Verified answer

Surface area of revolution about the x-axis = A = 2π∫y ds from a to b = 2π∫y√(1 + (dy/dx)^2) dx from a to b.

In this problem:

A = 2π∫(3√x) * √(1 + (3/(2√x))^2) dx from 4 to 10

= 2π∫3*√(x + 9/4) dx from 4 to 10

Let z = x + 9/4 => dz = dx

The integral becomes:

= 6π∫√z dz from 25/4 to 49/4 (Changed the limits after change of variable. Basically plug the old limits into z = x + 9/4.)

= 4π*((49/4)^(3/2) - (25/4)^(3/2))

= π/2*(7^3 - 5^3) = 109π