Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 14 - Vector Calculus - 14.6 Surface Integrals - 14.6 Exercises - Page 1123: 10

Answer

The upward flux is equal to the area , that is, projection in the $xy$ plane.

Work Step by Step

Let $Z$ denotes the orientation of a surface of a sphere and is given as: $Z=g(x,y)$ Here, $g(x, y)$ defines the continuous and differentiable function. Next, the normal vector for the given sphere can be expressed as: $(-Z_x, -Z_y, 1)$ in the upward direction. We are given that the vector field is: $F=\lt 0, 0, 1 \gt$ Thus, the Green's Theorem for the upward flux is: $F=\iint_S F \cdot n \ ds\\=\iint_R \lt 0, 0, 1 \gt \cdot (-Z_x, -Z_y, 1) \ dA \\= \iint_R dA\\=A$ This implies that the upward flux is equal to the area , that is, projection in the $xy$ plane.
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