Problem 43 - Thomas Calculus in SI Units

Question

Find the volume of the solid generated by revolving each region about the \(y\) -axis. The region in the first quadrant bounded above by the parabola \(y=x^{2},\) below by the \(x\) -axis, and on the right by the line \(x=2\)

Step-by-Step Solution

Answer

The volume is \( 8\pi \).

Step 1: Understand the region defined by curves

The region is bounded by the parabola \( y = x^2 \), the line \( x = 2 \), and the \( x \)-axis. We are in the first quadrant, so the region is above the \( x \)-axis and to the left of \( x = 2 \). It starts at the origin and runs up to \( x = 2 \).

Step 2: Set up the integral for volume

To find the volume of the solid of revolution, we use the method of cylindrical shells. The formula for volume is \( V = 2\pi \int_{a}^{b} x \, f(x) \, dx \) where \( f(x) \) is the height of the shell, which is \( y = x^2 \) in this case. Here, \( a=0 \) and \( b=2 \).

Step 3: Substitute and integrate

Substitute \( f(x) = x^2 \) into the integral: \[ V = 2\pi \int_{0}^{2} x(x^2) \, dx = 2\pi \int_{0}^{2} x^3 \, dx \]. Now, we need to integrate \( x^3 \).

Step 4: Perform the integration

Integrate \( x^3 \) with respect to \( x \): \[ \int x^3 \, dx = \frac{x^4}{4} \]. Evaluate this from \( 0 \) to \( 2 \): \[ \left. \frac{x^4}{4} \right|_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = \frac{16}{4} - 0 = 4 \].

Step 5: Compute the final volume

Substitute the result of the integration back into the expression for volume: \[ V = 2\pi \times 4 = 8\pi \]. This is the volume of the solid formed by revolving the region about the \( y \)-axis.

Key Concepts

Integration

Integration is a crucial mathematical tool that allows us to find areas, volumes, and other quantities that accumulate over a range. Here, we're specifically looking at volume, a concept used regularly in geometry and calculus. In this exercise, integration is used to calculate the volume of a solid. This solid is generated by rotating a given region around an axis, creating what's known as a 'solid of revolution.'
To find this volume, we integrate an expression that represents the accumulation of infinitely small elements, or slices, of the solid. By summing these elements over a defined range (from one boundary to another), we get the total volume. Key points about integration in this context:

• It transforms complex geometric shapes into manageable calculations.
• Applies the concept of adding up infinite tiny parts to get a whole.
• Necessary for calculating spaces under curves, and, by extension, the shapes they create when rotated.

Grasping integration is fundamental for understanding how we can use calculus to tackle real world problems.

Cylindrical Shells Method

The method of cylindrical shells is an effective approach for finding the volume of a solid of revolution, especially when dealing with functions of x and revolving them around the y-axis. It involves visualizing the solid as a collection of cylindrical shells, stacked together.
Think of a cylindrical shell as a thin hollow tube. When the region is rotated around the y-axis, each vertical slice of the region forms a cylindrical shell.Here's how to apply the method:

• Identify the boundaries: Define the height and radius of each shell.
• The height of a shell is determined by the function’s value at each point. Here, it's given by the parabola equation, \( y = x^2 \).
• The radius is the distance from the axis of rotation, which in this case is just \( x \).
• Set up the integral using the formula \( V = 2\pi \int_{a}^{b} x \cdot f(x) \cdot dx \).

By following these steps, you can derive the total volume by integrating over the region’s full width, from \( a \) to \( b \). By visualizing and breaking down the problem into simpler slices, calculating complex volumes becomes easier.

Parabolas

Parabolas are U-shaped curves that are the graph of quadratic functions. Here, the equation \( y = x^2 \) defines the parabola that forms part of the region we are revolving around the y-axis.
Properties of parabolas relevant to this problem:

• The vertex of the parabola (its lowest point) is at the origin (0,0).
• It is symmetric about the y-axis, making it convenient for rotation calculations.
• The arms of the parabola extend infinitely in the positive and negative x-direction, but for practical purposes, we are often concerned with a finite segment.

A parabola like \( y = x^2 \) is a simple function to work with when calculating areas and volumes because its curve is predictable and consistent. This simplicity allows for straightforward integration and the application of methods such as the cylindrical shells for calculating volumes of rotation.