Final answer:
To find the area of the surface generated when the curve y = 4 sqrt(x) for 60 ≤ x ≤ 96 is revolved about the x-axis, you can use the formula for the surface area of a solid of revolution.
Step-by-step explanation:
To find the area of the surface generated when the curve y = 4 sqrt(x) for 60 ≤ x ≤ 96 is revolved about the x-axis, we can use the formula for the surface area of a solid of revolution.
The formula is given by:
Surface Area = 2π ∫[a,b] f(x) √(1 + (f'(x))^2) dx
Plugging in our values, the integral becomes:
Surface Area = 2π ∫[60,96] 4 sqrt(x) √(1 + (1/2sqrt(x))^2) dx
Simplifying and solving the integral, we get:
Surface Area = 2π ∫[60,96] 4 sqrt(x) √(1 + 1/4x) dx
After evaluating the integral, the surface area is:
Surface Area = 517π sq units