   Chapter 16.7, Problem 28E

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.28. F(x, y, z) = yz i + zx j + xy k, S is the surface z = x sin y, 0 ≤ x ≤ 2, 0 ≤ y ≤ π, with upward orientation

To determine

To find: The value of SFdS for F(x,y,z)=yzi+zxj+xyk .

Explanation

Given data:

z=xsiny , 0x2 and 0yπ .

F(x,y,z)=yzi+zxj+xyk (1)

Formula used:

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

Consider g=z .

Substitute xsiny for z ,

g=xsiny

Apply limits and substitute yz for P , zx for Q , xy for R and xsiny for g in equation (2),

SFdS=0π02((yz)(xsiny)x(zx)(xsiny)y+xy)dxdy=0π02[yz(siny)zx(xcosy)+xy]dxdy

Substitute xsiny for z ,

SFdS=0π02[y(xsiny)(siny)(xsiny)x(xcosy)+xy]dxdy=0π02[xysin2yx3sinycosy+xy]dxdy=0π[12x2ysin2y14x4sinycosy+12x2y]02dy=0π[12(2202)ysin2y14(2404)sinycosy+12(2202)y]dy

Simplify the equation.

SFdS=0π[2ysin2y4sinycosy+2y]dy=0π2ysin2ydy0π4sinycosydy+0π2ydy=0π2y(1cos2y2)dy0π4sinycosydy+[2y22]0π=0πydy+0πycos2ydy0π4sinycosydy+π2

Modify the equation

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