   Chapter 16.7, Problem 7E

Chapter
Section
Textbook Problem

Evaluate the surface integral.7. ∬s y dS, S is the helicoid with vector equation r(u, v) = ⟨u cos v, u sin v, v⟩, 0 ≤ u ≤ 1, 0 ≤ v ≤ π

To determine

To find: The value of SydS .

Explanation

Given data:

r(u,v)=ucosv,usinv,v , 0u1 and 0vπ .

Formula used:

Sf(x,y,z)dS=Df(r(u,v))|ru×rv|dA (1)

ru=xui+yuj+zuk (2)

rv=xvi+yvj+zvk (3)

Find ru .

Substitute ucosv for x , usinv for y and v for z in equation (2),

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 (3),

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 |ru×rv| .

|ru×rv|=|sinvicosvj+uk|=(sinv)2+(cosv)2+(u)2=sin2v+cos2v+u2=(sin2v+cos2v)+u2

Simplify the equation

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