Question

Find the volume of the solid generated by revolving about the \(x\) -axis the region under the curve \(y=\frac{3}{x}\) from \(x=1\) to \(x=\infty\)

Short answer

The volume is \(9\pi\) cubic units.

Step-by-step solution

Set up the Volume Integral

To find the volume of the solid generated by revolving the region under the curve around the x-axis, we use the disk method. The formula for the volume of revolution about the x-axis is given by \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \]Here, \(f(x) = \frac{3}{x}\), \(a = 1\), and \(b = \infty\). Substitute into the integral to get:\[ V = \pi \int_{1}^{\infty} \left(\frac{3}{x}\right)^2 dx \]

Simplify the Integral

Simplify the integrand:\[ \left(\frac{3}{x}\right)^2 = \frac{9}{x^2} \]Thus, the volume integral becomes:\[ V = \pi \int_{1}^{\infty} \frac{9}{x^2} dx \]

Evaluate the Integral

Evaluate the integral:\[ V = \pi \int_{1}^{\infty} \frac{9}{x^2} dx = 9\pi \int_{1}^{\infty} \frac{1}{x^2} dx \]The antiderivative of \(\frac{1}{x^2}\) is \(-\frac{1}{x}\). So,\[ 9\pi \int_{1}^{\infty} \frac{1}{x^2} dx = 9\pi [-\frac{1}{x}]_{1}^{\infty} \]

Apply Limits of Integration

Substitute the limits of integration:\[ 9\pi \left([-\frac{1}{x}] \bigg|_{1}^{\infty} \right) = 9\pi \left(-0 + 1\right) \]Since \(\frac{1}{\infty} = 0\), the negative sign is already distributed, and you evaluate the limits as:\[ 9\pi (0 - (-1)) = 9\pi (1) = 9\pi \]

Finalize Volume Calculation

After calculating the integral and applying the limits, we find:\[ V = 9\pi \]Thus, the volume of the solid of revolution is \(9\pi\) cubic units.

Key concepts

Disk Method

The disk method is a technique used to find the volume of a solid of revolution. It is particularly useful when a region is revolved around a horizontal or vertical line, commonly the x-axis or y-axis. This method involves slicing the solid perpendicular to the axis of rotation, resulting in thin disk-shaped slices. The volume of these slices can then be summed using integration.

To apply the disk method, we consider the formula:

• Volume, \( V = \pi \int_{a}^{b} [f(x)]^2 dx \)

Here, \( f(x) \) represents the function defining the edge of the solid, and \([a, b]\) is the interval over which the solid extends.

In our exercise, the function \( y = \frac{3}{x} \) signifies the boundary that forms the shape when revolved around the x-axis, beginning from \( x=1 \) and extending to infinity. By substituting into the formula, the process starts with calculating the square of the function and integrating it over the desired interval.

Improper Integral

An improper integral is an integral that has at least one infinite limit or an integrand that becomes infinite within the interval of integration. When working with an improper integral, we typically consider such limits using techniques of integration and limits.

In the exercise on finding the volume of revolution, where the region under \( y = \frac{3}{x} \) is revolved about the x-axis, there is an infinite upper limit, \( x = \infty \). This makes the integral improper.

To tackle this, we solve it using a limit process:

• \( V = \pi \int_{1}^{\infty} \frac{9}{x^2} dx \)
• Evaluate the limit as \( b \to \infty \) of \( \int_{1}^{b} \frac{9}{x^2} dx \)

Here, the crucial step is recognizing that the function \( \frac{9}{x^2} \) decreases rapidly as \( x \) increases, allowing the integral to converge to a finite value.

Antiderivative

The antiderivative, also known as the indefinite integral, is a function whose derivative provides the original function within an integral. Finding the antiderivative is essential in solving definite integrals in calculus.

In our volume of revolution problem, once we set up the integral as \( \pi \int_{1}^{\infty} \frac{9}{x^2} dx \), our next task is to find the antiderivative of the integrand \( \frac{9}{x^2} \).

The antiderivative of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \). Therefore, the integral simplifies to:

• \( 9\pi \left[-\frac{1}{x}\right]_{1}^{\infty} \)

Calculating this, and using the limits of integration, involves substituting the upper limit (\( \infty \)), resulting in zero because \( \frac{1}{\infty} = 0 \), followed by substituting and evaluating the lower limit, \( x = 1 \). This process leads to the final volume result of \( 9\pi \).