Find the volume or the solid generated by revolving around the region bounded by the graphs of the equations about line x=6. xy=6, y=2, y=6, and x=6

y=2 intersects xy=6 at (3,2)

So we are rotating the region bounded by
(3,2) to (6,2), then down to (6,1) and joining the curve back to (3,2)

Vol = π∫(6-x)^2 dy from y = 1 to 2
= π∫(36 - 12x + x^2) dy
=π∫(36 - 12(6/y) + 36/y^2) dy from 1 to 2
= π[ 36y - 72lny - 36/y ] from 1 to 2
= π( (72 - 72ln2 - 18) - (36 - 72ln1 - 36) )
= π(54 - 72ln2)
= 18π(3 - 4ln2) or appr 12.86

check my arithmetic

To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x=6, we can use the method of cylindrical shells.

First, let's sketch the given region and the line x=6 to visualize the problem.

The region is bounded by the graphs of xy=6, y=2, y=6, and x=6. By solving the equation xy=6 for y, we get y=6/x. The graph of this equation is a hyperbola that intersects the y-axis at (0,6) and the x-axis at (6,0).

Furthermore, the line y=2 is a horizontal line passing through y=2, and x=6 is a vertical line passing through x=6.

By rotating this region about the line x=6, we will obtain a solid with a cylindrical shape.

To find the volume using cylindrical shells, we need to integrate the surface area of all the infinitely thin cylindrical shells formed by the rotation.

The height of each cylindrical shell can be expressed as the difference between the upper and lower y-values of the region, which is (6 - 2) = 4.

The radius of each cylindrical shell can be expressed as the distance between the line x=6 and any point on the curve. Since we are revolving around the line x=6, the distance will be the x-value of each point on the curve.

To set up the integral, we will integrate the circumference of each cylindrical shell multiplied by its height from the lower x-value to the upper x-value of the region.

Let's calculate the volume using the formula:

V = ∫[a,b] 2πrh dx

where r is the distance from the line x=6 to any point on the curve and h is the height of each cylindrical shell.

In this case, the lower x-value (a) is 2 and the upper x-value (b) is 6.

The distance from x=6 to any point on the curve is 6 - x.

Therefore, the volume can be expressed as:

V = ∫[2,6] 2π(6-x)(4) dx

Now, we can proceed to solve this integral to find the volume of the solid generated by revolving the region about the line x=6.

To find the volume of the solid generated by revolving around the line x = 6, we will use the method of cylindrical shells.

Step 1: Sketch the region bounded by the graphs of the equations.

The region is bounded by the curves xy = 6, y = 2, y = 6, and x = 6. It looks like a rectangle with one corner cut off and a hyperbola passing through it. The rectangle has sides parallel to the coordinate axes and vertices at (6,2) and (6,6).

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Step 2: Determine the limits of integration.

The region is bounded by x = 6 on the right and by the hyperbola xy = 6 on the left. The values of y range from 2 to 6.

Step 3: Set up the integral for the volume.

The volume of the solid can be calculated by integrating the area of the cylindrical shells. The height of each shell will be given by the difference in x-coordinates, while the radius will be given by the corresponding y-coordinate.

The volume can be expressed as:

V = ∫[a to b] 2πx * h(x) * dx

Where a is the lower limit of x (intersection point of xy=6 and y=2) and b is the upper limit of x (x=6).

Step 4: Find the height of each shell, h(x), and the limits of integration.

To find the height h(x), we need to express it in terms of x. From the equation xy = 6, we can solve for y:

y = 6 / x

The height h(x) is the difference between the y-values corresponding to the rectangle and hyperbola curves:

h(x) = 6 - 2 = 4

Given that the region is bounded by x = 6 and xy = 6, the limits of integration for x are 6 / 6 = 1 and 6 / 1 = 6.

Step 5: Evaluate the integral.

The integral to calculate the volume becomes:

V = ∫[1 to 6] 2πx * 4 * dx

V = 8π ∫[1 to 6] x dx

To evaluate the integral, we use the power rule:

V = 8π * [x^2 / 2] from 1 to 6

V = 8π * [6^2 / 2 - 1^2 / 2]

V = 8π * [18 - 0.5]

V = 8π * 17.5

Step 6: Simplify the result.

V ≈ 140π cubic units

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line x = 6 is approximately 140π cubic units.