Problem 33 - Thomas Calculus in SI Units

Question

The square region with vertices \((0,2),(2,0),(4,2),\) and (2,4) is revolved about the \(x\) -axis to generate a solid. Find the volume and surface area of the solid.

Step-by-Step Solution

Answer

The volume is \(\frac{128\pi}{3}\), and the surface area is \(16\sqrt{2}\pi\).

Step 1: Understand the Square Region

The given vertices represent the corners of a square. Plot these points: - (0,2) - (2,0) - (4,2) - (2,4). Connect the points to see that they form a square tilted at a 45-degree angle with each side of length \(2\sqrt{2}\).

Step 2: Determine the Boundaries of Rotation

The_square lies symmetric with respect to the line \(y=x\) and is centered at (2,2). Since the problem asks us to revolve this region around the x-axis, we'll observe how it would form a symmetrical shape.

Step 3: Calculate the Volume Using the Disk Method

To find the volume of the solid of revolution, use the disk method. Integrate the area from the lowest y-value \(y=0\) to the highest y-value \(y=4\). The radius of each disk is given by the distance from a point to the x-axis. For this case, the expression for the radius is \(r(y) = \sqrt{2} y\). The volume function becomes: \[ V = \int_0^4 \pi r(y)^2 \, dy = \int_0^4 \pi (\sqrt{2} y)^2 \, dy = \pi \int_0^4 2y^2 \, dy \].

Step 4: Execute the Volume Integral

Evaluate the integral to find the volume: \[ V = \pi \int_0^4 2y^2 \, dy \] Expanding the integrand gives: \[ V = 2\pi \int_0^4 y^2 \, dy \] Calculate: \[ V = 2\pi \left[ \frac{y^3}{3} \right]_0^4 = 2\pi \left[ \frac{4^3}{3} \right] = 2\pi \left[ \frac{64}{3} \right] = \frac{128\pi}{3} \].

Step 5: Find Surface Area Using the Surface of Revolution Formula

Utilize the formula for the surface area of a solid of revolution: \[ S = 2\pi \int y \, ds \] where \(\frac{dy}{dx}\) represents a derivative component of arc length. We have a boundary defined by: \(\frac{dy}{dx} = 1\), and considering the increase in lengths, the arc length is \(\sqrt{2}\). Thus, \[ S = 2\pi \int_0^4 y(\sqrt{2}) \, dy \].

Step 6: Calculate the Surface Area Integral

Complete the evaluation:\[ S = 2\sqrt{2}\pi \int_0^4 y \, dy = 2\sqrt{2}\pi \left[ \frac{y^2}{2} \right]_0^4 \] \[ S = 2\sqrt{2}\pi \left( \frac{16}{2} \right) = 2\sqrt{2}\pi \times 8 = 16\sqrt{2}\pi \].

Key Concepts

Surface Area of a Solid of Revolution

When we revolve a shape around an axis, we often want to know the extent of the surface we're creating, known as the surface area. Revolving a square around the x-axis turns it into a three-dimensional figure. To find the surface area of such a solid of revolution, we use the formula:

• \[ S = 2\pi \int y \, ds \]

Here, \( y \) is the distance from the x-axis to a curve, also known as the radius, and \( ds \) is an infinitesimal arc length element. Surface of revolution calculations require

• Identifying the functions that give us the boundaries of the region being revolved.
• Calculating the arc length differential, often involving the derivative of the function.

This particular problem revolves a square, creating a symmetric shape. With boundaries straight and slopes consistently the same, we find the arc length calculations simplified. Evaluating the integral step-by-step, you find the surface area to be \( 16\sqrt{2}\pi \). Break it into simple terms by imagining millions of tiny circles being added together along the x-axis.

Disk Method

The disk method is one of the primary techniques to find the volume of a solid of revolution. Imagine slicing the solid perpendicular to the axis of rotation into thin, disk-like slices. Each slice resembles a circle, and by summing up their volumes, we get the entire volume of the solid. It's much like stacking paper plates; you know the thickness and radius, so you can find out the full stack’s volume!To apply the disk method:

• Identify the radius of each disk, given by the distance from the curve to the axis of revolution. For this problem, it's \( r(y) = \sqrt{2} y \).
• Use the formula for the volume of a single disk, which is \( \pi \times \text{radius}^2 \times \text{thickness} \).

The integral for the volume is:

• \[ V = \int_{a}^{b} \pi \times [r(y)]^2 \, dy \]

For our shape, this becomes, \( V = 2\pi \int_0^4 y^2 \, dy \), which results in \( \frac{128\pi}{3} \).Step-by-step simplification and focus on the radius and thickness of each disk make using this method clear and approachable.

Integral Calculus

Integral calculus is the foundational mathematical tool you need to solve problems involving areas, volumes, and other related quantities. In integral calculus, the central idea is to sum up infinite infinitesimally small quantities to find total values over intervals or curves.The basics you need to understand here include:

• Definite integrals: These are used to calculate quantities like areas under a curve, with upper and lower bounds indicating the limits of integration. In this exercise, you integrate over \([0,4]\).
• Antiderivatives: Finding an integral involves reversing the derivative. For instance, if differentiating \( x^3 \) gives \( 3x^2 \), then the integral of \( x^2 \) would give you back \( \frac{x^3}{3} \).

When applying integral calculus to find the volume or surface area of solids of revolution, you transform a geometric problem into one of computation, evaluating integrals step by step. This way, you solve complex geometric problems through careful calculation, making integral calculus a powerful ally in mathematics.