Problem 70 - Calculus Early Transcendentals

Question

Find the area of the region bounded by the curve \(x=\frac{1}{2 y}-\sqrt{\frac{1}{4 y^{2}}-1}\) and the line \(x=1\) in the first quadrant. (Hint: Express \(y\) in terms of \(x\).)

Step-by-Step Solution

Answer

Answer: \(\frac{1}{2} \ln{3}\)

Step 1: Express y in terms of x

Isolate y from the given equation: \(x = \frac{1}{2y} - \sqrt{\frac{1}{4y^2} - 1}\) Move everything to one side of the equation and then square both sides to eliminate the square root: \((x + \sqrt{\frac{1}{4y^2} - 1})^2 = (\frac{1}{2y})^2\) Solve for \(y(x)\): \(y(x) = \frac{1}{2x+\sqrt{(2x-1)^2}}\)

Step 2: Find the intersection points

We need to find the intersections of the curve \(y(x)\) and the vertical line \(x = 1\). Set \(x = 1\) in the expression for \(y(x)\): \(y(1) = \frac{1}{2(1)+\sqrt{(2(1)-1)^2}}\) \(y(1) = \frac{1}{3}\) So the curve intersects the line \(x=1\) at \((1, \frac{1}{3})\) in the first quadrant.

Step 3: Set up the integral

Since we have expressed \(y\) in terms of \(x\), and we are given a vertical line with fixed x-value, we will integrate with respect to \(x\). We want to find the area between the curve and the vertical line \(x=1\) from \(y=0\) to \(y=\frac{1}{3}\). Using the expression we found earlier for \(y(x)\), we have: \(A = \int_0^1 y(x) dx\) \(A = \int_0^1 \frac{1}{2x+\sqrt{(2x-1)^2}} dx\)

Step 4: Evaluate the integral

Evaluate the integral by using substitution. Let \(u=2x-1\) and \(du=2dx\): \(A = \frac{1}{2} \int_{-1}^1 \frac{1}{u+\sqrt{u^2}} du\) Now, use another substitution, let \(v = u+\sqrt{u^2}\) and \(dv = \frac{u}{\sqrt{u^2}} du\): \(A = \frac{1}{2} \int_0^3 \frac{dv}{v}\) Now, integrate with respect to \(v\) and evaluate the definite integral: \(A = \frac{1}{2} \left[ \ln{v} \right]_0^3\) \(A = \frac{1}{2} \ln{3}\) So the area of the region bounded by the curve and the line \(x=1\) in the first quadrant is \(\frac{1}{2} \ln{3}\).

Key Concepts

Definite Integral

A definite integral is a concept in calculus that represents the total accumulation of quantities, like areas, over an interval. It's different from an indefinite integral, which represents a family of functions. In this problem, we use a definite integral to find the area between a curve and a line in the first quadrant.
To set up a definite integral, you need:

• An integrand, which in our case is the function expressed as \(y(x) = \frac{1}{2x+\sqrt{(2x-1)^2}}\).
• Limits of integration, which are the bounds on the x-axis or y-axis. Here, they are from \(0\) to \(1\).

Integrating this function over the specified interval gives the area under the curve between these limits. The calculations involve complex substitutions, making definite integrals powerful tools for handling intricate problems like this one.

Area between Curves

Finding the area between curves involves using calculus to measure the space enclosed by different functions. In this particular exercise, we are tasked with finding the area in the region bounded by a curve and a vertical line in the first quadrant.
Here's how you approach such problems:

• Identify the curves (in this case, a transformed curve and a vertical line).
• Find the intersection points. This ensures you're calculating the area over the correct segment.
• Set up the integral: The integral's function should go between these intersection points, giving a definite form to your area calculation.

Using definite integrals, we can capture the region between curves by integrating the top curve minus the bottom curve when both are functions of \(x\) or vice-versa. This way, we ensure that all the space between the curves is considered.

Function Transformation

Function transformation is a technique in mathematics that involves altering the position or shape of a graph of a function. In our problem, we deal with an unusual curve that needs manipulation for us to work with it easily.
Key transformations include:

• Translation: Shifting the graph up, down, left, or right.
• Reflection: Flipping the graph over a line, like the x-axis or y-axis.
• Stretching: Lengthening or shortening the graph in either direction.

To solve the integral problem, the given equation needed to be manipulated: it was restated from function \(x\) in terms of \(y\) to \(y\) in terms of \(x\). This transformation simplified the relationship, making applying calculus techniques more straightforward. Always consider transformations when faced with complex equations, as they can unveil an easier path to problem-solving.