   Chapter 16.6, Problem 56E

Chapter
Section
Textbook Problem

Find the area of the surface with vector equation r(u, v) = ⟨cos3u cos3v, sin3u cos3v, sin3v⟩, 0 ⩽ u ⩽ π, 0 ⩽ v ⩽ 2π. State your answer correct to four decimal places.

To determine

To find: The area of surface with the vector equation r(u,v)=cos3ucos3v,sin3ucos3v,sin3v,0uπ,0v2π .

Explanation

Given data:

The vector equation of the surface is given as follows.

r(u,v)=cos3ucos3v,sin3ucos3v,sin3v,0uπ,0v2π

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 cos3ucos3v,sin3ucos3v,sin3v for r(u,v) in equation (2),

ru=ucos3ucos3v,sin3ucos3v,sin3v=u(cos3ucos3v),u(sin3ucos3v),u(sin3v)=cos3vu(cos3u),cos3vu(sin3u),sin3vu(1)=cos3v(3cos2usinu),cos3v(3sin2ucosu),sin3v(0)

Simplify the equation.

ru=3cos2usinucos3v,3sin2ucosucos3v,0

Calculation of rv :

Substitute cos3ucos3v,sin3ucos3v,sin3v for r(u,v) in equation (3),

rv=vcos3ucos3v,sin3ucos3v,sin3v=v(cos3ucos3v),v(sin3ucos3v),v(sin3v)=cos3uv(cos3v),sin3uv(cos3v),3sin2vcosv=cos3u(3cos2vsinv),sin3u(3cos2vsinv),3sin2vcosv

Simplify the equation.

rv=3cos3ucos2vsinv,3sin3ucos2vsinv,3sin2vcosv

Calculation of ru×rv :

Substitute 3cos2usinucos3v,3sin2ucosucos3v,0 for ru and 3cos3ucos2vsinv,3sin3ucos2vsinv,3sin2vcosv for rv in the expression ru×rv ,

ru×rv=3cos2usinucos3v,3sin2ucosucos3v,0×3cos3ucos2vsinv,3sin3ucos2vsinv,3sin2vcosv

Rewrite and compute the expression as follows.

ru×rv=|ijk3cos2usinucos3v3sin2ucosucos3v03cos3ucos2vsinv3sin3ucos2vsinv3sin2vcosv|=(|3sin2ucosucos3v03sin3ucos2vsinv3sin2vcosv|i|3cos2usinucos3v03cos3ucos2vsinv3sin2vcosv|j+|3cos2usinucos3v3sin2ucosucos3v3cos3ucos2vsinv3sin3ucos2vsinv|k)

Simplify the expression as follows

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