Let S be the region of the xy-plane bounded above by the curve ((x^3)*y)=64, below by the line y=1, on the left by the line x=2, and on the right by the line x=4.

Find the volume of the solid obtained by rotating S around
(a) the x axis
(b) the y axis
(c) the line x=2

To find the volume of the solid obtained by rotating the region S around different axes, we can use the method of cylindrical shells.

(a) To rotate S around the x-axis, we integrate with respect to y.

Step 1: Find the limits of integration for y.
Since the region S is bounded below by the line y=1, the lower limit of integration for y is 1.
To find the upper limit of integration for y, we need to solve the equation ((x^3)*y)=64 for y when x=4.
Substitute x=4 into the equation: (4^3)*y=64
Simplify: 64y=64
Divide by 64: y=1
Therefore, the upper limit of integration for y is 1.

Step 2: Set up the integral.
The volume element of a cylindrical shell is given by dV=2πx*f(x)*dx, where f(x) is the height of the shell at x.

Let x be the radius of the shell, y be the height of the shell, and f(x) be the distance between the curves y=1 and ((x^3)*y)=64.

Since S is bounded on the left by x=2 and on the right by x=4, the limits of integration for x are 2 and 4.

The height of the shell at x is the difference between the upper curve ((x^3)*y)=64 and the lower curve y=1, which can be written as f(x)=((x^3)*64)-1.

The volume of the solid obtained by rotating S around the x-axis is given by the integral:
V = ∫[from 2 to 4] 2πx*((x^3)*64-1) dx

(b) To rotate S around the y-axis, we integrate with respect to x.

Step 1: Find the limits of integration for x.
Since the region S is bounded on the left by x=2 and on the right by x=4, the lower limit of integration for x is 2.
To find the upper limit of integration for x, we need to solve the equation ((x^3)*y)=64 for x when y=1.
Substitute y=1 into the equation: (x^3)*1=64
Simplify: x^3=64
Take the cube root: x=4
Therefore, the upper limit of integration for x is 4.

Step 2: Set up the integral.
The volume element of a cylindrical shell is given by dV=2πy*f(y)*dy, where f(y) is the distance between the curves x=2 and ((x^3)*y)=64 at a given y.

Let y be the radius of the shell, x be the height of the shell, and f(y) be the difference between the curves x=2 and ((x^3)*y)=64.

Since S is bounded above by ((x^3)*y)=64 and bounded below by y=1, the limits of integration for y are 1 and ((x^3)*y)=64.

The distance between the curves x=2 and ((x^3)*y)=64 can be written as f(y)=4-2=2.

The volume of the solid obtained by rotating S around the y-axis is given by the integral:
V = ∫[from 1 to 64] 2πy*2 dy

(c) To rotate S around the line x=2, we subtract volumes of two solids.

Step 1: Find the limits of integration for y.
Since the region S is bounded below by y=1, the lower limit of integration for y is 1.
To find the upper limit of integration for y, we need to solve the equation ((x^3)*y)=64 for y when x=2.
Substitute x=2 into the equation: (2^3)*y=64
Simplify: 8y=64
Divide by 8: y=8
Therefore, the upper limit of integration for y is 8.

Step 2: Set up the integral.
The volume element of a cylindrical disk is given by dV=π*r^2*dx, where r is the radius of the disk and dx is the thickness of the disk.

To find the volume of the solid obtained by rotating the region S around x=2, we subtract the volume of the hole inside S (formed by rotating the line x=2) from the volume of the solid obtained by rotating S around the x-axis.

The volume of the hole inside S is given by the integral:
V_hole = ∫[from 1 to 8] π*(2^2) dy

The volume of the solid obtained by rotating S around x=2 is given by the integral:
V = ∫[from 1 to 8] 2πx*((x^3)*64-1) dx - ∫[from 1 to 8] π*(2^2) dy

Please note that the exact expressions of the integrals need to be evaluated to get the final numerical values for the volume of the solids.

To find the volume of the solid obtained by rotating a region around an axis, we can use the method of cylindrical shells or the method of disks or washers. Let's use the method of cylindrical shells.

(a) The x-axis: To find the volume of the solid obtained by rotating the region S around the x-axis, we will integrate the formula for the volume of a cylindrical shell:

V = ∫ 2πx * h(x) * dx

Here, h(x) is the height of the cylindrical shell at a given x-value, and dx represents an infinitesimally small change in x.

In this case, the height of the cylindrical shell is given by the difference between the upper and lower boundaries of the region S. The upper boundary is the curve ((x^3)y) = 64, and the lower boundary is the line y = 1. Therefore, h(x) = ((x^3)y) - 1.

To find the limits of integration, we need to determine the x-values where the boundaries intersect. The curve ((x^3)y) = 64 intersects the line y = 1 when ((x^3)*1) = 64. Solving for x, we get x = 4.

Now, we can set up the integral:

V = ∫[2,4] 2πx * (((x^3)*y)-1) dx

To find the y-value corresponding to a given x-value, we can rearrange the equation ((x^3)*y) = 64 to solve for y:

y = 64/(x^3)

Substituting this into the integral, we get:

V = ∫[2,4] 2πx * (((x^3)*(64/(x^3)))-1) dx
= ∫[2,4] 2πx * (64/x^3 - 1) dx

Simplifying, we have:

V = 2π ∫[2,4] (64 - x^3) dx

Evaluating this integral will give us the volume of the solid obtained by rotating S around the x-axis.

(b) The y-axis: To find the volume of the solid obtained by rotating the region S around the y-axis, we will use a similar approach.

In this case, the cylindrical shells will have a radius equal to the x-value, as we are rotating around the y-axis. Therefore, the formula for the volume of a cylindrical shell becomes:

V = ∫ 2πy * h(y) * dy

Here, h(y) is the height of the cylindrical shell at a given y-value, and dy represents an infinitesimally small change in y.

The height of the cylindrical shell is given by the difference between the right and left boundaries of the region S. The right boundary is the line x = 4, and the left boundary is the line x = 2. Therefore, h(y) = 4 - 2 = 2.

To find the limits of integration, we need to determine the y-values where the boundaries intersect. The curve ((x^3)y) = 64 intersects the line y = 1 when ((x^3)*1) = 64. Solving for x, we get x = 4. Therefore, the upper limit of integration is y = 64/(4^3) = 1.

Now, we can set up the integral:

V = ∫[0,1] 2πy * 2 dy

Simplifying, we have:

V = 4π ∫[0,1] y dy

Evaluating this integral will give us the volume of the solid obtained by rotating S around the y-axis.

(c) The line x = 2: To find the volume of the solid obtained by rotating the region S around the line x = 2, we can use the method of cylindrical shells.

In this case, the cylindrical shells will have a radius equal to the distance between the line x = 2 and the x-value of the cylindrical shell, as we are rotating around the line x = 2. Therefore, the formula for the volume of a cylindrical shell becomes:

V = ∫2πr * h(r) * dr

Here, h(r) is the height of the cylindrical shell at a given radius r, and dr represents an infinitesimally small change in r.

The height of the cylindrical shell is still given by the difference between the upper and lower boundaries of the region S. The upper boundary is the curve ((x^3)y) = 64, and the lower boundary is the line y = 1. Therefore, h(r) = ((r+2)^3)y - 1.

To find the limits of integration, we need to determine the r-values where the boundaries intersect. The curve ((x^3)y) = 64 intersects the line y = 1 when ((x^3)*1) = 64. Solving for x, we get x = 4. Therefore, the upper limit of integration is r = 4 - 2 = 2.

Now, we can set up the integral:

V = ∫[0,2] 2πr * (((r+2)^3)y)-1) dr

To find the y-value corresponding to a given r-value, we can rearrange the equation ((x^3)*y) = 64 to solve for y:

y = 64/((r+2)^3)

Substituting this into the integral, we get:

V = ∫[0,2] 2πr * ((((r+2)^3)*(64/((r+2)^3)))) - 1) dr
= ∫[0,2] 2πr * (64 - 1) dr
= 63π ∫[0,2] r dr

Evaluating this integral will give us the volume of the solid obtained by rotating S around the line x = 2.