The problem asks to find the surface area of the solid generated by revolving the curve $y = \frac{1}{x}$ from $x = 1$ to $x = 5$ around the y-axis.

The formula for the surface area of a solid of revolution around the y-axis is given by:

$A = 2\pi \int_{x_1}^{x_2} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

Where:

• $y = \frac{1}{x}$
• $\frac{dy}{dx} = -\frac{1}{x^2}$
• The limits of integration are from $x = 1$ to $x = 5$.

Step 1: Setting up the integral

Substitute into the formula:

$A = 2\pi \int_1^5 x \sqrt{1 + \left(-\frac{1}{x^2}\right)^2} \, dx$

First, simplify the expression inside the square root:

$1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(-\frac{1}{x^2}\right)^2 = 1 + \frac{1}{x^4}$

Thus, the integral becomes:

$A = 2\pi \int_1^5 x \sqrt{1 + \frac{1}{x^4}} \, dx$

Step 2: Approximating the integral

Since this integral cannot be evaluated easily by hand, we will approximate it using a numerical method.

Let me calculate it for you.The approximate surface area of the solid generated by revolving the curve $y = \frac{1}{x}$ from $x = 1$ to $x = 5$ around the y-axis is approximately 76.8030 square units (rounded to four decimal places).

Would you like more details on the process or have any further questions?

Here are five related questions to explore:

1. How would the surface area change if the curve revolved around the x-axis instead of the y-axis?
2. What if the curve was $y = \frac{1}{x^2}$? How would the surface area integral differ?
3. How would the surface area be affected if the limits were from $x = 2$ to $x = 6$?
4. How can we compute the volume of this solid instead of its surface area?
5. What numerical methods can be used to approximate integrals like this one?

Tip: When setting up integrals for surface areas, always pay attention to whether the curve is revolving around the x-axis or the y-axis, as this changes the setup of the integral.