Problem 19 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves 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 Problem

The problem requires finding the volume of the solid formed when the area between the curves \(y = x^2\) and \(y = 0\) from \(x=0\) to \(x=2\) is revolved around the \(x\)-axis.

Step 2: Set Up the Integral for Disk Method

The volume of the solid can be determined using the disk method. The formula for the volume \(V\) is \[ V = \int_{a}^{b} \pi [f(x)]^2 \, dx \] where \(f(x) = x^2\), \(a = 0\), and \(b = 2\).

Step 3: Write Down the Integral Expression

Substitute the function and the limits into the integral. The expression becomes: \[ V = \int_{0}^{2} \pi (x^2)^2 \, dx \] which simplifies to \[ V = \pi \int_{0}^{2} x^4 \, dx. \]

Step 4: Integrate the Expression

Calculate the integral \(\int x^4 \, dx\). The antiderivative of \(x^4\) is \(\frac{x^5}{5}\). Hence, \[ \int_{0}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{0}^{2}. \]

Step 5: Evaluate the Integral at the Limits

Evaluate \(\frac{x^5}{5}\) from \(0\) to \(2\): \[ \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5}. \]

Step 6: Multiply by Pi

Finally, multiply the result by \(\pi\) to find the volume: \[ V = \pi \times \frac{32}{5} = \frac{32 \pi}{5}. \]

Step 7: Conclusion

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

Key Concepts

Disk Method

The Disk Method is a powerful technique in integral calculus used to find volumes of solids of revolution. When we revolve a region around the x-axis (or y-axis), it forms a 3D object.

This method visualizes the solid as a stack of infinitesimally thin disks (or washers) perpendicular to the axis of rotation.

• The formula for the volume of a solid using the disk method is \[ V = \int_{a}^{b} \pi [f(x)]^2 \, dx \]
• Here, \( f(x) \) represents the function defining the edge of the region being revolved.
• The limits \(a\) and \(b\) define the interval over which we are rotating.

This method is particularly effective when dealing with functions that are easy to integrate and symmetric. For the given exercise, we revolved the curve \(y = x^2\) from \(x = 0\) to \(x = 2\) around the x-axis. This led us to the volume formula based on the square of \(f(x)\), leveraging symmetry for a clean integral solution.

By visualizing disks along the axis, we approximate the volume by summing the volume of countless cylinders, each with a tiny height \(dx\). This intuition bridges the gap between 2D areas and 3D volumes.

Integral Calculus

Integral calculus is fundamental in calculating areas and volumes using the process of integration. It allows us to sum infinitely small components to find whole values, such as areas under curves or volumes of solids.

Key elements include:

• The integral symbol \(\int\), which signifies summing an infinite number of small parts.
• Limits of integration \(a\) and \(b\), defining the start and endpoints of our region of interest.
• The integrand, in our problem \(\pi (x^2)^2\), which includes both the function being integrated and any constants, like \(\pi\) in the disk method formula.

In the exercise, we set up the integral expression \(V = \pi \int_{0}^{2} x^4 \, dx\) to find the volume. The process starts with defining the curve and revolves it, creating a cylindrical component.Integration allows us to piece together these components to obtain a precise calculation of the total volume.

The power of integral calculus lies in its ability to handle complex shapes and patterns as long as they can be described mathematically.

Antiderivatives

An antiderivative is very much the reverse of differentiation. It's a function whose derivative is the original function or integrand. In integral calculus, finding an antiderivative is key to solving integrals.

Steps to find an antiderivative include:

• Recognizing the basic function that, once differentiated, results in the given function.
• Applying any rules of integration, such as the power rule: If \(F'(x) = x^n\), then \(F(x) = \frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

In the step-by-step solution, we found the antiderivative of \(x^4\), which is \(\frac{x^5}{5}\). To solve the volume problem, we evaluated this antiderivative from \(x = 0\) to \(x = 2\), using the Fundamental Theorem of Calculus to determine the exact volume.

Recognizing and correctly applying antiderivatives is essential for solving integrals, leading to solutions in areas, volumes, and other applications in calculus.