The curve y = √(10-x) ; 8 ≤ x ≤ 10 is revolved around the x-axis. Find the surface Area

Question

The curve y = √(10-x) ; 8 ≤ x ≤ 10 is revolved around the x-axis. Find the surface Area

If possible please explain the step by step process in a video.

Answer 1

Let y = f(x) = √(10 - x), 8 ≤ x ≤ 10

S = Surface area if f(x) is revolved about the x-axis

S = 2π∫_{(8 to 10)} f(x)√[1 + (f'(x))²] dx

= 2π∫_{(8 to 10)} √(10 - x) √ [ 1 + ( 1/(4(10-x))] dx

= 2π∫_{(8 to 10)} √(10 - x)√[(41 - 4x) / (4(10 - x))] dx

= π∫_{(8 to 10)} √(41 - 4x) dx

Let u = 41 - 4x. Then du = -4dx. So, dx = -(1/4)du

When x = 8, u = 9 and when x = 10, u = 1.

S = π∫_{(9 to 1)} √u (-1/4)du = (π/4)∫_{(1 to 9)} √udu = (π/6)u^3/2_{(1 to 9)} = (π/6)(27 - 1) = 13π/3