Problem 16 - Thomas Calculus in SI Units

Question

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$y=\sqrt{x+1}, \quad 1 \leq x \leq 5 ; \quad x \text { -axis }$$

Step-by-Step Solution

Answer

The surface area is \(4\pi\sqrt{5}\).

Step 1: Identify the Function and Interval

We are given the function \(y = \sqrt{x+1}\) and want to find the area of the surface created when this curve is revolved about the x-axis over the interval \(1 \leq x \leq 5\). This sets up the problem for integration.

Step 2: Surface Area Formula for Revolution

The formula for the surface area \(S\) of a curve \(y=f(x)\) revolved around the x-axis from \(a\) to \(b\) is \[ S = 2\pi \int_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx. \]

Step 3: Compute the Derivative

First, find the derivative of \(y\). \(y = \sqrt{x+1}\) implies \(\frac{dy}{dx} = \frac{1}{2\sqrt{x+1}}\).

Step 4: Set up the Integral

Substitute \(y\) and \(\frac{dy}{dx}\) into the surface area formula:\[ S = 2\pi \int_1^5 \sqrt{x+1} \sqrt{1 + \left(\frac{1}{2\sqrt{x+1}}\right)^2} \, dx.\]Simplifying the expression under the square root:\[ S = 2\pi \int_1^5 \sqrt{x+1} \sqrt{1 + \frac{1}{4(x+1)}} \, dx = 2\pi \int_1^5 \sqrt{x+1} \sqrt{\frac{5}{4(x+1)}} \, dx.\]

Step 5: Simplify the Integral

By simplifying further, the integral becomes:\[ S = 2\pi \int_1^5 \sqrt{\frac{5(x+1)}{4(x+1)}} \, dx = 2\pi \sqrt{\frac{5}{4}} \int_1^5 1 \, dx.\]Since \(\sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\), factor it out:\[ S = \pi \sqrt{5} \int_1^5 \, dx.\]

Step 6: Evaluate the Integral

Now compute the integral:\[ S = \pi \sqrt{5} [x]_1^5 = \pi \sqrt{5} (5 - 1) = 4\pi \sqrt{5}.\]

Step 7: Interpret the Result

The area of the surface generated by revolving the curve \(y = \sqrt{x+1}\) from \(x=1\) to \(x=5\) about the x-axis is \(4\pi\sqrt{5}\).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that focuses on finding the total size or value, namely finding areas underneath curves and computing accumulations. In this context, when we talk about the area of surfaces, integral calculus becomes an essential tool. It helps us not just measure areas on a flat surface, but also areas created by more complex shapes like the surface of a revolution.
When we say we're going to integrate a function, what we're really doing is summing up infinite tiny pieces of area that fit under a curve between two points. In our problem, we're revolving a curve around an axis, so we're not just adding up tiny areas, but rather building up a surface by stacking up these tiny bits that wrap around in a circular fashion. This requires setting up an integral using a special formula that incorporates the curve equation and its derivative to account for the curve's slope and shape as it revolves.

Surface Area Formula

The surface area formula is integral to figuring out areas of surfaces generated by curves. The specific formula we use when revolving a curve around the x-axis is:

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

Here, each term plays a role:

• \( 2\pi \) represents the full circle (360 degrees) that the curve forms as it revolves.
• \( \int_{a}^{b} \) signals we're summing continuous slices from point \( a \) to point \( b \).
• \( y \) is the function itself; we're essentially multiplying the "height" of each strip by the circumference of its rotation.
• \( \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \) accounts for the slope of the curve. This ensures that when the curve's surface is in motion, the incline of the curve is properly apportioned into the final surface area being computed.

This formula is elaborate, but each component ensures a precise calculation of the surface area.

Revolving Curves

Revolving curves is a fancy way of explaining how we create surfaces by rotating a 2D line or curve through 360 degrees around a straight line—the axis. The resulting 3D object is called a surface of revolution. Imagine holding a piece of string straight and then swinging it around a pencil you hold vertically; the string describes a surface as it swings around.
In our example, the curve \( y = \sqrt{x+1} \) is revolved around the x-axis. Each little length segment along this curve sweeps out a tiny ring, and all those rings together make a surface. When we use calculus, and particularly the integral we discussed, we are adding up the areas of all these rings or layers to get the total surface area.
Revolving curves can create beautiful shapes—like vases or horns. Understanding this concept not only develops your mathematical thinking but also your spatial awareness.