use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. xy=1, x=0, y=1, y=3

Using shells, you have thickness dy, so

v = ∫[1,3] 2πrh dy
where r = y and h = x = 1/y
v = 2π∫[1,3] y*1/y dy
= 2π∫[1,3] dy
= 2πy [1,3]
= 2π(3-1)
= 4π

That strikes me as odd. Let's try discs (washers) of thickness dx. The volume on [0,1/3] is just a hollow cylinder, of volume (1/3)π(3^2-1^2) = 8/3 π.

The remaining part has volume

v = ∫[1/3,1] π(R^2-r^2) dx
where R=y=1/x, r=1
v = π∫[1/3,1] (1/x^2-1) dx
= π(-1/x-x)[1/3,1]
= π((-1-1)-(-3-1/3))
= π(-2+10/3)
= 4/3 π
Add to that the 8/3 π from the constant part, and you have 4π

To find the volume of the solid obtained by rotating the region bounded by the curves about the x-axis using the method of cylindrical shells, follow these steps:

Step 1: Sketch the region

Sketch the region bounded by the curves xy = 1, x = 0, y = 1, and y = 3 on the xy-plane. This will help visualize the region you're going to rotate.

Step 2: Determine the interval for integration

Since we are rotating the region about the x-axis, the interval for integration will be determined by the x-values of the region. In this case, the region is bounded by x = 0 and xy = 1.

From xy = 1, we can solve for x: x = 1/y.

Thus, the interval for integration is from x = 0 to x = 1/3 (since y ranges from 1 to 3).

Step 3: Set up the integral using cylindrical shells

The volume of the solid can be calculated by integrating the circumference of each cylindrical shell times its height and thickness.

The circumference of each cylindrical shell at a given x-value is equal to 2πx. The height of each shell is the difference in y-values at that x-value (3 - 1 = 2). The thickness of each shell is dx.

So, the integral that represents the volume of the solid is:
V = ∫[from 0 to 1/3] (2πx)(2) dx

Step 4: Evaluate the integral

Integrate the above expression from 0 to 1/3:
V = ∫[from 0 to 1/3] 4πx dx

Integrating 4πx with respect to x gives:
V = 2πx^2 | [from 0 to 1/3]
V = 2π(1/9 - 0)
V = 2π/9

Therefore, the volume of the solid obtained by rotating the region bounded by the curves about the x-axis is (2π/9) cubic units.