Find the surface area of the volume generated when the following curve revolves around the x-axis. If you cannot evaluate the integral exactly, use your calculator to approximate it. (Round your answer to four decimal places.)
y =

x
from x = 2 to x = 6

To find the surface area of the volume generated when the curve \( y = x \) revolves around the x-axis from \( x = 2 \) to \( x = 6 \), we use the formula for the surface area of a solid of revolution:

\[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]

For the given function \( y = x \), the derivative with respect to \( x \) is:

\[ \frac{dy}{dx} = 1 \]

Substituting into the formula, we get:

\[ A = 2\pi \int_{2}^{6} x \sqrt{1 + (1)^2} \, dx \]

This simplifies to:

\[ A = 2\pi \int_{2}^{6} x \sqrt{2} \, dx \]

\[ A = 2\pi\sqrt{2} \int_{2}^{6} x \, dx \]

Now, we evaluate the integral:

\[ A = 2\pi\sqrt{2} \left[ \frac{x^2}{2} \right]_{2}^{6} \]

\[ A = 2\pi\sqrt{2} \left( \frac{6^2}{2} - \frac{2^2}{2} \right) \]

\[ A = 2\pi\sqrt{2} \left( \frac{36}{2} - \frac{4}{2} \right) \]

\[ A = 2\pi\sqrt{2} \left( 18 - 2 \right) \]

\[ A = 2\pi\sqrt{2} \cdot 16 \]

\[ A = 32\pi\sqrt{2} \]

Using a calculator, we find the numerical value:

\[ A \approx 142.1859 \]

Thus, the surface area of the volume generated is approximately **142.1859 square units**.

---

Would you like more details or have any questions? Here are some related questions:

1. How does the surface area formula change if the curve revolves around the y-axis?
2. What is the significance of the \(\sqrt{1 + (dy/dx)^2}\) term in the surface area formula?
3. How would the surface area calculation differ if the curve was \( y = x^2 \)?
4. Can you describe the general process of finding the surface area of a solid of revolution?
5. What are other methods to approximate integrals besides using a calculator?
6. What is the geometric interpretation of the integral in the surface area formula?
7. How does the surface area change if the limits of integration are altered?
8. What happens to the surface area if the function is rotated around a line other than the x-axis?

**Tip:** When solving for the surface area of a solid of revolution, ensure that you correctly identify the axis of rotation and apply the correct formula.

Learn how to find the surface area of a solid of revolution generated when the curve y = x revolves around the x-axis from x = 2 to x = 6. This problem involves integral calculus and the application of the surface area formula for solids of revolution. Approximate the answer to four decimal places.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation