   Chapter 16.7, Problem 22E

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.22. F(x, y, z) = z i + y j + x k, S is the helicoid of Exercise 7 with upward orientation

To determine

To find: The value of SFdS for F(x,y,z)=zi+yj+xk .

Explanation

Given data:

r(u,v)=ucosv,usinv,v , 0u1 and 0vπ .

F(x,y,z)=zi+yj+xk (1)

Formula used:

SFdS=DF(ru×rv)dA (2)

ru=xui+yuj+zuk (3)

rv=xvi+yvj+zvk (4)

Find ru .

Substitute ucosv for x , usinv for y and v for z in equation (3),

ru=u(ucosv)i+u(usinv)j+u(v)k=(cosv)u(u)i+(sinv)u(u)j+(v)u(1)k=cosvi+sinvj+0k=cosvi+sinvj

Find rv .

Substitute ucosv for x , usinv for y and v for z in equation (4),

rv=v(ucosv)i+v(usinv)j+v(v)k=(u)v(cosv)i+(u)v(sinv)j+v(v)k=u(sinv)i+ucosvj+k=usinvi+ucosvj+k

Find ru×rv .

ru×rv=(cosvi+sinvj)×(usinvi+ucosvj+k)=|ijkcosvsinv0usinvucosv1|=(sinv0)i(cosv0)j+(ucos2v+usin2v)k=sinvicosvj+[u(cos2v+sin2v)]k

ru×rv=sinvicosvj+uk {cos2θ+sin2θ=1}

Find F[r(u,v)]

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