Problem 43 - Calculus

Question

The slices in \(43-46\) are washers. Find the slice area and volume. The rectangle with sides \(x=1, x=3, y=2, y=5\) is rotated around the \(x\) axis.

Step-by-Step Solution

Answer

The slice area is \(21\pi\) and the volume is \(42\pi\).

Step 1: Identify the boundaries of the solid

The rectangle has boundaries at the lines: \(x=1\), \(x=3\), \(y=2\), and \(y=5\). This means the rectangle extends horizontally from \(x=1\) to \(x=3\) and vertically from \(y=2\) to \(y=5\).

Step 2: Determine the shape of the solid

When the rectangle is rotated around the \(x\)-axis, the resulting solid is a cylindrical shell with a hollow center, reminiscent of a washer. The outer radius of the washer is \(y=5\) and the inner radius is \(y=2\).

Step 3: Set up the washer method integral for volume

The volume of the solid can be found by integrating the cross-sectional area of each washer as it moves along the \(x\)-axis from \(x=1\) to \(x=3\). The volume \(V\) is given by the integral:\[ V = \int_{1}^{3} \pi \left((R_{ ext{outer}})^2 - (R_{ ext{inner}})^2 \right) \, dx \]where \(R_{ ext{outer}}=5\) and \(R_{ ext{inner}}=2\).

Step 4: Compute the integral for volume

Substitute the radii into the integral:\[ V = \int_{1}^{3} \pi (5^2 - 2^2) \, dx = \int_{1}^{3} \pi (25 - 4) \, dx = \int_{1}^{3} 21\pi \, dx \]\[ V = 21\pi \int_{1}^{3} \, dx = 21\pi [x]_{1}^{3} = 21\pi (3-1) = 42\pi \].

Step 5: Write the formula for the area of a slice

The area of each washer slice is given by the area of the outer circle minus the area of the inner circle:\[ A = \pi (R_{ ext{outer}}^2 - R_{ ext{inner}}^2) = \pi (5^2 - 2^2) = \pi (25 - 4) = 21\pi \].

Step 6: Conclusion

The area of each washer slice is \(21\pi\) square units, and the total volume of the solid is \(42\pi\) cubic units.

Key Concepts

Volume of Revolution

The volume of revolution refers to the volume of a three-dimensional object created by rotating a two-dimensional shape around an axis. In the case of the given exercise, we are rotating a rectangle around the x-axis. This technique is often used in integral calculus to find volumes that may not be easily measurable by simple geometric formulas.

When the rectangle with boundaries at lines \(x = 1\), \(x = 3\), \(y = 2\), and \(y = 5\) is rotated around the \(x\)-axis, it forms a solid that resembles a cylindrical shell with a hollow center. This method helps in visualizing and calculating the volume by understanding how the rotation forms the solid. The main challenge is often determining the right method of integration to accurately calculate that volume.

Cross-Sectional Area

The cross-sectional area in the washer method is crucial for determining the volume of the solid of revolution. Each slice of the solid can be thought of as a washer—a circle with a hole in the middle. The formula for finding the area of this washer slice is:

• Outer Area: \(\pi (R_{\text{outer}})^2\)
• Inner Area: \(\pi (R_{\text{inner}})^2\)

To find the area of the washer, you subtract the inner area from the outer area:\[ A = \pi (R_{\text{outer}}^2 - R_{\text{inner}}^2) \]In this exercise, the outer radius \(R_{\text{outer}}\) is 5 and the inner radius \(R_{\text{inner}}\) is 2. Therefore, the area of the washer slice is:\[ A = \pi (5^2 - 2^2) = 21\pi \]This formula gives the area for each slice that is revolved to create the volume.

Integral Calculus

Integral calculus is used to sum the infinite cross-sectional areas of the washers to find the total volume of the solid. We set up an integral along the axis of revolution, which in this case is the \(x\)-axis.

The integral calculus formula for volume using the washer method is:\[ V = \int_{a}^{b} \pi ((R_{\text{outer}})^2 - (R_{\text{inner}})^2) \, dx \]In simpler terms, this involves:

• Finding the area at a given point by using the washer formula.
• Integrating these areas over the entire span of the solid, from \(x = 1\) to \(x = 3\).

For this problem, the solution substitutes the radii in the integral:\[ V = \int_{1}^{3} 21\pi \, dx = 21\pi (3-1) = 42\pi \]The integral calculates the sum of these areas to find that the solid's volume is \(42\pi\) cubic units.

Cylindrical Shells

The concept of cylindrical shells can be a bit tricky to grasp but think of it as stacking rings to form a barrel-like structure. In this exercise, the solid formed by rotating the rectangle is essentially composed of such shells.

Each "shell" starts as a slice of the rectangle piece, revolves, and forms the washer. The outer and inner radii define the thickness and size of each shell. As these washer-like shells are revolved around the \(x\)-axis, they collectively generate the full solid. This visualization is crucial for solving problems related to twisting 2D shapes into 3D volumes using calculus.

Using the washer method helps distinguish the difference made by the hollow center, which cylindrical shells inherently possess, aiding in turning calculations from simple area to total volume.