   Chapter 16.6, Problem 48E

Chapter
Section
Textbook Problem

Find the area of the surface.48. The helicoid (or spiral ramp) with vector equation r(u, v) = u cos v i + u sin v j + v k, 0 ⩽ u ⩽ 1, 0 ⩽ v ⩽ π

To determine

To find: The area of helicoid with the vector equation r(u,v)=ucosvi+usinvj+vk,0u1,0vπ .

Explanation

Given data:

The vector equation of the helicoid is given as follows.

r(u,v)=ucosvi+usinvj+vk,0u1,0vπ

Formula used:

Write the expression to find the surface area of the plane with the vector equation r(u,v) .

A(S)=D|ru×rv|dA (1)

Here,

ru is the derivative of vector equation r(u,v) with respect to the parameter u and

rv is the derivative of vector equation r(u,v) with respect to the parameter v .

Write the expression to find ru .

ru=u[r(u,v)] (2)

Write the expression to find rv .

rv=v[r(u,v)] (3)

Calculation of ru :

Substitute ucosvi+usinvj+vk for r(u,v) in equation (2),

ru=u(ucosvi+usinvj+vk)=u(ucosv)i+u(usinv)j+u(v)k=cosvi+sinvj+(0)k

Calculation of rv :

Substitute ucosvi+usinvj+vk for r(u,v) in equation (3),

rv=v(ucosvi+usinvj+vk)=v(ucosv)i+v(usinv)j+v(v)k=usinvi+ucosvj+(1)k

Calculation of ru×rv :

Substitute cosvi+sinvj+(0)k for ru and usinvi+ucosvj+(1)k for rv in the expression ru×rv ,

ru×rv=(cosvi+sinvj+(0)k)×(usinvi+ucosvj+(1)k)

Rewrite and compute the expression as follows.

ru×rv=|ijkcosvsinv0usinvucosv1|=|sinv0ucosv1|i|cosv0usinv1|j+|cosvsinvusinvucosv|k=(sinv0)i(cosv0)j+[ucos2v(usin2v)]k=sinvicosvj+u(cos2v+sin2v)k

Simplify the expression as follows

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