91Ó°ÊÓ

To find the volume of a solid formed by rotating a region around an axis, we often use the cylindrical shells method. This method is very helpful when revolving around the y-axis. Think of it like peeling an onion, where each peel or shell has a radius, height, and thickness. Here's how it works:

• The radius is the distance from the axis of rotation, which for our setup is the variable x.
• The height is determined by the function we are considering, which in this case is the natural log function, \(\ln x\).
• The thickness is a small change in x, represented as \(\Delta x\).

The volume of each cylindrical shell can be calculated using the formula \(dV = 2 \pi x (\ln x) \Delta x\). By integrating this formula from the lower boundary to the upper boundary of x, you get the total volume of the solid. For the example problem, the limits of integration are from \(x = 1\) to \(x = 2\), making it \(\int_1^2 2\pi x \ln x\, dx\). This approach effectively captures the entire volume by summing up the infinitesimal shells.

When faced with integrals involving products of functions, like \(\int x \ln x\, dx\), the method of integration by parts becomes a powerful tool. Integration by parts is based on the product rule for differentiation and is expressed as: \(\int u \, dv = uv - \int v \, du\). Here’s how to apply it:

• Choose \(u\) and \(dv\) based on ease of differentiation and integration. In our case, we set \(u = \ln x\) and \(dv = x \, dx\).
• Differentiating \(u\) gives \(du = \frac{1}{x} \, dx\) and integrating \(dv\) gives \(v = \frac{x^2}{2}\).

Using the integration by parts formula, we have: \(\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \frac{1}{x} \, dx\). Simplifying further and substituting back provides the solution to the integral. This step-by-step approach is particularly efficient in solving more complex integrals and allows you to easily evaluate specific limits within a definite integral.

When rotating an area around the x-axis, the disk and washer method proves to be highly effective. By imagining the section spun around the axis, you can envision it transforming into a series of disks or washers. This method creates a different approach to volume calculations compared to cylindrical shells. Here’s a breakdown:

• The radius of each disk is the height of the function above the x-axis, here given by \(y = \ln x\).
• The thickness of the washer is again represented by \(\Delta x\).

The volume of a disk is calculated with \(dV = \pi (\ln x)^2 \, dx\). Integrating this formula over the specified bounds (from \(x = 1\) to \(x = 2\)) provides the total volume of the solid obtained by revolving the region. In complex cases, computing such integrals can be analytically tricky and numerical methods may be used to approximate results, as seen in our example where the integral \(\int_1^2 (\ln x)^2 \, dx\) was approximated. This demonstrates how the disk method provides a direct way to calculate volumes by integrating over the cross-sectional area formed by disks or washers.