Problem 46 - Thomas Calculus in SI Units

Question

Find the volume of the solid generated by revolving each region about the given axis. The region in the second quadrant bounded above by the curve \(y=-x^{3},\) below by the \(x\) -axis, and on the left by the line \(x=-1\) about the line \(x=-2\)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{3\pi}{5}\) cubic units.

Step 1: Identify the bounds and region

The region is bounded by the curve \(y = -x^3\), the x-axis, and the line \(x = -1\). In the second quadrant, the bounds for \(x\) are from \(-1\) to 0, and for \(y\) are from the curve to the x-axis.

Step 2: Set up the volume integral

Since we are revolving around the line \(x = -2\), use the shell method. The formula for the volume using the shell method is: \[ V = \int_{a}^{b} 2\pi (radius)(height) \, dx \] The radius from the axis of rotation is \(|x + 2|\), and the height is \(-x^3 - 0 = -x^3\).

Step 3: Determine the radius and height expressions

The radius from the axis of rotation to the shell is \(-x + 2\). Since the region is in the second quadrant and \(x\) is negative, \(\text{radius} = 2 + x\). The height of the shell is \(-x^3\).

Step 4: Write the integral for volume

Substitute the expressions into the shell method formula:\[ V = \int_{-1}^{0} 2\pi (2 + x)(-x^3) \, dx \]

Step 5: Simplify the integrand

First, expand the integrand:\[ V = \int_{-1}^{0} 2\pi (-x^3)(2 + x) \, dx \] \[ = \int_{-1}^{0} 2\pi (-2x^3 - x^4) \, dx \] \[ = 2\pi \left( \int_{-1}^{0} (-2x^3 - x^4) \, dx \right) \]

Step 6: Integrate the function

Integrate both terms:\[ \int_{-1}^{0} (-2x^3) \, dx = -\frac{1}{2}x^4 \Bigg|_{-1}^{0} \]\[ \int_{-1}^{0} (-x^4) \, dx = -\frac{1}{5}x^5 \Bigg|_{-1}^{0} \]

Step 7: Evaluate the definite integrals

Evaluate:\[ -\frac{1}{2}(0)^4 + \frac{1}{2}(-1)^4 = \frac{1}{2} \]\[ -\frac{1}{5}(0)^5 + \frac{1}{5}(-1)^5 = -\frac{1}{5} \] Combine results:\[ = 2\pi \left( \frac{1}{2} - \frac{1}{5} \right) = 2\pi \left( \frac{5}{10} - \frac{2}{10} \right) = 2\pi \left( \frac{3}{10} \right) = \frac{6\pi}{10} = \frac{3\pi}{5} \]

Step 8: Conclusion on volume

The volume of the solid is \(\frac{3\pi}{5}\) cubic units after evaluating the integrals and combining the results.

Key Concepts

Shell Method

The Shell Method is a powerful technique used in calculus to find the volume of solids of revolution. When a region in the plane is revolved around an axis not touching the region, we use this method. This involves imagining cylindrical "shells" that make up the solid.

• With the Shell Method, you revolve around a vertical or horizontal axis.
• The key part is finding the "shell radius" and "shell height" based on distances and functions involved.

To apply the Shell Method, visualize each layer of the shell being built from the vertical slices of the shape. For the exercise at hand, the region is described by the curve \(y = -x^3\) and is revolved around \(x = -2\). The radius of each shell is \(2 + x\), and the height is \(-x^3\). By setting up the integral with these components, you can calculate the volume.

Definite Integrals

Definite integrals are foundational in calculus, allowing us to calculate the "total" of quantities like area, volume, and more over a specific interval. The key features of definite integrals include:

• The integral is evaluated over a specific range, with limits \(a\) to \(b\).
• The result is a number representing the total accumulation.

Let's apply this to the problem of calculating the volume of the solid of revolution. The bounds are given by the domain of \(x\), from \(-1\) to 0. By setting up the integral as \(\int_{-1}^{0} 2\pi(2+x)(-x^3)\, dx\), we capture how the shell method's height and radius help describe the solid in terms of \(x\). Solving this integral computes the volume.

Calculus Applications

Applications of calculus allow us to solve real-world problems, including calculating the volume of shapes that would otherwise be complex to figure visually. Calculus not only aids in defining abstract mathematical concepts but also offers applicable techniques to measure tangible properties.

• Volume calculations are just one of several ways calculus can be applied to physical situations.
• By leveraging calculus, we can transform simple shapes through rotation into complex solids.

In practice, the shell method efficiently computes these volumes without needing complex 3D modeling. Understanding these applications underscores calculus as both a theoretical and practical tool, empowering us to address diverse real-world challenges, from engineering to physics, and beyond.