   Chapter 16.7, Problem 21E

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.21. F(x, y, z) = zexy i - 3zexy j + xy k, S is the parallelogram of Exercise 5 with upward orientation

To determine

To find: The value of SFdS for F(x,y,z)=zexyi3zexyj+xyk .

Explanation

Given data:

x=u+v , y=uv , z=1+2u+v , 0u2 and 0v1 .

F(x,y,z)=zexyi3zexyj+xyk (1)

Formula used:

SFdS=DF(ru×rv)dA (2)

ru=xui+yuj+zuk (3)

rv=xvi+yvj+zvk (4)

Find ru .

Substitute u+v for x , uv for y and 1+2u+v for z in equation (3),

ru=u(u+v)i+u(uv)j+u(1+2u+v)k=(1+0)i+(10)j+(0+2+0)k=i+j+2k

Find rv .

Substitute u+v for x , uv for y and 1+2u+v for z in equation (4),

rv=v(u+v)i+v(uv)j+v(1+2u+v)k=(0+1)i+(01)j+(0+0+1)k=ij+k

Find ru×rv .

ru×rv=(i+j+2k)×(ij+k)=|ijk112111|=(1+2)i(12)j+(11)k=3i+j2k

Find F[r(u,v)] .

Substitute u+v for x , uv for y and 1+2u+v for z in equation (1),

F[r(u,v)]={(1+2u+v)e(u+v)(uv)i3(1+2u+v)e(u+v)(uv)j+(u+v)(uv)k} {F(x,y,z)=F[r(u,v)]}={(1+2u+v)eu2v2i3(1+2u+v)eu2v2j+(u2v2)k} {(a+b)(ab)=a2b2}

As the z-component of ru×rv is negative, Modify the equation (2) as (ru×rv)

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