Problem 24 - Thomas Calculus

Question

The region between the curve \(y=2 / x\) and the \(x\) -axis from \(x=1\) to \(x=4\) is revolved about the \(x\) -axis to generate a solid. \begin{equation}\begin{array}{l}{\text { a. Find the volume of the solid. }} \\\ {\text { b. Find the center of mass of a thin plate covering the region if }} \\ {\text { the plate's density at the point }(x, y) \text { is } \delta(x)=\sqrt{x} \text { . }} \\ {\text { c. Sketch the plate and show the center of mass in your sketch. }}\end{array}\end{equation}

Step-by-Step Solution

Answer

a. The volume is \(3\pi\). b. Center of mass is \((\frac{7}{3},\frac{1}{2})\). c. Include these in the sketch of the region.

Step 1: Apply the Disk Method for Volume

To find the volume of a solid of revolution, we can use the disk method. The formula for the volume when revolving about the x-axis is: \[ V = \int_{a}^{b} \pi \, [f(x)]^2 \, dx \] In this case, \( f(x) = \frac{2}{x} \), and the limits of integration are from \( x = 1 \) to \( x = 4 \). This gives us: \[ V = \int_{1}^{4} \pi \left( \frac{2}{x} \right)^2 \, dx \] \[ = \int_{1}^{4} \frac{4\pi}{x^2} \, dx \].

Step 2: Evaluate the Integral for Volume

Now, we solve the integral: \[ V = \int_{1}^{4} \frac{4\pi}{x^2} \, dx \] This integral is basic calculus: \[ V = 4\pi [-\frac{1}{x}]_{1}^{4} \] \[ = 4\pi \left(-\frac{1}{4} + 1 \right) \] \[ = 4\pi \left(\frac{3}{4}\right) \] \[ = 3\pi \]. The volume of the solid is \( 3\pi \).

Step 3: Calculate the Center of Mass for the Plate

For the center of mass of a thin plate with density \( \delta(x) = \sqrt{x} \), we use the formulas for the x-coordinate and y-coordinate of the center of mass. Begin with the x-coordinate: \[ \bar{x} = \frac{1}{M} \int_{a}^{b} x \, \delta(x) \, f(x) \, dx \], where \( M \) is the mass of the region: \[ M = \int_{a}^{b} \delta(x) \, f(x) \, dx \]. First, compute \( M \):\[ M = \int_{1}^{4} \sqrt{x} \, \frac{2}{x} \, dx \]\[ = 2 \int_{1}^{4} \frac{1}{\sqrt{x}} \, dx \]\[ = 2 \left[ 2\sqrt{x} \right]_{1}^{4} \]\[ = 4 \times (2 - 1) \]\[ = 4 \]. Now, compute \( \bar{x} \):\[ \bar{x} = \frac{1}{4} \int_{1}^{4} x \times \sqrt{x} \times \frac{2}{x} \, dx \]\[ = \frac{1}{4} \times 2 \int_{1}^{4} \, x^{1/2} \, dx \]\[ = \frac{1}{2} [\frac{2}{3} x^{3/2}]_{1}^{4} \]\[ = \frac{1}{2} \times \left( \frac{2}{3} \times 8 - \frac{2}{3} \times 1 \right) \]\[ = \frac{1}{2} \times \frac{14}{3} \]\[ = \frac{7}{3} \]. Thus, \( \bar{x} = \frac{7}{3} \). For the y-coordinate: \[ \bar{y} = \frac{1}{M} \int_{a}^{b} \frac{1}{2} [f(x)]^2 \delta(x) \, dx \]\[ = \frac{1}{4} \int_{1}^{4} \frac{1}{2} \times \left( \frac{2}{x} \right)^2 \times \sqrt{x} \, dx \]\[ = \frac{1}{8} \int_{1}^{4} \frac{4}{x^2} \times \sqrt{x} \, dx \]\[ = \frac{1}{2} \int_{1}^{4} \frac{1}{x^{3/2}} \dx \]\[ = \frac{1}{2} \left[ -2x^{-1/2} \right]_{1}^{4} \]\[ = \frac{1}{2} \times (-2 \times \frac{1}{2} + 2) \]\[ = \frac{1}{2} \times (1) \]\[ = \frac{1}{2} \]. Thus, the center of mass is \( (\frac{7}{3}, \frac{1}{2}) \).

Step 4: Sketch the Plate with Center of Mass

Draw the curve \( y = \frac{2}{x} \) from \( x = 1 \) to \( x = 4 \), shading the area under the curve down to the x-axis. Mark the center of mass point \( (\frac{7}{3}, \frac{1}{2}) \) on this sketch; it would be slightly right of the middle of the interval along the x-axis and slightly above the x-axis. Ensure the density gradient shows increased density widthwise from left (\( x = 1 \)) to right (\( x = 4 \)).

Key Concepts

Volume of Revolution

When dealing with problems involving the volume of a solid formed by revolving a region around an axis, the concept of the **Volume of Revolution** is fundamental. One of the common techniques used is the Disk Method. This method helps us calculate the volume of a solid obtained when a particular region is revolved around an axis, in this case, the x-axis.

To use the Disk Method, imagine slicing the solid perpendicular to the axis of rotation into many fine, thin disks. Each disk's volume can be thought of as a cylinder with a small height. The formula to find the volume using this approach is given by:

• \[ V = \int_{a}^{b} \pi \,[f(x)]^2 \, dx \]

Here, \( f(x) \) is the function defining the curve, and \([a, b]\) represents the interval along the x-axis over which it's revolved. The square of \( f(x) \) finds the area of the cross-sectional circle, and integrating this over the interval gives us the total volume.

In our specific problem, the function is \( f(x) = \frac{2}{x} \), and it creates a solid when revolved between \( x = 1 \) and \( x = 4 \). By setting up and evaluating the integral \( \int_{1}^{4} \pi \left( \frac{2}{x} \right)^2 \, dx \), we calculate the volume as \( 3\pi \). This involves integrating the expression, which results in beautifully symmetrical structures around the axis.

Center of Mass

Finding the **Center of Mass** of a shape offers insight into its balance point. For a thin plate shaped like the region between a curve and an axis, considering density variations is crucial. The center of mass tells us where this region would balance perfectly if it were placed on a pin.

For a region defined by \( y = \frac{2}{x} \) and bounded by the x-axis from \( x = 1 \) to \( x = 4 \), we calculate both the x and y coordinates of the center of mass. This weighted average of points considers the varying density, \( \delta(x) = \sqrt{x} \), across the plate.

To find these coordinates, different integrals are set up based on the product of the position, function, and density:

• \[ \bar{x} = \frac{1}{M} \int_{a}^{b} x \, \delta(x) \, f(x) \, dx \]
• \[ \bar{y} = \frac{1}{M} \int_{a}^{b} \frac{1}{2} \left[f(x)\right]^2 \, \delta(x) \, dx \]

Here, \( M \) is the total mass, representing the integral of density across the interval. Solving these integrals, the center of mass is identified at \( (\frac{7}{3}, \frac{1}{2}) \). This precise point on the sketch indicates that the plate has a slightly weighted distribution to the right of the middle along the x-axis and reflects how objects with varying densities find equilibrium.

Integral Calculus

**Integral Calculus** plays a vital role in solving many geometric and physical problems related to continuous change. When we deal with volumes and centers of mass, integral calculus allows us to sum infinitesimally small elements over a region to determine an aggregate property.

The integral is a fundamental tool here, represented as \( \int \), which signifies the addition of an infinite number of quantities. Either as areas under curves, as in finding volumes, or the moments of weighted distributions, as in centers of mass, integrals help one transition from the very small to the whole.

Key components of applying integral calculus include:

• Identifying limits of integration [a, b], which denote the interval over which you integrate.
• Choosing the correct function \( f(x) \) that represents the curve or surface involved.
• Applying specific formulas, like disk or shell methods, as appropriate to the symmetry and arrangement of the problem.

By transforming these seemingly small calculations into a robust analysis, integrals open up a world of precise modeling, as clearly illustrated by the solutions obtained for the volume and center of mass in our problem.

Density Function

A **Density Function** describes how mass is distributed across an object or region. When calculating the center of mass, it's crucial to account for variations in density, which, in this context, depend on position along the x-axis.

The density function \( \delta(x) = \sqrt{x} \) signifies that mass differs at different points of the curve defined by \( y = \frac{2}{x} \). Higher density indicates more mass per unit area at that point, subtly influencing the center of mass location.

When such variances arise, they are integrated into the calculation of the total mass:\[ M = \int_{a}^{b} \delta(x) \, f(x) \, dx \]
This formula encapsulates how density overlaps with the physical extent of the object. Integrating over the interval \( [1, 4] \), we can determine how these weighted differences contribute to the object's overall characteristics.

Understanding density functions thus allows holistic control over structural dynamics, illuminating how various parameters affect balance and stability within different conditions.