How to Find the Surface Area of a Rotated Curve y1/x; 1/2≤x≤1

Calculus is a powerful tool in understanding the properties of curves and their transformations. One intriguing problem involves finding the surface area generated by rotating a curve around the x-axis. In this article, we will explore how to calculate the surface area of the surface obtained by rotating the curve y1/x within the interval 1/2≤x≤1 about the x-axis.

Understanding the Problem

The curve in question is y1/x, defined from x1/2 to x1. We need to find the surface area of the surface generated by rotating this curve around the x-axis.

The Formula and Steps

The surface area of a surface generated by rotating a curve around the x-axis can be found using the formula:

Area 2π∫[y ds], where ds is the differential arc length of the curve.

Step 1: Finding the Arc Length Element

The arc length element of the curve y1/x is given by:

ds √(1 (dy/dx)2) dx

For the given curve, we have:

y 1/x

Therefore, the derivative with respect to x is:

dy/dx -1/x2

Substituting this into the arc length formula:

ds √(1 (-1/x2)2) dx

This simplifies to:

ds √(1 1/x4) dx

Step 2: Setting Up the Integral

The surface area element when the curve is rotated about the x-axis is given by:

2πy ds 2π(1/x) √(1 1/x4) dx

To find the total surface area, we need to integrate this expression from 1/2 to 1:

Surface Area 2π ∫1/21 (1/x) √(1 1/x4) dx

Step 3: Simplifying the Integral

To simplify the integral, we can make a substitution. Let:

x 1/y

Then:

dx -1/y2 dy

When x 1/2, y 2, and when x 1, y 1. Also, the integrand transforms according to this substitution:

2π ∫21 y √(1 y4) dy/y2

This simplifies to:

2π ∫21 √(1 y4) dy/y

Conclusion and Further Insights

The integral, 2π ∫21 √(1 y4) dy/y, can be evaluated using numerical methods or advanced calculus techniques. It represents the surface area of the surface formed by rotating the curve y1/x from x1/2 to x1 about the x-axis.

Understanding and applying these concepts can provide deeper insights into the properties of curves and their transformations. For further learning, you can explore more complex integrals and their applications in real-world scenarios. Whether you're a mathematician, engineer, or student, mastering these concepts can unlock new avenues of exploration in calculus and geometry.

Remember that the key to solving these problems effectively lies in breaking down the steps and using appropriate substitutions. By following these steps, you can tackle similar problems and deepen your understanding of calculus and its applications.