The region bounded by y=3/x, y=0, x=3, and x=5 is rotated about the x-axis. Find the volume of the resulting solid.

V = π ∫(9/x^2) dx from x=3 to 5

= π( -9/x) | 3 to 5
= π(-9/5 - (-9/3) )
= π(6/5)
= 1.2π cubic units

To find the volume of the solid obtained by rotating the region bounded by y=3/x, y=0, x=3, and x=5 about the x-axis, we need to use the method of cylindrical shells.

First, we need to determine the limits of integration for x. The region is bounded by x=3 and x=5, so we will integrate from x=3 to x=5.

Next, we need to express y in terms of x. From the equation y=3/x, we can rewrite it as x=3/y. This gives us x=3/y.

Now, we can start the integration process. The volume of the solid is given by the formula:

V = 2π∫[a to b] x * f(x) * dx

where a and b are the limits of integration, f(x) is the function representing the height of the solid at each x-value, and dx represents an infinitesimally small width element.

In this case, f(x) is the distance from the curve y=3/x to the x-axis, which is simply y=3/x.

Therefore, the volume V can be calculated as follows:

V = 2π∫[3 to 5] x * (3/x) * dx

Simplifying, we get:

V = 2π∫[3 to 5] 3 dx

Integrating, we have:

V = 2π [3x]∣[3 to 5]

Evaluating the limits, we get:

V = 2π [3(5) - 3(3)]

V = 2π (15 - 9)

V = 2π (6)

V = 12π

Therefore, the volume of the resulting solid is 12π cubic units.

To find the volume of the solid obtained by rotating the given region about the x-axis, we can use the method of cylindrical shells.

First, let's visualize the given region. The graph of the function y = 3/x gives a hyperbola that extends from y = 0 to infinity and from x = 3 to x = 5. The region bounded by y = 3/x, y = 0, x = 3, and x = 5 is the shaded area below:

_________
| |
| |
_|_ |
| | |
| | |
| | |
_|___|_______|
(3,0) (5,0)

Next, we need to consider a vertical line segment of width dx (infinitesimally small width) at a given x-coordinate. When this line segment is rotated about the x-axis, it forms a cylindrical shell with thickness dx, height equal to the y-coordinate, and radius equal to the x-coordinate.

The volume of each cylindrical shell is given by the formula:

dV = 2πrh dx

where r is the radius, h is the height, dx is the width of the shell, and dV is the volume element.

To find the volume of the entire solid, we need to integrate the volume element over the entire region. We integrate from x = 3 to x = 5, since the region is bounded by these x-values.

V = ∫[3,5] 2πrh dx

Now, let's express the radius (r) and height (h) in terms of x.

Since the region is bounded by y = 3/x and y = 0, the height of each cylindrical shell is given by y = 3/x, and the radius is given by x.

Therefore, r = x and h = 3/x.

Substituting these values into the integral formula, we have:

V = ∫[3,5] 2π(3/x)(x) dx
= 2π ∫[3,5] 3 dx
= 2π [3x] | [3,5]
= 2π [15 - 9]
= 12π

Hence, the volume of the resulting solid is 12π cubic units.