Determine the resulting volume (in cu. units) when enclosed area between the functions below are revolved around the y-axis: 𝑦=𝑥2+1, 𝑦=𝑥2, 𝑦=1 and 𝑦=4

To find the volume of the solid formed by revolving the area between the curves around the y-axis, we can use the method of cylindrical shells. The formula for the volume of a cylindrical shell is V = 2π ∫ [r(h) dr] from a to b, where r is the radius and h is the height of the cylindrical shell.

The radius of the cylindrical shell is the x-coordinate, and the height is the difference between the two functions.

First, we need to find the points of intersection of the curves y=x^2+1 and y=x^2. Setting these equal to each other, we get x^2+1 = x^2, which simplifies to 1=0. This has no solution, so these two curves do not intersect.

Next, we find the points of intersection of the curves y=x^2 and y=1. Setting these equal to each other, we get x^2 = 1, which gives x = -1 and x = 1.

Finally, we find the points of intersection of the curves y=x^2 and y=4. Setting these equal to each other, we get x^2 = 4, which gives x = -2 and x = 2.

So, the region we are interested in is bounded by x=-1, x=1, y=1, and y=4.

The volume of the solid is then given by:

V = 2π ∫ from -1 to 1 [x * ((x^2+1) - x^2) dx] + 2π ∫ from -2 to -1 [x * (4 - x^2) dx] + 2π ∫ from 1 to 2 [x * (4 - x^2) dx]

Solving these integrals, we get:

V = 2π [∫ from -1 to 1 x dx] + 2π [∫ from -2 to -1 (4x - x^3) dx] + 2π [∫ from 1 to 2 (4x - x^3) dx]

V = 2π [0] + 2π [4*(-1) - (-1)^3/4 - (4*(-2) - (-2)^3/4)] + 2π [42 - 2^3/4 - (41 - 1^3/4)]

V = 2π [0] + 2π [-4 - 1/4 - (-8 - 2)] + 2π [8 - 2 - (4 - 1/4)]

V = 2π [0] + 2π [-4.25 + 10] + 2π [6 - 3.75]

V = 2π [0] + 2π [5.75] + 2π [2.25]

V = 2π [0 + 5.75 + 2.25]

V = 2π [8]

V = 16π cubic units.