Problem 11 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises 9–14 give the areas of regions in the xy-plane. Sketch each region, label each bounding curve with its equation, and give the coordinates of the points where the curves intersect. Then find the area of the region. $$ \int_{0}^{\pi / 4} \int_{\sin x}^{\cos x} d y d x $$

Step-by-Step Solution

Answer

The area of the region is \( \sqrt{2} - 1 \).

Step 1: Set Up the Problem

We're given the double integral \( \int_{0}^{\pi / 4} \int_{\sin x}^{\cos x} d y \, d x \), which represents the area between two curves, \( y = \sin x \) and \( y = \cos x \), from \( x = 0 \) to \( x = \frac{\pi}{4} \).

Step 2: Sketch the Region

Draw the functions \( y = \sin x \) and \( y = \cos x \) on a graph from \( x = 0 \) to \( x = \frac{\pi}{4} \). These curves intersect when \( \sin x = \cos x \), which occurs at \( x = \frac{\pi}{4} \), and originate together at \( y = 0 \) when \( x = 0 \). The region of interest is bounded by these curves vertically.

Step 3: Identify Intersection Points

The equations \( y = \sin x \) and \( y = \cos x \) intersect when \( \sin x = \cos x \). Solving \( \tan x = 1 \) gives \( x = \frac{\pi}{4} \). Thus, the intersection point is \( (\frac{\pi}{4}, \sin(\frac{\pi}{4})) = (\frac{\pi}{4}, \frac{\sqrt{2}}{2}) \). The other boundary point, where \( x = 0 \), is \( (0,0) \).

Step 4: Set Up the Integral

To find the area, resolve the double integral. The limits for \( y \) are from \( \sin x \) to \( \cos x \), and for \( x \) from 0 to \( \frac{\pi}{4} \). The integral becomes \( \int_{0}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx \).

Step 5: Evaluate the Integral

Perform the integration: \[ \int ( \cos x - \sin x ) \, dx = \sin x + \cos x \]. Evaluate at the boundaries: \[ [ \sin(x) + \cos(x) ]_{0}^{\frac{\pi}{4}} = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (\sin(0) + \cos(0)) \].

Step 6: Calculate the Result

Substitute the boundary points: \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \) and \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \), so \[ \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \]. At \( x = 0 \), \( \sin(0) + \cos(0) = 1 \). Thus, the area is \( \sqrt{2} - 1 \).

Key Concepts

Area between curves

When dealing with double integrals that calculate the area between curves, it's key to visualize the region they cover. In this exercise, the curves are given by the equations \( y = \sin x \) and \( y = \cos x \). The double integral \( \int_{0}^{\pi / 4} \int_{\sin x}^{\cos x} d y \, d x \) represents the vertical area between these two curves from \( x = 0 \) to \( x = \frac{\pi}{4} \). Understanding this setup can simplify visualizing complex areas. The region is effectively a vertical slice bounded by the given limits. By integrating with respect to \( y \) first, the area between these two curves is isolated. The result gives the physical area in the plane, offering insight into how these mathematical structures behave graphically.

Intersection Points

Finding the points where curves intersect is critical for setting correct integration limits. In this problem, \( y = \sin x \) and \( y = \cos x \) intersect where \( \sin x = \cos x \). Solving for \( x \), we find \( x = \frac{\pi}{4} \). At this \( x \)-value, both \( \sin x \) and \( \cos x \) are equal to \( \frac{\sqrt{2}}{2} \). Therefore, the intersection point is \( (\frac{\pi}{4}, \frac{\sqrt{2}}{2}) \). The other point of intersection is apparently at the origin \( (0, 0) \) because both functions achieve zero at \( x = 0 \). These points serve as essential guides when sketching the graph and choosing integration limits. They help ensure that the double integral only covers the desired section of the plane.

Integration Limits

Setting integration limits correctly is crucial for successfully evaluating a double integral. In this context, finding the area between \( \sin x \) and \( \cos x \) involved two sets of limits: for \( y \) and for \( x \). Here, \( y \) ranges from \( \sin x \) to \( \cos x \), representing the vertical span between the two curves at each \( x \). The limits for \( x \) are from 0 to \( \frac{\pi}{4} \), defining the horizontal interval over which we compute the area. For double integrals covering area, the order of limits reflects a step-by-step approach, first nesting integration within vertical boundaries before addressing the horizontal scope. Correct limits ensure the calculated area is precise and represents only the intended section between the curves.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are fundamental in many calculus problems, especially those involving periodic or circular phenomena. These functions exhibit particular symmetry and periodicity, impacting how intersections and integrations are calculated. For instance, \( \sin x \) and \( \cos x \) intersect at precise points due to their unique waveforms. Moreover, evaluating integrals involving these functions requires an understanding of their derivatives and antiderivatives: \( \int \cos x \, dx = \sin x \) and \( \int \sin x \, dx = -\cos x \). In this case, using trigonometric identities and derivatives speeds up calculations, helping to solve integrals where these functions serve as boundaries or components within more complex expressions. Appreciating these mathematical behaviors is essential when approaching calculus problems involving these functions.