Question

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.)

Answer 1

Answer:

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:

                           $S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx$

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

                            $f'(x) = \frac{21-x}{\sqrt{42x-x^2} }$

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

                            $f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }$

- Now use the surface area formula:

                           $S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi$

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

Answer 2

Answer:

210π m^3

Step-by-step explanation:

Area of surface of revolution about the x axis = $\int\limits^p_q {2\pi y(x)\sqrt{1 + y'(x)^2} } \, dx$

$y(x) =\sqrt{42x-x^2} \\\\y'(x) =\frac{\frac{1}{2}( 42-2x)}{\sqrt{42x-x^2} } \\\\y'(x) =\frac{21-x}{\sqrt{42x-x^2} }$

Area of surface of revolution about the x axis =

$\int\limits^p_q {2\pi y(x)\sqrt{1 + y'(x)^2} } \, dx \\\\=\int\limits^6_1 {2\pi * \sqrt{42x-x^2}\sqrt{1 + (\frac{21-x}{\sqrt{42x-x^2} })^2} } \, dx =\int\limits^6_1 {2\pi * \sqrt{42x-x^2}\sqrt{ (\frac{42x-x^2 + (21-x)^2}{42x-x^2 })} } \, dx\\\\=\int\limits^6_1 {2\pi * \sqrt{ ({42x-x^2 + 441-42x+x^2 })} } \, dx\\\\=\int\limits^6_1 {2\pi * \sqrt{ 441 } } \, dx\\\\=\int\limits^6_1 {2\pi * 21 } \, dx\\\\ =42\pi (6-1) = 210\pi$