   Chapter 16.7, Problem 23E

Chapter
Section
Textbook Problem

Evaluate the surface integral ∫∫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.23. F(x, y, z) = xy i + yz j + zx k, S is the part of the paraboloid z = 4 - x2 - y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

To determine

To find: The value of SFdS for F(x,y,z)=xyi+yzj+zxk .

Explanation

Given data:

z=4x2y2 , 0x1 and 0y1 .

F(x,y,z)=xyi+yzj+zxk (1)

Formula used:

SFdS=D(PgxQgy+R)dA (2)

Consider g=z .

Substitute 4x2y2 for z ,

g=4x2y2

Substitute xy for P , yz for Q , zx for R and 4x2y2 for g in equation (2),

SFdS=0101((xy)(4x2y2)x(yz)(4x2y2)y+zx)dydx=0101((xy)(2x)(yz)(2y)+zx)dydx=0101(2x2y+2y2z+zx)dydx

Substitute 4x2y2 for z ,

SFdS=0101[2x2y+2y2(4x2y2)+(4x2y2)x]dydx=0101[2x2y+8y22y2x22y4+4xx3xy2]dydx=01[

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