time to objects that move in a straight line at a constant speed. The objects move at the speed of light. When an object approaches light speed, its mass is endless and cannot move faster than light. The most crucial postulate of Einstein's special relativity is that the laws of nature do not change in each frame of reference.

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How to slove this question

For Problems 4 9, calculate the flux of the vector field through the given surface.
4. F = zk through a square of side length 5 in the plane z = 2. The square is centered on the z-axis, has sides
parallel to the axes, and is oriented in the positive z-direction.
5. F
(x2 – 2) i through a square of side length 3 in the yz-plane, oriented in the negative r-direction.
6. F = z(x2 + y²) k through the disk of radius 4 in the plane z = 2, centered on the z-axis, oriented upward.
7. F = (e-x²
-y')k through the disk of radius 2 in the ry-plane, oriented upward.
8. F = 7+2xi+ 4y k, and S is the part of the surface z = 2x2 + y + 3 above the square 0 < x < 1, 0 < y < 1,
oriented upward.
9. F = xỉ+ (x + 2y – 8)3+ 2z k, and S is the part of the surface z = 4 – x? – y
in the first octant (see figure to the right), oriented upward.

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