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In the figure below, set-up the integral expression when the area bound by the functions 𝑦=1𝑥, 𝑦=𝑥, and 𝑥=4 is revolved around the 𝑥=4 axis using the method of disks.

Question

In the figure below, set-up the integral expression when the area bound by the functions 𝑦=1𝑥, 𝑦=𝑥, and 𝑥=4 is revolved around the 𝑥=4 axis using the method of disks.

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Solution

The area bounded by the functions y = 1/x, y = x, and x = 4 is a region in the first quadrant of the xy-plane. When this region is revolved around the line x = 4, it forms a solid of revolution. The volume of this solid can be found using the method of disks (or washers).

The method of disks involves slicing the solid into thin disks perpendicular to the axis of rotation, and then summing up the volumes of these disks. The volume of each disk is given by the formula πr²h, where r is the radius of the disk and h is its thickness.

The radius of each disk is the distance from the disk to the line x = 4. For a disk at position x, this distance is given by 4 - x. The height of each disk is the difference between the y-values of the two functions at position x. This is given by x - 1/x.

Therefore, the volume of the solid is given by the integral from 1 to 4 of π(4 - x)²(x - 1/x) dx.

This integral can be simplified to π ∫ from 1 to 4 of (16x - 8/x - x³ + 1) dx.

This is the integral expression for the volume of the solid formed when the area bounded by the functions y = 1/x, y = x, and x = 4 is revolved around the line x = 4.

This problem has been solved

Similar Questions

In the figure below, the area bound by the functions  𝑦=1𝑥, 𝑦=4, 𝑥=4 and  is to be revolved around the 𝑥=4 axis. Which of the following methods would be valid in determining the volume of the resulting revolved plane area if the differential strip to be revolved is horizontal?

Determine the resulting volume (in cu. units) when enclosed area between the functions below are revolved around the y-axis: 𝑦=𝑥2+1, 𝑦=𝑥2, 𝑦=1 and 𝑦=4

Use volume by disk method to Determine the volume of the surface of revolution about1. 𝒙 = 𝟎2. 𝒚 = 𝟎3. 𝒙 = +𝒂

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The area bounded by 𝑥=𝑦2−2 and 𝑥=𝑒𝑦  between  𝑦=−1 and 𝑦=1 is revolved about the line 𝑦=1. The Volume integral for the solid generated is

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