Problem 18 - Essential Calculus

Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line \(x=6\). $$ x y=6, \quad y=2, \quad y=6, \quad x=6 $$

Step-by-Step Solution

Answer

The volume of the solid generated is \(9\pi\) units.

Step 1: Sketch the region

Start by sketching the region bounded by the given equations. This region is bounded by the rectangular hyperbola \(xy=6\), the lines \(y=2\), \(y=6\), and \(x=6\). Sketch all of these curves and identify the region.

Step 2: Identify the radii for the washer method

The solid generated by revolving the region around the line \(x=6\) has the shape of a donut or washer, so use the washer method. The outer radius \(R(x)\) of the washer is the distance from the line \(x=6\) to the line \(y=6\), which is constant and equal to 6. The inner radius \(r(x)\) is the distance from the line \(x=6\) to the curve \(xy=6\), so \(r(x)=6-x\). This is because for points on the curve \(xy=6\), the y-coordinate is equal to \(6/x\).

Step 3: Set up the integral

The volume \(V\) of the solid is given by the integral \(V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] dx\). In this case, \(a=3\), \(b=6\), \(R(x)=6\), and \(r(x)=6-x\). Therefore, the integral becomes \(V = \pi \int_{3}^{6} [(6)^2 - (6-x)^2] dx\).

Step 4: Evaluate the integral

Evaluate the integral by first simplifying the integrand to get \(V = \pi \int_{3}^{6} [36 - (36 - 12x + x^2)] dx = \pi \int_{3}^{6} (12x - x^2) dx\). Then calculate the antiderivative at the endpoints to find \(V = \pi [6x^2 - x^3/3]_{3}^{6} = \pi [(6(6)^2 - (6)^3/3) - (6(3)^2 - (3)^3/3)] = \pi [72 - 72 -18 + 9] = \pi [18 - 9] = 9\pi\).

Key Concepts

Solid of Revolution

A solid of revolution is a 3D figure obtained by rotating a 2D shape around an axis. Imagine cutting out a flat shape from a piece of paper, and then spinning it around a line. The 3D shape formed by this spinning is called a solid of revolution.

In our exercise, we revolve the region bounded by the equations around the line \( x=6 \). This line becomes our axis of rotation. Visualize this as stretching and wrapping the 2D region into a donut shape.

• The outer boundary forms the outer surface of the solid.
• The inner boundary forms the hollow part of the solid.

This hollow structure is why the washer method is employed for finding the volume.

Volume by Integration

Integration is used to calculate areas and volumes, particularly when the shapes are irregular. To find the volume of a solid of revolution, we use integration to add up tiny disk-shaped slices of the formed solid. Picture slicing the 3D solid into many thin discs. By summing up the volume of each disc, we find the total volume.

The washer method is a specific technique for integration when finding the volume of solids containing a hollow region. It calculates the volume by integrating over the difference between the square of the outer and inner radii:

• The formula for the washer method is \( V = \pi \int_{a}^{b} [R(x)^2 - r(x)^2] dx \).
• \( R(x) \) is the outer radius, and \( r(x) \) is the inner radius of the washer.

The outer radius represents the boundary line, and the inner radius tracks the curve's distance from the axis.

Rectangular Hyperbola

A rectangular hyperbola is a type of curve where the product of the x-coordinates and y-coordinates is constant. For example, the graph of the equation \( xy = 6 \) is a rectangular hyperbola. This implies that any point on the curve will have coordinates that multiply to give 6.

In our problem, this curve is one of the boundaries of the region we revolve. It's this unique equation that causes the curve to approach the axes asymptotically—meaning it gets closer and closer but never touches them.

• It results in a hyperbolic shape when plotted on a graph.
• In this exercise, it defines the curved boundary that affects the inner radius \( r(x) \) for the washer method.
• As \( y = \frac{6}{x} \), this relationship determines the distance from the axis of revolution.

Understanding these properties helps in setting up the correct integral limits and radii for finding the volume.