Consider the solid obtained by rotating the region bounded by the given curves about the x-axis.?

1 Answer

• y = 1 – 9x^2

  y = 0

  finding limits:

  1 – 9x^2 = 0

  1 = 9x^2

  (1/9) = x^2 =====> x = +/- (1/3)

  about x-axis

  A (x) = π ( outer radius )^2 – π ( inner radius )^2

  A (x) = π ( 1 – 9x^2 )^2 – π ( 0 )^2

  A (x) = π ( 1 – 18x^2 + 81x^4 )

  (1/3)

  ∫ π [ ( 1 – 18x^2 + 81x^4 ) ] dx

  -(1/3)

  . . .. .. . . .. . . . . . . … . . . (1/3)

  π [ x – 6x^3 + (81/5)x^5 ) ]

  . . .. . .. . . . . . .. .. . .. . . .(-1/3)

  π [ [ 1/3 – (-1/3) ] – 6 [ (1/3)^3 – (-1/3)^3 ] + (81/5) [ (1/3)^5 – (-1/3)^5 ) ]

  π [ [ 1/3 + (1/3) ] – 6 [ (1/27) – (-1/27) ] + (81/5) [ (1/243) – (-1/243) ) ]

  π [ 2/3 – 6 [ (1/27) + (1/27) ] + (81/5) [ (1/243) + (1/243) ) ]

  π [ 2/3 – 6 [ (2/27) ] + (81/5) [ (2/243) ]

  π [ 2/3 – (12/27) + (2/5) ]

  π (28/45)