Find the volume of the solid obtained by rotating the region enclosed by the graphs x=y^(3) and x= y^(1/3) about the y-axis over the interval [0,1].

⌠pi[(y^1/3)^2 - (y^3)^2] dy



= pi INT [y^2/3 - y^6)dy

= pi[(3/5)y^5/3 - (1/7)y^7]│ from 0 to 1

=pi (3/5 - 1/7)
= 16pi/7

To find the volume of the solid obtained by rotating the region enclosed by the graphs x=y^(3) and x=y^(1/3) about the y-axis over the interval [0,1], we can use the method of cylindrical shells.

First, we need to determine the bounds of integration. In this case, the region is enclosed by the two functions x=y^(3) and x=y^(1/3). To find the points of intersection, we set these two functions equal to each other:

y^(3) = y^(1/3)

Raise both sides to the power of 1/3:

(y^(3))^(1/3) = (y^(1/3))^(1/3)

y = y^(1/9)

To find the points of intersection, we then solve the equation:

y = y^(1/9)

Raise both sides to the power of 9:

y^9 = y

Rearranging:

y^9 - y = 0

Factoring out y:

y(y^8 - 1) = 0

Setting each factor equal to zero:

y = 0 or y^8 - 1 = 0

For the second factor, we have:

y^8 - 1 = 0

(y^4 - 1)(y^4 + 1) = 0

Setting each factor equal to zero:

y^4 - 1 = 0 or y^4 + 1 = 0

which gives us:

y = 1, -1

Since we are considering the interval [0,1], the only relevant point is y = 1.

Now, we can set up the integral to calculate the volume using the cylindrical shell method. The volume of each shell is given by the formula:

dV = 2πrh * dx

where r is the distance from the axis of rotation (y-axis) to the shell (which is x in this case), h is the height of the shell, and dx is the thickness of the shell.

In our case, the height of the shell, h, is given by the difference between the upper and lower functions:

h = y^(3) - y^(1/3)

The radius, r, is the x-coordinate of the shell, which is y.

So, the volume of each shell is:

dV = 2πy(y^(3) - y^(1/3)) * dx

To find the total volume, we integrate this expression over the given interval [0,1]:

V = ∫[0,1] 2πy(y^(3) - y^(1/3)) dx

Now we need to express everything in terms of x, so we use the equation y = x^(1/3) as a substitute:

V = ∫[0,1] 2π(x^(1/3))(x^(3) - x^(1/3)) dx

Simplifying:

V = ∫[0,1] 2π(x^(4/3) - x^(2/3)) dx

Integrating term by term:

V = 2π ∫[0,1] (x^(4/3)) dx - 2π ∫[0,1] (x^(2/3)) dx

Evaluating the integrals:

V = 2π (3/7)x^(7/3) - 2π (3/5)x^(5/3) | from 0 to 1

V = 2π [(3/7)(1^(7/3)) - (3/5)(1^(5/3))] - 0

V = 2π [(3/7) - (3/5)]

V = 2π (15/35 - 21/35)

V = 2π (-6/35)

So, the volume of the solid obtained by rotating the region enclosed by the graphs x=y^(3) and x=y^(1/3) about the y-axis over the interval [0,1] is -12π/35 or approximately -1.08 cubic units.