Problem 19 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves in Exercises \(19-28\) about the \(x\) -axis. \begin{equation} y=x^{2}, \quad y=0, \quad x=2 \end{equation}

Step-by-Step Solution

Answer

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

Step 1: Identify the region to be revolved

The region is bounded by the curves and lines: \(y = x^2\), \(y = 0\) (the x-axis), and \(x = 2\). This is a region located in the first quadrant from \(x = 0\) to \(x = 2\).

Step 2: Set up the integral for the volume

To find the volume of the solid formed by revolving the region around the x-axis, use the disk method. The formula for the volume \( V \) is given by \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]. Here, \( f(x) = x^2 \), \(a = 0\), and \(b = 2\). So, \[ V = \pi \int_{0}^{2} (x^2)^2 \, dx \].

Step 3: Simplify the integral

The integral becomes \( \pi \int_{0}^{2} x^4 \, dx \), as \((x^2)^2 = x^4\).

Step 4: Integrate

Compute the integral \( \int_{0}^{2} x^4 \, dx \). The antiderivative of \( x^4 \) is \( \frac{x^5}{5} \). Thus, \[ \pi \left[ \frac{x^5}{5} \right]_{0}^{2} \].

Step 5: Evaluate the definite integral

Substitute the limits into the antiderivative: \[ \pi \left( \frac{2^5}{5} - \frac{0^5}{5} \right) = \pi \left( \frac{32}{5} \right) = \frac{32\pi}{5} \].

Step 6: Conclude the solution

The volume of the solid generated by revolving the given region about the x-axis is \( \frac{32\pi}{5} \).

Key Concepts

Disk Method

The Disk Method is a powerful technique used to find the volume of a solid of revolution. It involves rotating a region around an axis to form a three-dimensional solid. The method is specifically advantageous when dealing with solids obtained by revolving a region around the x-axis or y-axis.

In the Disk Method:

• Imagine slicing the solid into thin disks perpendicular to the axis of rotation.
• Each disk has a circular face, and the radius of the disk is determined by the function value at a particular x-coordinate.
• The volume of a single disk can be approximated as the area of the circle (π times radius squared) times the thickness of the disk (an infinitesimally small change in x, denoted as dx).

Mathematically, the volume of the entire solid is given by integrating the volumes of all these disks across the interval of interest. This leads to the formula:\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]The integration essentially sums up the volumes of all these tiny disks across the given range from \(a\) to \(b\). The solid's revolved cross-section is visualized as a series of stacked disks, hence the name 'Disk Method.'

It's crucial to clearly define the region to be revolved, as shown in the original problem, which uses bounds \(y = x^2\), \(y = 0\), and \(x = 2\).

Definite Integral

A Definite Integral is a key tool in calculating the volume of solids formed by rotation, as seen in the Disk Method. It is used to sum the continuous product of function values and tiny increments over a specified interval.

The general form of a definite integral is expressed as \[ \int_{a}^{b} f(x) \, dx \], where:

• \(f(x)\) represents the function you're integrating.
• \(a\) and \(b\) are the limits of integration, indicating the interval over which to integrate.

In our example problem:

• The function \(f(x)\) is \(x^4\).
• The integration bounds are \(0\) to \(2\).

Definite Integrals involve evaluating the antiderivative of the function at the upper limit (b) and subtracting the antiderivative at the lower limit (a). This allows us to capture the total accumulation of values over the range. Visually, it provides the total area under the curve from \(x = 0\) to \(x = 2\) in the context of the volume exercise. In essence, the definite integral translates the concept of adding up infinite thicknesses of disks to find the total volume of a rotated solid.

Antiderivative

Finding the Antiderivative is an essential step in solving a definite integral. It refers to the process of reverse-differentiation, where you determine a function whose derivative matches the given function.

In our original step-by-step solution:

• The antiderivative of \(x^4\) was determined as \(\frac{x^5}{5}\).
• This function is found by increasing each power of \(x\) by one and dividing by the new power.

The beauty of an antiderivative is its ability to simplify the process of evaluating definite integrals. Once the antiderivative is calculated, computation involves simply substituting the boundary values and calculating the difference. This substitution gives a concrete number, representing, in our case, the volume of the solid.

Remember, every continuous function has an infinite number of antiderivatives, differing only by a constant. However, in definite integrals, this constant cancels out and does not affect the result. This assures that determining the antiderivative correctly is crucial in providing a clear pathway to the exact solution.