A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume that x and y are measured in meters. The spherical zone generated when the curve y = Squareroot 42x - x^2 on the interval [1, 6] is revolved about the x-axis. The volume of paint needed is m^3. (Type an exact answer, using pi as needed.)

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

- The derivative of the function f'(x) is as follows:

- The square of derivative of f(x) is:

- Now use the surface area formula:

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

210π m^3

Step-by-step explanation:

Area of surface of revolution about the x axis =

Area of surface of revolution about the x axis =

what does x eaqual

step-by-step explanation: