Problem 43 - Thomas Calculus

Question

In Exercises 43 and \(44,\) find the volume of the solid generated by revolving each region about the given axis. The region in the first quadrant bounded above by the curve \(y=x^{2},\) below by the \(x\) -axis, and on the right by the line \(x=1\) about the line \(x=-1\)

Step-by-Step Solution

Answer

The volume is \( \frac{7\pi}{6} \).

Step 1: Identify the solid of revolution

The region to be revolved is bounded by the curve \( y = x^2 \), the \( x \)-axis, and the line \( x = 1 \). We will revolve this region around the vertical line \( x = -1 \).

Step 2: Determine the method

Since we are revolving around a vertical line, we choose the shell method. This involves integrating along \( x \), which means we express the formula in terms of \( x \).

Step 3: Set up the shell method formula

The shell method formula is \( V = 2\pi \int_a^b x \, h(x) \, dx \), where \( h(x) \) is the height of the shell and \( x \) is the radius from the axis of rotation. Here, the radius from \( x = -1 \) to \( x \) is \( x - (-1) = x + 1 \), and the height is \( h(x) = x^2 \).

Step 4: Define the integration bounds and expression

The bounds of integration are from \( x = 0 \) to \( x = 1 \). Substitute into the shell formula: \[ V = 2\pi \int_0^1 (x+1)(x^2) \, dx = 2\pi \int_0^1 (x^3 + x^2) \, dx. \]

Step 5: Integrate the function

Integrate the expression: \[ \int_0^1 (x^3 + x^2) \, dx = \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^1. \]

Step 6: Evaluate the definite integral

Evaluate the antiderivative at the bounds: \[ \left[ \frac{x^4}{4} + \frac{x^3}{3} \right]_0^1 = \left(\frac{1^4}{4} + \frac{1^3}{3}\right) - \left(\frac{0^4}{4} + \frac{0^3}{3}\right) = \frac{1}{4} + \frac{1}{3}. \]

Step 7: Compute the exact volume

Calculate \( \frac{1}{4} + \frac{1}{3} \): \[ = \frac{3}{12} + \frac{4}{12} = \frac{7}{12}. \] Multiply by \( 2\pi \): \[ V = 2\pi \times \frac{7}{12} = \frac{7\pi}{6}. \]

Step 8: Conclusion

The volume of the solid obtained by revolving the given region around the line \( x = -1 \) is \( \frac{7\pi}{6} \).

Key Concepts

Shell Method

The shell method is a powerful technique for finding the volume of a solid of revolution. When we have a region that we need to revolve around a vertical line, the shell method comes into play.

In simple terms, imagine cutting the solid into thin hollow cylinders or "shells." The method is especially helpful for cases where regions are revolving around lines that are not coordinate axes.

The formula for the shell method is:

• \[ V = 2\pi \, \int_a^b x \, h(x) \, dx \]
• Here, \( x \) is the radius of the shell (distance from the axis of rotation), and \( h(x) \) is the height of the shell.

Choosing the right method is crucial. When revolving around a horizontal axis, it's often easier to use the disk or washer method. For a vertical axis, the shell method often simplifies the process.

Definite Integral

A definite integral is a fundamental concept in calculus used to find the accumulation of quantities, such as areas under a curve. In the context of finding volumes, it helps in summing up an infinite number of small quantities, like the volumes of tiny shells or disks.

The definite integral has two limits (the bounds of integration), which define the interval over which we are integrating.

• In our case, these limits are from \( x=0 \) to \( x=1 \).
• This ensures that we consider the entire area of the region under the curve that we are revolving.

In our problem, we set up the integral \[ \int_0^1 (x^3 + x^2) \, dx \]and compute it to determine the total volume. Evaluating this will give us the net signed area, which in turn contributes to the calculated volume of the solid of revolution.

Solid of Revolution

The concept of a solid of revolution involves creating a 3D shape by revolving a 2D area or region around an axis. Imagine spinning a shape really fast like a windmill, forming volumes in the process.

This method is often used in calculus to model real-world objects like vases or bells, thanks to the symmetry and clarity that rotational methods provide.

Our specific exercise focuses on revolving the area bounded above by the curve \( y = x^2 \), below by the \( x \)-axis, and on the right by the line \( x = 1 \) around the line \( x = -1 \). By performing such rotations, we can find volumes of complex 3D shapes using integration techniques, like the shell method. These calculations are essential for fields such as engineering and physics.

Antiderivative

An antiderivative is a function whose derivative is the given function. While differentiation is like splitting numbers or shrinking, finding an antiderivative, or integration, is about combining them back together.

In the volume of revolution problems, we use antiderivatives to evaluate definite integrals. The antiderivative gives us the accumulated 'sum' of the function over an interval.

• Consider the function \( x^3 + x^2 \). Its antiderivative is \( \frac{x^4}{4} + \frac{x^3}{3} \).
• This step is crucial because it allows us to determine the volume by evaluating the antiderivative at the boundaries of our definite integral.

In our case, the antiderivative is evaluated at \( x = 0 \) and \( x = 1 \) to compute the total volume, and it's from the antiderivative that we find the actual volume: \( \frac{7\pi}{6} \), which is the final answer.