Problem 74 - Thomas Calculus

Question

The infinite paint can or Gabriel's horn As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\) -axis, di- verges also. By comparing the two integrals, we see that, for every finite value \(b>1\) , $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (GRAPH NOT COPY) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. (a) Calculate it. ( \(\mathbf{b} )\) This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

Step-by-Step Solution

Answer

The volume is \(\pi\). Finite volume fills infinite surface due to different convergence properties.

Step 1: Understand the Problem

We are asked to calculate the volume of the solid of revolution of the curve \(y = \frac{1}{x}\) from \(x = 1\) to \(x = \infty\) about the x-axis, using the integral \(\int_{1}^{\infty} \pi \left(\frac{1}{x}\right)^{2} dx\). Then, explain why this solid is like a paint can that cannot cover its own interior yet can fill with paint.

Step 2: Set Up the Volume Integral

The formula for the volume of a solid of revolution about the x-axis involves \(\pi\) times the integral of the square of the function. Here, the function is \(\frac{1}{x}\), so we set up the integral as \(\int_{1}^{\infty} \pi \left(\frac{1}{x}\right)^{2} dx\).

Step 3: Simplify the Integrand

Simplify the integrand \(\pi \left(\frac{1}{x}\right)^{2}\) to \(\pi \frac{1}{x^2}\). The integral then becomes \(\int_{1}^{\infty} \pi \frac{1}{x^2} dx\).

Step 4: Integrate

Integrate \(\pi \frac{1}{x^2}\) with respect to \(x\). The integral of \(\frac{1}{x^2}\) is \(-\frac{1}{x}\), so the antiderivative of \(\pi \frac{1}{x^2}\) is \(-\pi \frac{1}{x}\).

Step 5: Evaluate the Definite Integral

Evaluate \([-\pi \frac{1}{x}]_{1}^{\infty}\). - As \(x\) approaches infinity, \(-\pi \frac{1}{x}\) approaches 0.- At \(x = 1\), \(-\pi \frac{1}{1} = -\pi\).Thus, the volume is \(0 - (-\pi) = \pi\).

Step 6: Address the Apparent Contradiction

The paint can analogy describes a paradox where a finite volume can fill the solid, yet the surface area is infinite. This happens because as you integrate \(\frac{1}{x^2}\) to find volume, the terms converge, but integrating \(\frac{1}{x}\) (for surface area) diverges. Filling with paint refers to filling space, which is finite; covering surface refers to endlessly spreading paint, which is infinite.

Key Concepts

Improper Integrals

An improper integral is used when dealing with integrals that have infinite limits or integrands with infinite discontinuities. This happens typically when the integral extends over an infinite range, or when the integrand is undefined at some points within the range. For instance, the exercise involves the integral \( \int_{1}^{\infty} \frac{1}{x} \, dx \), which diverges, meaning it doesn't result in a finite value.

• Divergent Integrals: These are integrals that do not converge. When you work with \( \int_{1}^{\infty} \frac{1}{x} \, dx \), over an infinite interval, it indicates the area under the curve of \( y = \frac{1}{x} \) from 1 to infinity keeps increasing: hence the divergence.
• Comparison Test: It helps us determine the convergence or divergence of integrals. When comparing \( \int_{1}^{\infty} 2\pi\frac{1}{x}\sqrt{1+\frac{1}{x^{4}}} \, dx \) with \( \int_{1}^{\infty} \frac{1}{x}dx \), we see both share convergence or divergence properties because the latter diverging means the former does too.

These integrals help in finding areas and volumes in cases like these but can lead to paradoxes, as seen with the solid of revolution in this exercise.

Surface Area of Revolution

Surface area of revolution is calculated when a curve is revolved around a given axis. The formula typically depends on the curve's equation and requires setting up an integral. In the case of the curve \( y = \frac{1}{x} \) being revolved around the x-axis, the integral that measures this surface area is \( \int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} \, dx \). This shows that although this integral diverges, meaning it heads towards infinity, it represents an intriguing phenomenons:

• Formula Derivation: The surface area formula \( 2 \pi \int_{a}^{b} f(x)\sqrt{1+[f'(x)]^{2}}\, dx \) is derived based on calculus, involving the arc length formula.
• Appearance of Divergence: The integral diverges as it assesses an infinite surface area, which aligns with the paradox of Gabriel's Horn – infinitely large surfaces, despite finite limits.

The takeaway is the concept of divergence in certain integrals, which starkly contrasts with finite volume in solids of revolution.

Volume of Solids of Revolution

To find the volume of a solid that's created when a curve is revolved around an axis, you use the disk or washer method. For the curve \( y = \frac{1}{x} \) revolved around the x-axis, the volume is computed using the integral \( \int_{1}^{\infty} \pi \left( \frac{1}{x} \right)^{2} \, dx \). This specific integral surprisingly converges, meaning it results in a finite value, which illustrates a famous mathematical paradox:

• Disk Method: This method involves slicing the solid into thin disks perpendicular to the axis of rotation, evaluating the volume of each disk, and totalizing.
• Convergence in Action: Despite the infinite surface, the volume \( \pi \int_{1}^{\infty} \frac{1}{x^2} \, dx \) converges to \( \pi \), pointing out how infinitude in height contribution doesn't affect a limited volume build-up.
• Gabriel's Horn Paradox: This horn has a finite volume (meaning it can be filled with a finite amount of paint) yet an infinite surface area (meaning it theoretically requires an infinite amount of paint to cover).

This is key in calculus for understanding concepts of finite versus infinite regions and explains how we mathematically interpret and work through these paradoxes.