The given curve is rotated about the y-axis. Find the area of the resulting surface.

Question

y = 1/4x^2 − 1/2ln x,
4 ≤ x ≤ 5

The answer is supposed to be 64/3*pi

Answer 1

The surface area is the sum of the circumferences of all the little slices between x=4 and 5.

C = 2 pi y

so, the surface area is
Int[2 pi 1/4 x^2 - 1/2 ln x]dx |4:5

Int ln x = x ln x - x

so The area integral F(x) = 2 pi [1/12 x^3 - 1/2(x ln x - x)]
= pi x^3 - pi x ln x + pi x

F(5) = 125pi - 5pi ln 5 + 5pi = 5pi(26 - ln5)

F(4) = = 64pi - 4pi ln 4 + 4pi = 4pi(17 - ln4)

So the surface area is pi[62 - (5ln5 - 4ln4)]

Answer 2

Rats - that was 1/12 x^3, not 1/2. Oh well, make the change and proceed.