Problem 20 - Thomas Calculus

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\) -axis. \begin{equation} y=x^{3}, \quad y=0, \quad x=2 \end{equation}

Step-by-Step Solution

Answer

The volume is \(\frac{128\pi}{7}\).

Step 1: Identify the Bounded Region

The region is bounded by the curve \(y = x^3\), the line \(y = 0\) (the horizontal axis), and the line \(x = 2\). This region is in the first quadrant and extends from \(x = 0\) to \(x = 2\).

Step 2: Set Up the Disk Method Formula

The volume of a solid of revolution about the x-axis can be found using the disk method, which is given by the formula \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\). Here, \(f(x) = x^3\), and \(a = 0\) and \(b = 2\).

Step 3: Write Down the Integral

Substitute \(f(x) = x^3\) into the disk method formula. The integral becomes \(\pi \int_{0}^{2} (x^3)^2 \, dx = \pi \int_{0}^{2} x^6 \, dx\).

Step 4: Evaluate the Integral

To find the integral \(\int_{0}^{2} x^6 \, dx\), apply the power rule for integration: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\). So, \(\int_{0}^{2} x^6 \, dx = \left[\frac{x^7}{7}\right]_{0}^{2}\).

Step 5: Calculate the Antiderivative

Evaluate \(\left[\frac{x^7}{7}\right]_{0}^{2} = \left.\frac{x^7}{7}\right|_{0}^{2} = \frac{2^7}{7} - \frac{0^7}{7} = \frac{128}{7}\).

Step 6: Compute the Volume

Substitute back into the volume formula: \(V = \pi \cdot \frac{128}{7} = \frac{128\pi}{7}\).

Key Concepts

Volume of Solids of Revolution

To understand the volume of solids of revolution, imagine a 2D region being spun around an axis to create a 3D object, like turning a flat disc into a cone by spinning it around its edge. The Disk Method helps us calculate the volume of such objects when they're formed by rotating a region about an axis, typically the x-axis.
If you picture a flat slice of the solid as a thin disk or a circle, the volume involves stacking an infinite number of these infinitesimally thin disks along the axis. Each disk has a radius equal to the function value at a point and thickness equal to an infinitely small change in the x-direction, noted as \(dx\).
Here's the process:

• The formula \(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\) is used to sum up the volumes of these thin disks. \(V\) represents the volume of the solid of revolution.\(a\) and \(b\) are the interval bounds along the axis of rotation.
• \(f(x)\), the function describing the curve, acts as the radius of each disk. So \([f(x)]^2\) gives the area of the cross-section of the disk.
• The integration process adds up all these infinitesimally small areas to find the total volume.

Forming a solid of revolution takes practice, but once you're familiar with setting up and understanding the integral, it turns into a straightforward calculation!

Integration using Power Rule

Integration, the mathematical process of finding the area under a curve, becomes essential in calculating the volume of solids of revolution. The Power Rule for integration simplifies this process significantly. The rule states: \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), where \(n\) is a constant and \(C\) is the integration constant (usually omitted in definite integrals).
This rule is powerful when dealing with polynomial functions, like the one in our example: \(x^3\). Applying this rule, we can handle each term in the polynomial separately and combine the results:

• First, increase the exponent by one: For \(x^3\), become \(x^4\).
• Divide the new term by the new exponent: \(\frac{x^4}{4}\).

When computing definite integrals from \(a\) to \(b\), the constant \(C\) cancels out, leaving us with the evaluated results at these bounds. As such, for our original problem of finding \(\int_{0}^{2} x^6 \, dx\), applying the power rule results in \(\frac{x^7}{7}\), evaluated from \(0\) to \(2\). This gives \(\frac{128}{7}\), which, when multiplied by \(\pi\), provides the volume of the solid.

Bounded Regions in Calculus

Understanding bounded regions helps us establish where integration takes place. In the context of volume of solids of revolution, defining these bounds is crucial to setting up correct integrals.
A bounded region in a graph indicates the area enclosed by curves and lines, and for our purposes, it extends along specific intervals on the x-axis or y-axis. When a curve like \(y = x^3\) is involved, it creates a boundary with other lines or curves like \(y=0\) and \(x=2\).
Steps to identify these bounded regions include:

• Graphing the functions to visually see the enclosed area. Here, \(y = x^3\) defines one bound, \(y = 0\) (the x-axis) defines another, and the line \(x=2\) marks the vertical boundary.
• Determining the point of intersections or the endpoints. In this case, the region extends from \(x=0\) to \(x=2\).

Once these boundaries are clear, they are used as the lower and upper limits in the integral, ensuring we calculate the volume exactly for the desired solid.