Question

Find the area of the surface generated when the given curve is revolved about the \(y\) -axis. The part of the curve \(y=\frac{1}{2} \ln (2 x+\sqrt{4 x^{2}-1})\) between the points \(\left(\frac{1}{2}, 0\right)\) and \(\left(\frac{17}{16}, \ln 2\right)\)

Short answer

Use the finding that after simplification: \((y')^2 + 1 = 1 + \frac{16x^2}{(2x+\sqrt{4x^2-1})^2}\).

Step-by-step solution

Find the derivative of y with respect to x

To find the derivative, we apply the chain rule to the given function, \(y=\frac{1}{2} \ln (2 x+\sqrt{4 x^{2}-1})\). Differentiating with respect to \(x\), we get: $$ y' = \frac{1}{2} \cdot \frac{1}{2x+\sqrt{4x^2-1}} \cdot \frac{d(2x+\sqrt{4x^2 -1})}{dx} $$ Now, we need to find the derivative of \(2x + \sqrt{4x^2 - 1}\) with respect to \(x\). $$ \frac{d(2x +\sqrt{4x^2 -1})}{dx} = 2 + \frac{d(\sqrt{4x^2 -1})}{dx} $$ Differentiating the square root term: $$ \frac{d(\sqrt{4x^2 -1})}{dx} = \frac{4x}{\sqrt{4x^2-1}} $$ So, our derivative becomes: $$ y' = \frac{1}{2} \cdot \frac{1}{2x+\sqrt{4x^2-1}} \cdot \left(2 + \frac{4x}{\sqrt{4x^2-1}}\right) $$

Square the derivative and add 1

Now that we have our derivative, we need to square it and add \(1\). When we do this, we need to simplify the expression as much as possible. $$ (y')^2+1 = \left[\frac{1}{2} \cdot \frac{1}{2x+\sqrt{4x^2-1}} \cdot \left(2 + \frac{4x}{\sqrt{4x^2-1}}\right)\right]^2 + 1 $$ After some algebraic simplifications, it should look something like this: $$ (y')^2+1 = 1 + \frac{16x^2}{(2x+\sqrt{4x^2-1})^2} $$

Multiply the function x with the square root of the squared derivative plus one

Now we have the expression for \((y')^2+1\). Next, we should multiply the function \(x\) with the square root of this expression: $$ x \sqrt{1+ (y')^2} = x\sqrt{1 + \frac{16x^2}{(2x+\sqrt{4x^2-1})^2}} $$

Set up the integral and evaluate the surface area

With the expression above, now we can set up our integral and calculate the surface area. We integrate from the x coordinates of our given points: \(a = \frac{1}{2}\) and \(b = \frac{17}{16}\). $$ A = 2 \pi \int_{\frac{1}{2}}^{\frac{17}{16}} x\sqrt{1 + \frac{16x^2}{(2x+\sqrt{4x^2-1})^2}} \, dx $$ This integral can be solved using numerical integration methods or using a graphing calculator. This will provide us with the surface area when the curve is revolved about the \(y\)-axis. Note: The integral does not have a simple closed form, so an approximation is required for the evaluation.

Key concepts

Calculus

Calculus is a branch of mathematics focused on the study of rates of change and accumulation. In this problem, we explore how calculus helps to find the surface area of a curve revolution around an axis. Understanding calculus profoundly involves mastering two main ideas: differentiation and integration. Differentiation gives us the rate at which something changes—here, it is the derivative of the function that defines the curve. Integration, on the other hand, accumulates or sums up values over a range. It is key to solving surface area problems because we need to accumulate the infinitesimally small areas that make up an overall shape. The line integration from point to point on a curve and around an axis is what gives rise to the surface area when a curve is revolved.

Definite Integral

A definite integral is a fundamental concept in calculus, representing the accumulation of quantities over a certain interval. In the context of this exercise, it helps us find the exact surface area of the curve as it revolves around the y-axis. The definite integral we encounter can be expressed as:

• It involves the limits of integration; here from \( x = \frac{1}{2} \) to \( x = \frac{17}{16} \), reflecting where along the x-axis the area is measured.
• This integral takes the form \( \int_{a}^{b} f(x) \, dx \), where \( f(x) \) represents the function describing the curve revolved.
• The definite integral essentially sums up all infinitely small areas within these limits, compiling them together to produce one total surface area.

The result is a single value, indicating the total area of the surface created. The difficulty lies in solving this integral; often, numerical methods or specific software help us estimate the value, especially when the integral lacks a straightforward antiderivative.

Chain Rule

The chain rule is a technique in calculus used to find the derivative of a composition of functions. When differentiating functions within functions, like the curve given, the chain rule becomes a valuable tool. Here's how the chain rule fits into our problem:

• First, the function \( y = \frac{1}{2} \ln(2x + \sqrt{4x^2-1}) \) is not straightforward. It involves a natural logarithm and a nested square root function.
• The chain rule allows us to deal with these layers appropriately, first targeting the outer function and then multiplying by the derivative of the inner function.
• The sequence of differentiation is the hallmark of the chain rule: differentiate outer, then inner, and multiply them together to get the result.

By applying the chain rule carefully, we obtain the derivative \( y' \), which we need to calculate the surface area of the curve revolved around the y-axis. This method of differentiation ensures we do not miss any steps and maintain the integrity of the function's growth rate as we work through it.