Find the volume of the solid obtained by rotating the region bounded by y= x^(1/3), and y= x about the line y=1

here is a picture of what we are doing

http://www.wolframalpha.com/input/?i=plot+y%3D+x%5E%281%2F3%29+%2C+y+%3D+x%2C+y+%3D+1

clearly y = x and y = x^(1/3) intersect at (0,0) and (1,1)
so

outer radius = R = x-1 -->R^2 = x^2-2x+1
inner radius = r = x^(1/3) - 1) --> r^2 = x^(2/3)-2x^(1/3) + 1)

V = π∫(x^2 - 2x + 1 - x^(2/3) + 2x^(1/3) - 1) dx from 0 to 1
= π∫(x^2 - 2x - x^(2/3) + 2x^(1/3) dx from 0 to 1
= π [ x^3/3 - x^2 - (3/5)x^(5/3) + (3/2)x^(4/3) ] from 0 to 1
= π( 1/3 - 1 - 3/5 + 3/2 - 0 ...)
= 7π/30

check my steps, I should have written them out on paper first.

To find the volume of the solid obtained by rotating the region bounded by the curves y = x^(1/3) and y = x about the line y = 1, we will use the method of cylindrical shells.

Step 1: Find the points of intersection.
Set the two equations equal to each other:
x^(1/3) = x
Raise both sides to the power of 3 to get rid of the fractional exponent:
x = x^3
Rearrange the equation:
x^3 - x = 0
Factor out x:
x(x^2 - 1) = 0
Solve for x:
x = 0 and x^2 - 1 = 0
Simplifying further:
x = 0 and x = ±1
Therefore, the curves intersect at the points (0,0), (1,1), and (-1,1).

Step 2: Set up the integral.
The volume of the solid can be found using the formula:
V = ∫(2πrh) dx
where r refers to the radius and h refers to the height of each cylindrical shell.

To find the radius, we need to calculate the distance between the line of rotation (y = 1) and the curve y = x^(1/3). This is given by r = 1 - x^(1/3).

The height of each cylindrical shell is given by h = y2 - y1, which is the difference in y-values between the two curves at the corresponding x-values.

Step 3: Determine the limits of integration.
Since the curves intersect at x = -1 and x = 1, the limits of integration for x are -1 to 1.

Step 4: Calculate the volume.
The volume can be expressed as:
V = ∫(2πrh) dx
= ∫(2π(1 - x^(1/3))(x^(1/3))) dx
= ∫(2π(x^(1/3) - x^(2/3))) dx
= 2π ∫(x^(1/3) - x^(2/3)) dx

To find this integral, let's integrate each term separately:

∫x^(1/3) dx:
Using the power rule of integration, add one to the exponent and divide by the new exponent:
= (3/4)x^(4/3)

∫x^(2/3) dx:
Again, using the power rule of integration:
= (3/5)x^(5/3)

Putting everything together, the integral becomes:
V = 2π [(3/4)x^(4/3) - (3/5)x^(5/3)]

Step 5: Evaluate the integral.
To find the volume, we need to evaluate the integral between the limits of integration -1 and 1:

V = 2π ∫[-1 to 1] [(3/4)x^(4/3) - (3/5)x^(5/3)] dx

Plugging in the limits of integration and simplifying, we get:

V = 2π [(3/4)(1)^(4/3) - (3/5)(1)^(5/3)] - [(3/4)(-1)^(4/3) - (3/5)(-1)^(5/3)]
V = 2π [(3/4) - (3/5) - (3/4) + (3/5)]
V = 2π [(3/5) - (3/5)]
V = 0

Therefore, the volume of the solid obtained by rotating the region bounded by y = x^(1/3) and y = x about the line y = 1 is 0 units cubed.

To find the volume of the solid obtained by rotating the region bounded by y = x^(1/3) and y = x about the line y = 1, we can use the method of cylindrical shells.

First, let's sketch the region and the axis of rotation:

1. Plot the graphs of y = x^(1/3) and y = x on a coordinate plane.
2. Draw the line y = 1.
3. Identify the region bounded by these curves.

Now, let's set up the integral to calculate the volume using cylindrical shells:

1. Consider an infinitesimally thin vertical strip with thickness dx at a distance x from the y-axis.
2. The height of the strip is the difference between the y-values of the curves y = x^(1/3) and y = x. So, the height is (x^(1/3) - x).
3. The circumference of the cylindrical shell is 2π times the distance of the strip from the axis of rotation, which is (1 - x) since we are rotating about y = 1.
4. The volume of this cylindrical shell is given by the formula: dV = 2π(x^(1/3) - x)(1 - x)dx.

To find the total volume, we need to integrate the formula for dV from the appropriate range of x-values:

1. Set up the integral: V = ∫[a, b] 2π(x^(1/3) - x)(1 - x) dx, where [a, b] is the x-range of the region.
2. Determine the limits of integration, which are the x-values where the two curves intersect. Solve the equation x^(1/3) = x for x: x = 0 and x = 1.
3. Rewrite the integral with the appropriate limits: V = ∫[0, 1] 2π(x^(1/3) - x)(1 - x) dx.
4. Evaluate the integral using calculus techniques to find the volume.

By following these steps, you should be able to determine the volume of the solid obtained by rotating the region bounded by y = x^(1/3) and y = x about the line y = 1.

we can check using shells.

V = ∫[0,1] 2πrh dy
where r = 1-y and h = y-y^3
V = 2π∫[0,1] (1-y)(y-y^3) dy
= 2π∫[0,1] y^4-y^3-y^2+y dy
= 2π(1/5 y^5 - 1/4 y^4 - 1/3 y^3 + 1/2 y^2) [0,1]
= 2π(1/5 - 1/4 - 1/3 + 1/2)
= 2π(7/60)
= 7π/30

So, looks like Reiny's done it again!