Problem 25 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises \(25-30\) about the \(y\) -axis. The region enclosed by \(x=\sqrt{5} y^{2}, \quad x=0, \quad y=-1, \quad y=1\)

Step-by-Step Solution

Answer

The volume of the solid is 0.

Step 1: Understand the Problem

We need to find the volume of the solid formed by rotating the area enclosed by the curve \(x = \sqrt{5}y^2\), the line \(x = 0\), and the lines \(y = -1\) and \(y = 1\) around the \(y\)-axis.

Step 2: Set Up the Integral

The solid of revolution volume using the shell method is given by \(V = 2\pi \int_{a}^{b} p(y)h(y) \, dy\), where \(p(y)\) is the radius from the \(y\)-axis and \(h(y)\) is the height of the cylindrical shell. In this problem, \(p(y) = y\) and \(h(y) = \sqrt{5}y^2\).

Step 3: Determine the Limits of Integration

The limits of integration for \(y\) are from \(-1\) to \(1\). This is derived from the given lines \(y = -1\) and \(y = 1\).

Step 4: Substitute into the Integral

Substitute \(p(y) = y\) and \(h(y) = \sqrt{5}y^2\) into the integral formula: \[ V = 2\pi \int_{-1}^{1} y(\sqrt{5}y^2) \, dy = 2\pi \sqrt{5} \int_{-1}^{1} y^3 \, dy. \]

Step 5: Solve the Integral

Calculate \(\int_{-1}^{1} y^3 \, dy\). The antiderivative of \(y^3\) is \(\frac{1}{4}y^4\). Evaluating from \(-1\) to \(1\) gives: \[ \left[ \frac{1}{4}y^4 \right]_{-1}^{1} = \frac{1}{4}(1^4 - (-1)^4) = \frac{1}{4}(1 - 1) = 0.\]

Step 6: Interpret the Result

Since the evaluated integral is zero, this means that the solid generated by revolving this region has no volume. It indicates that the region above and below the \(x\)-axis cancels out symmetrically.

Key Concepts

Shell Method

The shell method is a technique for finding the volume of a solid of revolution. In this method, we revolve a region around an axis to create a 3D shape. The idea is to break the solid into thin cylindrical shells and then add up their volumes. This method is often useful when revolving around the y-axis, especially when the function is already solved in terms of y.

To apply the shell method effectively:

• Identify the region that will be rotated.
• Determine the radius and height of a typical shell.
• Set up the integral that computes the volume of the entire solid.

In the exercise provided, we're revolving around the y-axis which makes the shell method an appropriate choice. We use the function given, treat our shells as cylindrical in shape, and then use basic integration to find the volume.

Integration

Integration is the mathematical procedure of adding up infinitely small quantities to find an area under a curve or the volume of a solid. In our case, integration helps sum up the volumes of cylindrical shells over the interval we've defined.

To integrate using the shell method:

• Identify the function for the volume element, often in terms of y for y-axis revolutions.
• Determine the appropriate limits of integration.
• Compute the integral to summarize the total volume.

For the given formula, the function to integrate is derived from shell components. Once the integral is set up, solving it gives us the volume of the solid formed by revolving the region. For our exercise, however, the integral results in zero, reflecting a symmetry that cancels the volume out.

Cylindrical Shell

A cylindrical shell is essentially a hollow tube with a certain radius and height. When using the shell method, each little piece of the solid we're interested in is structured as a thin cylindrical shell.

The definitions for a cylindrical shell used in our example are:

• Radius: The distance from the axis of rotation to the shell. Here, it's simply the y-coordinate, so we use \( p(y) = y \).
• Height: This is the length of the shell, which relates to how far the curve reaches when rotated. In our example, the height \( h(y) \) is given by \( \sqrt{5}y^2 \).

Each shell has a small thickness defined as \( \Delta y \), which gets infinitely small when we integrate. The full volume is the sum of all these shells from one limit to the other.

Limits of Integration

Limits of integration define the bounds over which you sum your elements in the integration process. These should capture the full region you want to revolve around an axis, ensuring you have the complete volume calculation.

In the exercise, these limits are crucial:

• Here, limits are from \( y = -1 \) to \( y = 1 \), encompassing the full height of our region of interest.
• Such symmetry in this particular example means that parts of the volume cancel out, resulting in no total volume when integrated fully.

By setting your limits correctly, you include the entire area from bottom to top. This completely describes the slice of the region you are revolving, ensuring accurate calculation of volume.