Problem 15 - Thomas Calculus

Question

The integrals and sums of integrals in Exercises \(13 - 18\) give the areas of regions in the \(x y\) -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: Identify the Boundaries

From the given double integral \( \int_{0}^{\pi/4} \int_{\sin x}^{\cos x} dy\, dx \), we identify that the outer limits of integration for \(x\) are \(0\) to \(\pi/4\), and the inner limits for \(y\) are from \(\sin x\) to \(\cos x\). This means the region is bounded by \(y = \sin x\) and \(y = \cos x\) for \(x\) in \([0, \pi/4]\).

Step 2: Sketch the Region and Identify Intersection Points

Plot the curves \(y = \sin x\) and \(y = \cos x\) over the interval \([0, \pi/4]\). The curves intersect where \(\sin x = \cos x\), which occurs when \(x = \pi/4\). At \(x = 0\), \(\sin x = 0\) and \(\cos x = 1\). At \(x = \pi/4\), \(\sin x = \cos x = \frac{\sqrt{2}}{2}\). The region is therefore bounded by these two curves between the limits of \(x = 0\) and \(x = \pi/4\).

Step 3: Set Up the Integral for the Area

To find the area of the region, we use the double integral \(\int_{0}^{\pi/4} \int_{\sin x}^{\cos x} dy\, dx\). This integral evaluates the area between \(y = \sin x\) and \(y = \cos x\) as \(x\) runs from \(0\) to \(\pi/4\).

Step 4: Evaluate the Inner Integral

The inner integral \(\int_{\sin x}^{\cos x} dy\) is simply \((\cos x - \sin x)\), as it represents the calculation of the difference in the \(y\)-values between \(\cos x\) and \(\sin x\) for a constant value of \(x\).

Step 5: Evaluate the Outer Integral

Now, plug the result of the inner integral into the outer integral: \(\int_{0}^{\pi/4} (\cos x - \sin x)\, dx\). Separate this integral into two parts: \(\int_{0}^{\pi/4} \cos x\, dx - \int_{0}^{\pi/4} \sin x\, dx\).

Step 6: Compute the Integral of \(\cos x\)

Compute \(\int \cos x \ dx\), which is \(\sin x + C\). Evaluating from 0 to \(\pi/4\) gives \(\sin(\pi/4) - \sin(0)=\frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}\).

Step 7: Compute the Integral of \(\sin x\)

Compute \(\int \sin x \ dx\), which is \(-\cos x + C\). Evaluating from 0 to \(\pi/4\) gives \(-\cos(\pi/4) + \cos(0)= -\frac{\sqrt{2}}{2} + 1\).

Step 8: Calculate the Final Area

Subtract the results from Steps 6 and 7: \(\frac{\sqrt{2}}{2} - (-\frac{\sqrt{2}}{2} + 1) = \sqrt{2} - 1\). Therefore, the area of the region is \(\sqrt{2} - 1\).

Key Concepts

Area of Regions

Understanding the concept of area in double integrals is key to solving problems like the one presented. For regions in the coordinate plane, the double integral calculates the area by accumulating infinitesimally small vertical slices of the region. These slices add up to give the full area.
In this exercise, we used the double integral \( \int_{0}^{\pi/4} \int_{\sin x}^{\cos x} dy\, dx \). The integration bounds set the limits for the region in the x-y plane. This method allows us to consider regions that are not necessarily rectangular. The given double integral provides the measure of a non-rectangular area bounded by the curves \( y = \sin x \) and \( y = \cos x \) between \( x = 0 \) and \( x = \pi/4 \).
By using this approach, complicated shapes and areas can be calculated more easily.

Bounding Curves

Bounding curves are the perimeter lines that define a region in the coordinate plane when analyzing integrals. In this exercise, the bounding curves are given by the equations \( y = \sin x \) and \( y = \cos x \). These curves act as the upper and lower limits for the inner integral.
For \( x \) in the interval \([0, \pi/4]\), the curve \( y = \cos x \) is always greater than \( y = \sin x \). Thus, when setting up the integral, the function \( y = \cos x \) serves as the upper limit, while \( y = \sin x \) is the lower limit.
These bounding curves determine the vertical spread of the slices, giving a clear bracket for the y-values for any fixed x-value within the integration range.

Intersection Points

Intersection points occur where two bounding curves cross each other on a graph. Finding these points is crucial because they may affect the limits of integration. In our problem, the intersection occurs when \( \sin x = \cos x \). This equality holds true at the point \( x = \pi/4 \).
Thus, the intersection point is \( x = \pi/4 \), where both functions take on the value \( y = \frac{\sqrt{2}}{2} \). Understanding this point helps confirm the limits of integration for \( x \) as well as ensuring the calculation of the integral is done over the correct region.
Determining these intersection points offers a deeper understanding of how and where the curves relate to one another within the specified interval.

Coordinate Plane

The coordinate plane is the 2D space where graphs of functions are plotted with an x-axis and y-axis. It allows us to visualize functions and their interactions, such as boundaries and intersections. For this exercise, our x-axis span is \([0, \pi/4]\) and the region of interest lies between the curves \( y = \sin x \) and \( y = \cos x \).
On the coordinate plane, any point can be defined by an ordered pair \((x, y)\), where 'x' is a projection on the x-axis and 'y' is on the y-axis. It is instrumental in visually analyzing the bounded region and where functions meet, thus setting the stage for finding the desired area using integration.
This concept is fundamental for bringing abstract mathematical ideas into a visual and comprehensible form, which is essential for problem-solving and understanding the relationships between curves.