Problem 15 - Thomas Calculus in SI Units

Question

The region bounded by the curves \(y=\pm 4 / \sqrt{x}\) and the lines \(x=1\) and \(x=4\) is revolved about the \(y\) -axis to generate a solid. a. Find the volume of the solid. b. Find the center of mass of a thin plate covering the region if the plate's density at the point \((x, y)\) is \(\delta(x)=1 / x\). c. Sketch the plate and show the center of mass in your sketch.

Step-by-Step Solution

Answer

a. Volume is \( \frac{224\pi}{3} \). b. Center of mass \(x, y\) is \( \left( \frac{224}{3}, 0 \right) \).

Step 1: Set up the Volume Integral using Shell Method

We need to revolve the region around the y-axis. Using the shell method, the formula is \ V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx \where \( h(x) \) is the height of the shell. In our problem, \( h(x) = 2\times\frac{4}{\sqrt{x}} = \frac{8}{\sqrt{x}} \). The limits are from \(x = 1\) to \(x = 4\).Therefore, the expression for the volume is \ V = 2 \pi \int_{1}^{4} x \cdot \frac{8}{\sqrt{x}} \, dx = 16\pi \int_{1}^{4} \sqrt{x} \, dx \

Step 2: Evaluate the Volume Integral

We need to integrate \(\sqrt{x}\) to find the volume. The antiderivative of \( \sqrt{x} \) is \( \frac{2}{3}x^{3/2} \).Evaluating the definite integral:\[16\pi \cdot \left. \left( \frac{2}{3}x^{3/2} \right) \right|_{1}^{4} = 16\pi \cdot \left[ \frac{2}{3}(4^{3/2}) - \frac{2}{3}(1^{3/2}) \right] \] Simplifying, we get:\[16\pi \cdot \left[ \frac{2}{3}(8) - \frac{2}{3}(1) \right] \] \[= 16\pi \cdot \frac{14}{3} = \frac{224\pi}{3} \]

Step 3: Set up the Mass Integral for the Thin Plate

To find the mass of the plate, integrate the density function. Given \(\delta(x) = \frac{1}{x}\), the mass \(m\) is:\[ m = \int_{1}^{4} \delta(x) \cdot 2\cdot\frac{4}{\sqrt{x}} \, dx = \int_{1}^{4} \frac{8}{x\sqrt{x}} \, dx = \int_{1}^{4} \frac{8}{x^{3/2}} \, dx \]

Step 4: Evaluate the Mass Integral

Integrate \(\frac{8}{x^{3/2}}\):We know the antiderivative of \(x^{-3/2}\) is \(-2x^{-1/2}\). Therefore:\[ m = 8 \cdot \left. \left( -2x^{-1/2} \right) \right|_{1}^{4} = -16 \cdot \left[ x^{-1/2} \right]_{1}^{4} \]\[= -16 \times \left( \frac{1}{2} - 1 \right) = -16 \times \left( -\frac{1}{2} \right) = 8 \]

Step 5: Determine the Center of Mass Coordinates

The coordinates \((\bar{x}, \bar{y})\) of the center of mass are given by:\[\bar{x} = \frac{1}{m} \int_{1}^{4} \left( x \cdot \delta(x) \cdot 2f(x) \right) dx = \frac{1}{8} \int_{1}^{4} \frac{8}{\sqrt{x}} \cdot x \, dx \] Since \(\frac{8}{\sqrt{x}} \cdot x = 8\sqrt{x}\):\[= \frac{1}{8} \cdot 16 \left. \left( \frac{2}{3}x^{3/2} \right) \right|_{1}^{4} = \frac{32}{3} \cdot \left( 8 - 1 \right) = \frac{224}{3} \]\[\bar{y} = 0 \] because the plate is symmetric about the x-axis.

Key Concepts

Revolution around the y-axis

When a region in a plane is revolved around a line (in this case, the y-axis), it creates a three-dimensional solid. Imagine taking a flat shape and spinning it around a straight line to form a solid. This process is fundamental in calculus and is essential for finding volumes of various shapes.
The formula for the volume of a solid of revolution around the y-axis using the shell method is neat and clever. It involves creating cylindrical shells from the original shape.

• Consider a thin vertical strip of the area, which gets revolved around the y-axis.
• This strip forms a shell, with its radius being the distance from the y-axis (the x-coordinate here).
• The height of this shell is determined by the shape or function.
• The volume of this shell is given by the expression:
  \[ V = 2\pi \ imes \text{radius} \ imes \text{height} \ imes \text{thickness}\]

It's fascinating because this method allows us to move into the third dimension and calculate volumes precisely by using integration.

Integration

Integration is a mathematical process used to calculate areas, volumes, and other related quantities. It’s the backbone of calculus and helps in "adding up" tiny pieces to find a whole. In our problem, integration is used twice:
First, to find the volume of the revolution, and second, to find the mass of the plate.

• The first integral computes the volume of the solid created by revolving around the y-axis. It integrates the function \(\sqrt{x}\) over the interval from 1 to 4. The result is multiplied by \(2\pi\) to account for the revolution.
• The second integral computes the mass, accounting for the changing density \(\delta(x) = \frac{1}{x}\). Here, the function \(\frac{8}{x^{3/2}}\) is integrated over the same interval.

This process of integration helps in understanding how changes in space or density affect the total values, like volume and mass.

Center of Mass

The center of mass represents the point in a body or system where all of its mass can be considered to be concentrated. For a thin plate, it’s the point where you could balance it perfectly, given its density.
Finding the center of mass involves evaluating coordinates \((\bar{x}, \bar{y})\).

• For the x-coordinate \(\bar{x}\), we integrate the product of \(x\) and the density function over the given region, dividing by the total mass \(m\) to allocate average position.
• Here, \(\bar{y} = 0\) due to the symmetry about the x-axis, indicating a balance along the horizontal midpoint.

This highlights how calculus can be used to determine even complex distributions of mass in a given shape.

Shell Method

The shell method is a powerful way to find the volume of solids of revolution. Unlike the disc or washer method, which slices the solid perpendicular to the axis of rotation, the shell method works by slicing parallel.
Here's how the shell method works in our context:

• Each shell is like a hollow pipe or cylinder, distinct in that it has thickness, height, and radius – calculated from the x-value and function height.
• The integral formula \( V = 2\pi \int_{a}^{b} x \cdot h(x) \, dx \) is used to sum the volumes of these shells over the interval.
• This method is particularly effective here because the region revolves around the y-axis, making a cylindrical approach intuitive and easier to integrate.

This approach simplifies calculations and is perfect for problems where the area is bounded by functions involving variables of the dimension being revolved around.