Find the surface area of the volume generated when the curve 

*y* = 

*x*

 revolves around the *x*-axis from 

(1, 1)

 to 

(4, 2).

To find the surface area of the volume generated when the curve \( y = x \) revolves around the x-axis from the point \((1, 1)\) to \((4, 4)\), 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 \( y = x \), we find the derivative with respect to \( x \):

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

Substituting into the formula, we have:

\[ A = 2\pi \int_{1}^{4} x \sqrt{1 + (1)^2} \, dx \]

This simplifies to:

\[ A = 2\pi \int_{1}^{4} x \sqrt{2} \, dx \]

\[ A = 2\pi\sqrt{2} \int_{1}^{4} x \, dx \]

Next, we evaluate the integral:

\[ A = 2\pi\sqrt{2} \left[ \frac{x^2}{2} \right]_{1}^{4} \]

\[ A = 2\pi\sqrt{2} \left( \frac{4^2}{2} - \frac{1^2}{2} \right) \]

\[ A = 2\pi\sqrt{2} \left( \frac{16}{2} - \frac{1}{2} \right) \]

\[ A = 2\pi\sqrt{2} \left( 8 - 0.5 \right) \]

\[ A = 2\pi\sqrt{2} \cdot 7.5 \]

\[ A = 15\pi\sqrt{2} \]

Calculating the numerical value:

\[ A \approx 15 \times 3.1416 \times 1.4142 \]

\[ A \approx 66.3514 \]

Thus, the surface area of the volume generated is approximately **66.3514 square units**.

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

1. What changes occur in the surface area formula if the curve rotates around the y-axis instead?
2. How do you find the surface area if the function \( y = \sin(x) \) is revolved around the x-axis?
3. What is the significance of the differential element in the surface area formula?
4. How would you set up the integral for the surface area if the function were \( y = \frac{1}{x} \)?
5. What is the role of the function's derivative in the surface area formula?
6. How does the choice of limits affect the calculation of the surface area?
7. Can this method be used for curves that are not continuous over the interval?
8. How can numerical methods be used to approximate surface areas for more complex functions?

**Tip:** Always ensure the function and its derivative are continuous and well-defined over the interval of integration when using these formulas.

Learn how to calculate the surface area of revolution when the curve y = x revolves around the x-axis from (1, 1) to (4, 2). Detailed step-by-step solution using integral calculus and surface area formulas. Suitable for undergraduate students studying calculus.

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