   Chapter 16.7, Problem 24E

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.24. F(x, y, z) = -x i - y j + z3 k, S is the part of the cone z   =   x 2 + y 2 between the planes z = 1 and z = 3 with downward orientation

To determine

To find: The value of SFdS for F(x,y,z)=xiyj+z3k .

Explanation

Given data:

z=x2+y2 , plane lies between z=1 and z=3 .

F(x,y,z)=xiyj+z3k (1)

Formula used:

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

Consider g=z .

Substitute x2+y2 for z ,

g=x2+y2

The D is annular region {1x2+y29} . Since S has downward orientation, modify equation (2).

SFdS=D(PgxQgy+R)dA

Substitute x for P , y for Q , z3 for R and x2+y2 for g ,

SFdS=D[(x)(x2+y2)x(y)(x2+y2)y+z3]dA=D[(x)(xx2+y2)(y)(yx2+y2)+z3]dA=D[(x2x2+y2)+(y2x2+y2)+z3]dA=D[(x2+y2x2+y2)+z3]dA

Substitute x2+y2 for z ,

SFdS=D[(x2+y2x2+y2)+(x2+y2)3]dA (3)

Let x=rcosθ , y=rsinθ and limits for r and θ are 1 to 3 and 0 to 2π .

Modify the equation (3) as follows

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