Problem 81 - Calculus Early Transcendentals

Question

Find the area of the region bounded by \(y=\operatorname{sech} x, x=1,\) and the unit circle.

Step-by-Step Solution

Answer

Answer: The area of the region is given by \(\frac{\pi}{4} - \frac{1}{2}\left(\frac{e^2 - 1}{e^2 + 1}\right)^2\).

Step 1: Find the Intersection Points

To find the intersection points, we need to solve the equations \(y = \operatorname{sech} x, x = 1,\) and \(x^2 + y^2 = 1\) simultaneously. First, notice that \(x = 1\) is a vertical line, so we just need to find the points where it intersects with \(y = \operatorname{sech} x\) and with the unit circle. For the unit circle, substitute x=1 into the equation of the unit circle: \(1^2 + y^2 = 1\). That gives us \(y^2 = 0\), so \(y = 0\). So the intersection point is \((1,0)\). For the curve \(y = \operatorname{sech} x\), substitute x=1 into the equation: \(y =\operatorname{sech}(1)\). So the second intersection point is \((1, \operatorname{sech}(1))\).

Step 2: Set up the Integrals

To find the desired area, we will integrate along the x-axis from 0 to 1. Write the integral for each curve. For the unit circle: \(A_1 = \int_0^1 \sqrt{1-x^2}dx\) For the curve \(y = \operatorname{sech} x\): \(A_2 = \int_0^1 \operatorname{sech} x dx\) The desired area, \(A, is the difference between these two areas: \)A = A_1 - A_2$.

Step 3: Compute the Integrals

Calculate the individual integrals: For the unit circle: \(A_1 = \frac{\pi}{4}\), since this is the quarter of the unit circle. For \(y = \operatorname{sech} x\), we can make a substitution. Let \(u = \tanh(x)\). Thus, \(du = \operatorname{sech}^2(x) dx\) and \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\). So, our integral becomes, \(A_2 = \int \frac{1}{\cosh(x)} dx = \int \frac{\operatorname{sech}^2(x)}{\operatorname{sech}(x)} dx = \int u \cdot du\). From \(0\) to \(1\), the bounds of \(u\) change to \(0\) to \(\frac{e^2 - 1}{e^2 + 1}\). Thus, \(A_2 = \frac{1}{2}\left(\frac{e^2 - 1}{e^2 + 1}\right)^2\).

Step 4: Calculate the Desired Area

We have found the individual areas \(A_1\) and \(A_2\). Now we can calculate the desired area \(A\): \(A = A_1 - A_2 = \frac{\pi}{4} - \frac{1}{2}\left(\frac{e^2 - 1}{e^2 + 1}\right)^2\). The desired area of the region bounded by \(y = \operatorname{sech} x, x = 1,\) and the unit circle is \(\frac{\pi}{4} - \frac{1}{2}\left(\frac{e^2 - 1}{e^2 + 1}\right)^2\).

Key Concepts

Area Under a Curve

Finding the area under a curve is a fundamental concept in calculus. Imagine you have a curve defined on a graph, and you want to know how much space is underneath it between certain points. This is what is referred to as the "area under a curve."

Using integrals, you can calculate this area precisely. The process involves integrating the function of the curve over a specific interval.

• The integral of a function gives you the accumulated area from the starting point to the end point.
• This is equivalent to summing an infinite number of infinitely small rectangles under the curve.

In our problem, we used this concept to determine the area between the curve given by \(y = \operatorname{sech} x\) and the unit circle from \(x = 0\) to \(x = 1\). Understanding how to set up the integral is crucial for solving such problems, and helps in visualizing how different curves interact with each other.

Definite Integral

The definite integral is a specific type of integral used to calculate the exact area under a curve between two points. It has set limits, unlike an indefinite integral, which represents a family of functions.

In the context of this exercise, we set up definite integrals to find the areas under two different curves.

• For the unit circle, the integral was \(\int_0^1 \sqrt{1-x^2} \, dx\) which describes the area under the curve \(\sqrt{1-x^2}\).
• For the function \(y = \operatorname{sech} x\), the integral was \(\int_0^1 \operatorname{sech} x \, dx\).

These definite integrals give us precise numerical values, illustrating how much space each curve occupies over the specified interval. Understanding definite integrals as a tool to measure area helps in numerous applications of calculus, from physics to economics.

Unit Circle

The unit circle is a basic concept in mathematics where a circle has a radius of 1. It's centered at the origin of the coordinate system, described by the equation \(x^2 + y^2 = 1\). This simple structure offers profound insights into trigonometry and calculus.

In this exercise, the unit circle serves as a boundary for part of the region whose area we wish to find. Rather than a complete circle, we're concerned with the part from \(x = 0\) to \(x = 1\), effectively making use of a quarter of the circle.

• This section forms part of the area calculation involving integration.
• Integrating under the curve \(\sqrt{1-x^2}\) gives the quarter-circle area \(\frac{\pi}{4}\).

The unit circle is frequently employed in problems involving cyclic or periodic phenomena, due to its symmetry and foundational links to trigonometric functions.