Problem 44 - Calculus Volume 1

Question

For the following exercises, find the exact area of the region bounded by the given equations if possible. If you are unable to determine the intersection points analytically, use a calculator to approximate the intersection points with three decimal places and determine the approximate area of the region. $$ [T]y=\sqrt{1-x^{2}} \text { and } y^{2}=x^{2} $$

Step-by-Step Solution

Answer

The exact area is \(\frac{\pi}{2} - \frac{\sqrt{2}}{2}\).

Step 1: Rewrite the Equations for Intersection Points

The given equations are \(y = \sqrt{1-x^2}\) and \(y^2 = x^2\). Start by rewriting the second equation as two separate equations: \(y = x\) and \(y = -x\). This is because \(y^2 = x^2\) implies either \(y = x\) or \(y = -x\).

Step 2: Find Intersection Points

Find the points where the two sets of equations intersect.1. **For \(y = x\):** Set \(x = \sqrt{1-x^2}\). - Square both sides: \(x^2 = 1-x^2\) which simplifies to \(2x^2 = 1\) and further to \(x^2 = \frac{1}{2}\). - Thus, \(x = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}\). - Correspondingly, \(y = \pm\frac{\sqrt{2}}{2}\).2. **For \(y = -x\):** Set \(-x = \sqrt{1-x^2}\). - Square both sides: \(x^2 = 1-x^2\) which simplifies the same way as above. - The intersection points remain \(x = \pm\frac{\sqrt{2}}{2}\), and \(y = \mp\frac{\sqrt{2}}{2}\).

Step 3: Set Up the Integral for the Area

The region bounded by the equations forms a symmetric shape around the origin. Use the equation \(y = \sqrt{1-x^2}\) as the upper boundary from \(-x\) to \(x\), and the lines \(y = x\) and \(y = -x\) as lower boundaries.The integral for one segment is:\[\int_{0}^{\frac{\sqrt{2}}{2}} \left( \sqrt{1-x^2} - x \right) \, dx + \int_{-\frac{\sqrt{2}}{2}}^{0} \left( \sqrt{1-x^2} + x \right) \, dx\]

Step 4: Evaluate the Integral

The area is naturally symmetrical, doubling the integral from 0 to \(\frac{\sqrt{2}}{2}\) suffices:- \[2 \int_{0}^{\frac{\sqrt{2}}{2}} \left( \sqrt{1-x^2} - x \right) \, dx\]- Apply substitution \(x = \sin(\theta)\), so \(dx = \cos(\theta)\, d\theta\):- \[\theta = \frac{\pi}{4}, 0 \Rightarrow 2 \int_{0}^{\frac{\pi}{4}} (\cos(\theta) - \sin(\theta)\cos(\theta)) \, d\theta\]- Calculate the integral using trigonometric identities and fundamental calculus to find it equals \(\frac{\pi}{4} - \frac{\sqrt{2}}{4}\).

Step 5: Present the Final Calculation and Answer

Doubling this value, given the symmetry, results in the area of:\[2 \left( \frac{\pi}{4} - \frac{\sqrt{2}}{4} \right)\]Which simplifies to \(\frac{\pi}{2} - \frac{\sqrt{2}}{2}\).

Key Concepts

Intersection Points

Finding intersection points is crucial for understanding the regions where two or more equations meet. In this context, we're examining the curves described by the equations \( y = \sqrt{1-x^2} \) and \( y^2 = x^2 \). The key is to solve for where these equations overlap.
To start, interpret \( y^2 = x^2 \) as two separate lines: \( y = x \) and \( y = -x \). This implies that the solutions must satisfy either of these lines simultaneously with the curve \( y = \sqrt{1-x^2} \).
To find the points of intersection, equate \( y = x \) and \( y = -x \) with \( y = \sqrt{1-x^2} \), leading to \( x^2 = 1-x^2 \), which simplifies to \( x^2 = \frac{1}{2} \). Consequently, \( x = \pm \frac{\sqrt{2}}{2} \). Substituting these \( x \)-values back gives \( y = \pm \frac{\sqrt{2}}{2} \). With this, the intersection points are found at \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) \) and \((-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}) \). Understanding these points helps set boundaries for any definite integrals calculated between these intersections.

Integration Techniques

Integrating functions helps find the area under a curve between two points. In our case, the integral of \( \sqrt{1-x^2} - x \) over the interval \(-\frac{\sqrt{2}}{2}\) to \(\frac{\sqrt{2}}{2} \) gives the required area.
We start by splitting the integral into segments based on symmetry: one integral from \( 0 \) to \( \frac{\sqrt{2}}{2} \) and another is similar due to the symmetry from \(-\frac{\sqrt{2}}{2} \) to \( 0 \). Using symmetry, however, allows us to simply focus on doubling the result of one section.

• By setting up the expressions, we solve: \[ 2 \int_{0}^{\frac{\sqrt{2}}{2}} (\sqrt{1-x^2} - x) \, dx \]
• This facilitates easier computation without repeated calculations.

Understanding when to exploit symmetry and how to properly set up integral bounds are valuable integration techniques. It saves time and effort while maintaining accuracy.

Trigonometric Substitution

Trigonometric substitution is a useful method to simplify integrals involving square roots of quadratic expressions. In this exercise, dealing with \( \sqrt{1-x^2} \) suggests using the substitution \( x = \sin(\theta) \), which is effective due to the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).
This substitution transforms the integral into easier-to-manage terms:

• \( dx = \cos(\theta) \, d\theta \), which matches the integral's form.
• The bounds change according to \( x=\sin(\theta) \), so the initial bound, \( x=0 \), leads to \( \theta=0 \), and \( x=\frac{\sqrt{2}}{2} \) corresponds to \( \theta=\frac{\pi}{4} \).

The integral becomes \[ 2 \int_{0}^{\frac{\pi}{4}} (\cos(\theta) - \sin(\theta)\cos(\theta)) \, d\theta \], which simplifies with trigonometric identities to compute the exact area. This technique is particularly helpful in handling complex integrals effortlessly.

Symmetry in Calculus

Symmetry in calculus plays a pivotal role in simplifying problems, especially those involving areas and integrals. For the curves defined by \( y = \sqrt{1-x^2} \) and lines \( y = x \), \( y = -x \), symmetry about the origin means calculations can often be halved in effort yet accurately doubled in result.
The beauty of symmetry is in reducing the complexity of integrations and solutions:

• The problem's geometrical nature indicates the curves form a symmetric shape about both the x-axis and y-axis.
• This symmetry allows one to calculate the integral over one-quarter of the region and then either multiply by four or double an entire symmetrical half.

By embracing symmetry, you not only simplify the computations through reduced bounds but also gain a deeper understanding of the problem's inherent geometry. This concept empowers solutions to be more efficient and insightful.