Problem 38 - Single Variable Calculus

Question

In Exercises \(9-40\), sketch the region bounded by the graphs of the given equations and find the area of that region. $$ y=\sec ^{2} x, \quad y=\cos x, \quad x=-\frac{\pi}{3}, \quad x=\frac{\pi}{3} $$

Step-by-Step Solution

Answer

The area of the region bounded by the given graphs \(y = \sec^2{x}\), \(y = \cos{x}\), \(x = -\frac{\pi}{3}\), and \(x = \frac{\pi}{3}\) can be found by first determining the intersection point of the functions as \(x = 0\) and \(y = 1\). Then, we can calculate the area by integrating the difference between the functions over the given intervals: \[ Area = \int_{-\frac{\pi}{3}}^0 (\sec^2{x} - \cos{x}) dx + \int_{0}^\frac{\pi}{3} (\sec^2{x} - \cos{x}) dx. \] After calculating the integrals, we find the area to be \(\sqrt{3}\) square units.

Step 1: Sketch the graphs of the given equations

First we need to sketch the graphs of the given equations, namely \(y = \sec^2{x}\), \(y = \cos{x}\), \(x = -\frac{\pi}{3}\), and \(x = \frac{\pi}{3}\). The graphs will help us determine the region bounded by all these equations.

Step 2: Determine the intersection points

In order to calculate the area, we need to find the intersection points of the functions \(y = \sec^2{x}\) and \(y = \cos{x}\). We do this by solving the following equation for x: \(\sec^2{x} = \cos{x}\) We can change this to sine and cosine functions: \(\frac{1}{\cos^2{x}} = \cos{x}\) After multiplying both sides by \(\cos^2{x}\), we get: \(1 = \cos^3{x}\) From this equation, we can find the intersections of these two functions. Since we know that the intersection points must lie between \(x=-\frac{\pi}{3}\) and \(x=\frac{\pi}{3}\), we can find that there is only one intersection point that lies in the given interval, which is \(x=0\) with the corresponding \(y\) value being \(y=1\).

Step 3: Find the area of the bounded region

Now that we have found the intersection points, we need to find the area of the region bounded by the graphs. We can do this by taking the difference between the two functions and integrating with respect to x. Since the functions intersect at x = 0, we will have to split the integral into two parts, one for the interval \(-\frac{\pi}{3} \leq x \leq 0\) and another for \(0 \leq x \leq \frac{\pi}{3}\). \[ Area = \int_{-\frac{\pi}{3}}^0 (\sec^2{x} - \cos{x}) dx + \int_{0}^\frac{\pi}{3} (\sec^2{x} - \cos{x}) dx \]

Step 4: Calculate the integrals

Now let's find the integral of each function over the given intervals: \( \int_{-\frac{\pi}{3}}^0 \sec^2{x} dx = \left[ \tan{x} \right]_{-\frac{\pi}{3}}^0 = 0 - (-\sqrt{3}) = \sqrt{3} \) \( \int_{0}^\frac{\pi}{3} \sec^2{x} dx = \left[ \tan{x} \right]_{0}^{\frac{\pi}{3}} = \sqrt{3} - 0 = \sqrt{3} \) \( \int_{-\frac{\pi}{3}}^0 \cos{x} dx = \left[ \sin{x} \right]_{-\frac{\pi}{3}}^0 = 0 - (-\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} \) \( \int_{0}^\frac{\pi}{3} \cos{x} dx = \left[ \sin{x} \right]_{0}^{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} - 0 = \frac{\sqrt{3}}{2} \) Now we can substitute our results back into the area formula: \[ Area = \left( \sqrt{3} - \frac{\sqrt{3}}{2} \right) + \left( \sqrt{3} - \frac{\sqrt{3}}{2} \right) = 2\left(\frac{\sqrt{3}}{2} \right) = \sqrt{3} \] So the area of the region bounded by the given graphs is \(\sqrt{3}\) square units.

Key Concepts

Integration

Integration is a fundamental concept in calculus, and it is especially useful in finding the area under curves. When you have functions represented on a graph, like in our exercise, integrating them gives you the accumulated area between the curve and the x-axis.
In our specific case, where we have multiple functions such as

• \(y = \sec^2{x}\)
• \(y = \cos{x}\)

we need to calculate the area within specific bounds: \(x = -\frac{\pi}{3}\) and \(x = \frac{\pi}{3}\). This is done by computing definite integrals over those intervals and finding the difference. If you have two functions, integration allows you to determine the area between them by subtracting the integral of one from the other.
So essentially, integration helps bridge the gap between static equations and dynamic calculations of areas, allowing us to visualize and compute the space enclosed by curves.

Intersection Points

Finding intersection points is crucial because they help identify the boundaries of regions enclosed by curves. In our exercise, we had two curves

• \(y = \sec^2{x}\)
• \(y = \cos{x}\)

These intersection points occur where the curves meet, meaning the y-values of each function are equal at a particular x-value. By equating the two functions and solving for x, you will determine exactly where they overlap.
A pivotal step involved transforming \(\sec^2{x} = \cos{x}\)into a polynomial equation that would allow us to solve for x. In this particular scenario, the functions intersect at only one point within the given interval—specifically at \(x = 0\) with \(y = 1\). This single intersection point is essential for figuring out the limits of definite integration required for area calculations.

Graph Sketching

Before we dive into computing areas, sketching graphs is a helpful visual guide for understanding our functions and seeing the area we need to calculate. Drawing the graphs of

• \(y = \sec^2{x}\)
• \(y = \cos{x}\)

between the bounds \(x = -\frac{\pi}{3}\) and \(x = \frac{\pi}{3}\) allows us to observe how each curve behaves over the interval. The graph of \(y = \sec^2{x}\) has a different shape than \(y = \cos{x}\), and sketching lets us figure out where they intersect and enclose the region whose area we are about to calculate.
It is particularly useful when matching known shapes or patterns to see error opportunities in our calculations. Once sketched, you can label critical points, such as intersections and where each curve peaks or drops, especially if these coincide with limits of integration.

Definite Integrals

Definite integrals calculate the net area between a function and the x-axis across a particular interval. Mathematically, it is represented as: \[\int_{a}^{b} f(x) \, dx\]This denotes the integral of a function \(f(x)\) from point \(a\) to point \(b\). It is especially important when seeking the area between two curves as you have to integrate the top curve minus the bottom curve over the given interval.
In our exercise example, using definite integrals, we took \(\int_{a}^{b} (\sec^2{x} - \cos{x}) \, dx\) and split it into two integrals from each limit, calculating net areas separately for left and right portions with respect to our intersection point at x = 0. Through this application, definite integrals not only solved for area but illustrated the practicality of calculus in real-world problems.