Problem 217 - Calculus Volume 1

Question

Explain why the surface area is infinite when \(y=1 / x\) is rotated around the \(x\) -axis for \(1 \leq x<\infty\) , but the volume is finite.

Step-by-Step Solution

Answer

The surface area is infinite due to divergence, while the volume is finite due to convergence.

Step 1: Understand the Problem

We are asked about the surface area and the volume when the curve \(y = \frac{1}{x}\) is rotated around the \(x\)-axis from \(x=1\) to \(x=\infty\). The goal is to determine why the surface area is infinite while the volume is finite for this solid of revolution.

Step 2: Formula for Surface Area

The formula for the surface area \(S\) of a solid obtained by rotating a curve \(y = f(x)\) about the \(x\)-axis is\[ S = \int_a^b 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx.\]For our function \(y = \frac{1}{x}\), we need to calculate \( \frac{dy}{dx} \) and substitute into this formula with the limits \(a = 1\) and \(b = \infty\).

Step 3: Calculate Derivative and Integrate for Surface Area

The derivative \(\frac{dy}{dx} = -\frac{1}{x^2}\). Substituting in the surface area formula gives:\[ S = \int_1^\infty 2\pi \frac{1}{x} \sqrt{1 + \left( \frac{-1}{x^2} \right)^2} \, dx = \int_1^\infty 2\pi \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} \, dx. \]This integral diverges as \(x \to \infty\), leading to an infinite surface area.

Step 4: Formula for Volume

The formula for the volume \(V\) of the solid is\[ V = \pi \int_a^b y^2 \, dx. \]For \(y=\frac{1}{x}\), this becomes:\[ V = \pi \int_1^\infty \left(\frac{1}{x}\right)^2 \, dx. \]

Step 5: Integrate for Volume

Substituting the function into the volume formula, we have:\[ V = \pi \int_1^\infty \frac{1}{x^2} \, dx = \pi \left[ -\frac{1}{x} \right]_1^\infty = \pi \left(0 + 1\right) = \pi. \]Since the integral converges, the volume is finite.

Step 6: Conclusion

The surface area is infinite because its corresponding integral diverges, while the volume is finite because its integral converges.

Key Concepts

Surface Area

The concept of surface area involves measuring the total extent of a 3D object's outer layer. When we rotate a curve around an axis, we create a solid of revolution, and calculating the surface area can become quite complex. To find the surface area when a curve like \(y = \frac{1}{x}\) is rotated about the \(x\)-axis, we use a specific formula:

• \(S = \int_a^b 2\pi y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx\)

Here, \(y=\frac{1}{x}\) and its derivative \(\frac{dy}{dx} = -\frac{1}{x^2}\) need to be plugged into this formula. The limits of integration are given as \(a = 1\) and \(b = \infty\). Substituting these into the formula results in an integral:

• \(S = \int_1^\infty 2\pi \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} \, dx\)

As \(x\) approaches infinity, this integral diverges, which means the surface area becomes infinitely large. This is why we say the surface area is infinite when revolving the curve \(y=\frac{1}{x}\) around the \(x\)-axis.

Volume

Volume deals with the amount of space enclosed within a solid. For solids of revolution, the formula to calculate volume is:

• \(V = \pi \int_a^b y^2 \, dx\)

Applying this to our function \(y = \frac{1}{x}\), the formula becomes:

• \(V = \pi \int_1^\infty \left(\frac{1}{x}\right)^2 \, dx\)

Simplifying further, the integral is:

• \(V = \pi \int_1^\infty \frac{1}{x^2} \, dx\)

When we evaluate this integral, it is found to converge, resulting in a finite value:

• \(V = \pi \left[ -\frac{1}{x} \right]_1^\infty = \pi (0 + 1) = \pi\)

This means the volume is finite, even as the shape extends infinitely along the \(x\)-axis. The behavior of the function \(\frac{1}{x^2}\) ensures the integral doesn't diverge, leading to a finite volume measure.

Solid of Revolution

A solid of revolution is created when a two-dimensional curve is rotated around an axis. This process generates a three-dimensional shape. To visualize, imagine taking a flat curve and spinning it around the \(x\)-axis like spinning a hula hoop. This rotation creates a symmetrical 3D object.

• The curve \(y = \frac{1}{x}\) rotated about the \(x\)-axis forms such a solid.
• The portion of the curve from \(x = 1\) to \(x = \infty\) results in a shape that extends infinitely along the \(x\)-axis.

Understanding the properties of these solids, particularly in terms of surface area and volume, provides deep insights into the behavior of certain functions and integrals. In our case, the peculiar property is having infinite surface area but finite volume.

Improper Integrals

Improper integrals are a special type of integral used when the limits of integration approach infinity or when the modulus of the integrand increases without bound. In simple terms, they help calculate the area under curves that stretch endlessly.

• The instructions solve the integral from 1 to \(\infty\) to find the surface area and volume of the solid of revolution.
• For the surface area, the integral diverges due to the behavior of \(2\pi \frac{1}{x}\sqrt{1 + \frac{1}{x^4}}\) as \(x\) grows, making the result infinite.
• Conversely, for the volume, the behavior of \(\frac{1}{x^2}\) as \(x\) increases ensures convergence, resulting in a finite outcome.

Improper integrals thus allow us to manage infinities and assess whether certain areas or volumes are finite or infinite. These are crucial for analyzing curves and solids that extend to infinity.