Problem 17 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises \(17-22\) about the \(x\) -axis. $$ y=x^{2}, \quad y=0, \quad x=2 $$

Step-by-Step Solution

Answer

The volume is \( \frac{32\pi}{5} \).

Step 1: Understand the Region

The region is bounded by the curve \(y = x^2\), the line \(y = 0\) (the x-axis), and the vertical line \(x = 2\). This means the area of interest is under the parabola \(y = x^2\) from \(x = 0\) to \(x = 2\).

Step 2: Use Disk Method Formula

The volume of the solid of revolution can be found using the disk method. The formula is \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]where \(f(x)\) is the function representing the radius of the disks, \(a\) and \(b\) are the bounds of \(x\). Here, \(f(x) = x^2\), \(a = 0\), and \(b = 2\).

Step 3: Set Up the Integral

Substitute \(f(x) = x^2\) and the bounds \(0\) to \(2\) into the disk method formula:\[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \].

Step 4: Simplify the Integral Expression

Simplify the integral:\[ V = \pi \int_{0}^{2} x^4 \, dx \].

Step 5: Compute the Integral

Integrate \(x^4\) with respect to \(x\):\[ \int x^4 \, dx = \frac{x^5}{5} + C \].

Step 6: Evaluate the Definite Integral

Evaluate the integral from \(0\) to \(2\):\[ V = \pi \left[ \frac{2^5}{5} - \frac{0^5}{5} \right] = \pi \left[ \frac{32}{5} \right] \].

Step 7: Calculate the Volume

Calculate the volume:\[ V = \frac{32\pi}{5} \].

Key Concepts

Disk Method

The disk method is a powerful technique for finding the volume of a solid of revolution. Imagine slicing the solid into thin, circular disks (like a stack of coins). Each disk's volume is like a cylinder's: a small height and a circular base. The disk method uses an integral to sum up the volumes of infinitely many thin disks. It's especially handy when the solid is formed by rotating a region around an axis.

Here's how it works:

• First, identify the function representing the radius of each disk. This is often the vertical distance from the axis of rotation to the curve.
• Next, determine the bounds of integration by finding where the region starts and ends along the axis of rotation.
• Apply the formula for the disk method: \[ V = \pi \int_{a}^{b} [f(x)]^2 \ dx \]This formula calculates the volume by integrating from the starting point \(a\) to the ending point \(b\).
• In the exercise, the radius is \(x^2\), and the area rotates around the \(x\)-axis between \(x = 0\) and \(x = 2\).
• Substitute the given function and bounds into the disk method formula to find the volume.

Understanding the disk method is crucial for working with volumes of solids of revolution and can be applied to various types of functions and regions.

Definite Integral

The definite integral is a fundamental concept in calculus. It calculates the net area under a curve between two specific values, often representing physical quantities like area, distance, or in this case, volume.

To use a definite integral effectively:

• Identify the function \(f(x)\), which represents the curve.
• Determine the integration limits \(a\) and \(b\), which are the specific x-values where the area calculation begins and ends.
• Carry out the integration process, which often involves finding the antiderivative of \(f(x)\).
• Finally, evaluate the antiderivative at the upper and lower bounds, \(b\) and \(a\), and subtract: \[\int_a^b f(x) \, dx = F(b) - F(a).\]

In exercises involving the disk method, the definite integral helps to precisely calculate the volume by accounting for every infinitesimally small disk from start to finish. In our specific problem, the definite integral \[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \] was used to neatly sum up the volumes of all disks from \(x = 0\) to \(x = 2\). Understanding this process turns what can seem like a complex problem into a methodical and solvable one.

Parabola

A parabola is a specific type of curve defined by quadratic equations like \(y = x^2\). It's symmetrical, U-shaped, and has important properties that make it straightforward to analyze in geometry and calculus.

Some key characteristics of parabolas include:

• The vertex, which is the highest or lowest point on the curve depending on orientation. For \(y = x^2\), the vertex is at the origin \((0,0)\), making it easy to work with.
• The symmetry about the vertical line through the vertex, which simplifies calculations and analyses since behavior on one side mirrors the other side.
• The fact that it opens upwards when \(a > 0\) in the general form \(y = ax^2 + bx + c\), creating a bowl-like shape.

In terms of problems on volumes of solids of revolution, parabolas serve as common function shapes. They help visualize the region being revolved and can determine areas of interest under the curve.

In the exercise, the parabola \(y = x^2\) was bounded by \(x = 2\) and the \(x\)-axis, creating a shape well-suited for rotation using the disk method. Recognizing such parabolic forms and their boundaries greatly aids in setting up integrals and ultimately solving problems efficiently.