Question

Sketch the region \(R\) bounded by the graphs of the given equations and show a typical horizontal slice. Find the volume of the solid generated by revolving \(R\) about the \(y\) -axis. \(x=\frac{2}{y}, y=2, y=6, x=0\)

Short answer

The volume is \(\frac{4\pi}{3}\).

Step-by-step solution

Understand the Boundaries

Identify the curves and lines that form the boundaries of the region. Here, the region is bounded by the curve \(x = \frac{2}{y}\), the horizontal lines \(y = 2\) and \(y = 6\), and the vertical line \(x = 0\).

Sketch the Region

Draw the lines and curve on the coordinate plane. The curve \(x = \frac{2}{y}\) is a decreasing hyperbola that doesn't cross \(x = 0\). Draw horizontal boundaries at \(y = 2\) and \(y = 6\) to finalize the region \(R\).

Visualize a Horizontal Slice

Envision a typical horizontal slice of thickness \(dy\) across the region \(R\). This slice extends horizontally from \(x = 0\) to \(x = \frac{2}{y}\), across the range of \(y\) values from 2 to 6.

Setup the Integral for Volume

The area of the typical horizontal slice is \( \text{width} = \frac{2}{y} \). The circumference of the slice when revolved around the \(y\)-axis is \(2\pi x = 2\pi \frac{2}{y} = \frac{4\pi}{y} \). The volume element \(dV\) can be expressed as \(dV = \pi\left(\frac{2}{y}\right)^2 dy\).

Integrate to Find Volume

Integrate the volume element from \(y = 2\) to \(y = 6\):\[V = \int_{2}^{6} \pi \left(\frac{2}{y}\right)^2 dy\].Calculate the integral step by step:

Simplify and Compute the Integral

Simplify the integral:\[V = \int_{2}^{6} \pi \left(\frac{4}{y^2}\right) dy = 4\pi \int_{2}^{6} \frac{1}{y^2} dy = 4\pi \left[ -\frac{1}{y} \right]_{2}^{6}\].Compute the result:\[V = 4\pi \left( -\frac{1}{6} + \frac{1}{2} \right) = 4\pi \left( \frac{1}{2} - \frac{1}{6} \right) = 4\pi \left( \frac{3}{6} - \frac{1}{6} \right) = 4\pi \cdot \frac{2}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}\].

Key concepts

Horizontal Slice Method

The horizontal slice method is a powerful way to find the volume of a solid of revolution. By using this technique, you imagine slicing the solid into thin, horizontal pieces. Each piece has a very small thickness, denoted as \(dy\), which signifies a change in the \(y\)-direction.

Here's how it works: these slices are perpendicular to the axis of rotation, the \(y\)-axis in our case. Each horizontal slice can be visualized as a thin disc or washer. In our problem, the boundaries indicate that the region is enclosed horizontally from \(x = 0\) to \(x = \frac{2}{y}\).

Understanding the region's bounds helps visualize how the volume accumulates. By integrating these slices, you capture the entire volume as the region rotates around the \(y\)-axis.

Integral Calculus

Integral calculus plays a pivotal role in calculating areas and volumes, especially in complex figures. In the context of our problem, we use integral calculus to determine the volume of the solid formed by revolving a region around an axis.

For the given exercise, the volume is found by setting up an integral that accumulates the volume of each horizontal slice. The integral itself, \( V = \int_{2}^{6} \pi \left(\frac{2}{y}\right)^2 dy \), represents a continuous sum of these small contributions from each slice.

The integration bounds, from \(y = 2\) to \(y = 6\), ensure we account for the entire height of the region being revolved. Through evaluating this integral, integral calculus allows the computation of the exact total volume.

Volume Calculation

Volume calculation is the goal of combining the horizontal slice method and integral calculus in this scenario. The slices provide the building blocks of volume, while integrals sum these infinitesimal parts into a whole.

In this problem, each horizontal slice contributes a small volume \(dV\), given as \(dV = \pi\left(\frac{2}{y}\right)^2 dy\). When these slices revolve around the \(y\)-axis, they form the volume of the solid.

The integration over the interval \([2, 6]\) sums these contributions to find the full volume, which in this exercise was calculated as \(\frac{4\pi}{3}\). By understanding both the setup and computation process, working with volumes of revolution through calculus becomes intuitive and manageable.