Q.311. Surface area generated by revolving a curve.

Question 311.
Find the area of the surface generated by revolving x = (1/2√2) (y^2 - lny)
about y-axis between y = 1 and y = 2.

Answer 311.
x = (1/2√2) (y^2 - lny)
=> dx/dy = (1/2√2) (2y - 1/y)

Surface area
= 2π ∫ (y=1 to 2) x √[1 + (dx/dy)^2] dy
= 2π ∫ (y=1 to 2) * (1/2√2) (y^2 - lny) * √[1 + (1/8)(2y - 1/y)^2] dy
= (π/4) ∫ (y=1 to 2) (y^2 - lny) √[8 + (2y - 1/y)^2] dy
= (π/4) ∫ (y=1 to 2) (y^2 - lny) (2y + 1/y) dy
= (π/4) ∫ (y=1 to 2) (2y^3 - 2ylny + y - lny/y) dy
= (π/4) [y^4/2 - 2(y^2/2 lny - y^2/4) + y^2/2 - (lny)^2/2] ... (y=1 to 2)
= (π/4) [y^4/2 - y^2lny + y^2 - (lny)^2/2] ... (y=1 to 2)
= (π/4) [ (8 - 4ln2 + 4 - (ln2)^2 / 2) - (1/2)]
= (π/8) [21 - 8ln2 - (ln2)^2]
= (π/8) [21 - 5.5452 - 0.4805]
= 5.8804.

Link to YA!