Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

y=x^4,y=1; about y=5

To find the volume of the solid obtained by rotating the region bounded by the curves y = x^4 and y = 1 about the line y = 5, we can use the method of cylindrical shells.

First, let's sketch the region bounded by the curves y = x^4 and y = 1. This region is bounded on the x-axis by x = 0 and x = 1, and is enclosed between the curves and the line y = 5.

Since we are rotating the region about the line y = 5, we need to find the height of each individual shell. The height of each shell is the distance between the line y = 5 and the curve at that specific x-value.

Let's denote the radius of each shell as r. The radius of each shell will be the x-coordinate of the point on the curve y = x^4. Therefore, r = x.

The height of each shell, h, is given by the difference between the lines y = 5 and y = x^4. Therefore, h = 5 - x^4.

Now, the volume of each individual cylindrical shell is given by the formula:

V = 2πrhΔx

where Δx represents an infinitesimally small width of each shell.

To find the total volume, we need to integrate this formula over the interval [0,1] with respect to x:

V = ∫[0,1] 2πrh dx

Substituting the values for r and h, we get:

V = ∫[0,1] 2π(x)(5 - x^4) dx

Now, we can integrate this expression to find the volume of the solid:

V = 2π ∫[0,1] (5x - x^5) dx

Evaluating the integral, we get:

V = 2π [ (5/2)x^2 - (1/6)x^6 ] evaluated from x=0 to x=1

V = 2π [ (5/2)(1)^2 - (1/6)(1)^6 ] - [ (5/2)(0)^2 - (1/6)(0)^6 ]

V = 2π [ (5/2) - (1/6) ] - 0

V = 2π [(15/6) - (1/6)]

V = 2π (14/6)

Simplifying this expression, we get:

V = (28/3)π

Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x^4 and y = 1 about the line y = 5 is (28/3)π cubic units.