Question

Let R be the region bounded by the following curves. Use the method of your choice to find the volume of the solid generated when R is revolved about the​ x-axis. y = 6x^2 & y = 28 - x^2

Answer 1

The volume of the solid formed by revolving the region between y = 6x^2 and y = 28 - x^2 around the x-axis can be calculated using the disk method, by setting up an integral concerning x for x ranging from -2 to 2, and integrating the square of the difference between the functions times pi.

Step-by-step explanation:

To find the volume of the solid generated when the region bounded by the curves y = 6x^2 and y = 28 - x^2 revolves around the x-axis, we can use the disk method.

First, we need to find the points of intersection of the two curves by setting them equal to each other:

6x^2 = 28 - x^2

x^2 = 4

x = -2 or x = 2

This defines our limits of integration from x = -2 to x = 2. The volume V of the solid of revolution is then given by the integral

V = ∫_{-2}^{2} π (28 - x^2 - 6x^2)^2 dx

After simplifying,

V = ∫_{-2}^{2} π (28 - 7x^2)^2 dx

The resulting integral can be computed to find the volume of the solid.

Answer 2

V = 7168π /3

Step-by-step explanation:

R ( region ) bounded by curves ; y = 6x^2 , y = 28 - x^2

determine the volume of the solid generated when R is revolved about the x-axis

Volume = 7168π /3

attached below is the remaining part of the detailed solution to the problem above

Given that :

y = 6x^2----- ( 1 ) , y = 28 - x^2 ------ ( 2 )

equate ( 1 ) and ( 2 )

6x^2 = 28 - x^2

7x^2 = 28

∴ x = √4 = ± 2