Engineering Board Exam Reviewer · Applications: Volume, Arc Length, Surface Area · medium

Find the area of the surface generated by revolving y=x3, 0≤ x≤1, about the x-axis.

  1. π ≤ft(-1 + 10 sqrt(10) )27 3.563
  2. π ≤ft(-1 + 10 sqrt(10) )9 10.689
  3. 2 π ≤ft(-1 + 10 sqrt(10) )27 7.126
  4. π ≤ft(1 + 10 sqrt(10) )27 3.796

Answer

A. π ≤ft(-1 + 10 sqrt(10) )27 3.563

Solution

S=2π 01x3sqrt(1+9x4),dx. Let u=1+9x4, du=36x3dx: S=2π/36 110u1/2du=π/18· 23≤ft(103/2-1 )=π/27≤ft(10sqrt(10)-1 ).

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