Calculating Double Integral: |cos(x+y)| Over Region D

Hey guys! Today, we're diving into a fascinating problem in multivariable calculus: calculating the double integral of the absolute value of the cosine of (x+y) over a specific region. This might sound intimidating, but we'll break it down step by step. So, grab your thinking caps, and let's get started!

The problem we're tackling is:

$$\iint_D|\cos(x+y)| \ dx \ dy, \quad D=[0,\pi]\times[-\pi,\pi]$$

Where D represents the rectangular region defined by 0 ≤ x ≤ π and -π ≤ y ≤ π. Essentially, we want to find the volume under the surface defined by the absolute value of cos(x+y) over this rectangle. The absolute value makes things interesting because it means we're always dealing with a non-negative function, which geometrically translates to the volume always being "above" the xy-plane. This problem beautifully combines trigonometry, absolute values, and multivariable integration, making it a fantastic exercise for solidifying our understanding of these concepts. The use of absolute value with the cosine function introduces a periodic "folding" effect, creating a wavy surface with crests and troughs. Understanding this behavior is crucial for setting up the integral correctly. We'll also leverage the symmetry and periodicity of the cosine function to simplify our calculations. The rectangular region D provides a well-defined boundary for our integration, making the problem manageable. Visualizing this region in the xy-plane is an essential first step. Remember, double integrals are powerful tools for calculating volumes, areas, and other quantities in two dimensions. Mastering these techniques is fundamental for many applications in physics, engineering, and computer graphics. So, let's delve into the solution and see how we can conquer this integral!

Visualizing the Function and Region

Before we jump into the calculations, let's visualize what we're dealing with. Imagine the function |cos(x+y)|. Because of the absolute value, the cosine wave, which normally oscillates between -1 and 1, is now "folded" upwards, oscillating between 0 and 1. This creates a series of crests. Now, picture the region D in the xy-plane: a rectangle with x ranging from 0 to π and y ranging from -π to π. We are trying to find the volume under the surface created by |cos(x+y)| and above this rectangle. Thinking about the geometry helps a ton! Visualizing the function and the region of integration is a crucial step in solving double integrals. It allows us to develop intuition about the problem and anticipate potential challenges or simplifications. For example, in this case, noticing the symmetry of the cosine function and the rectangular region can guide us towards breaking the integral into smaller, more manageable parts. Software like Desmos can be invaluable for visualizing functions and regions, especially in multivariable calculus. By plotting |cos(x+y)| over the region D, we can observe the four “leaves” or humps that the function forms, as mentioned earlier. This visual confirmation helps us understand the periodic behavior of the function and how it interacts with the integration region. Moreover, visualization can reveal potential pitfalls in setting up the integral. For instance, we might realize that we need to split the integral into different regions based on where the cosine function is positive or negative before applying the absolute value. So, always remember to visualize the problem before diving into calculations. It can save you time and effort in the long run.

Breaking Down the Integral

The key to solving this integral lies in understanding where cos(x+y) is positive and negative within our region D. Remember that |cos(x+y)| is equal to cos(x+y) when cos(x+y) is non-negative, and -cos(x+y) when cos(x+y) is negative. We need to divide the region D into subregions where cos(x+y) has a consistent sign. To do this, we consider where cos(x+y) = 0. This occurs when x+y = (2n+1)π/2, where n is an integer. Within our region D, this gives us the lines x+y = -π/2, x+y = π/2, and x+y = 3π/2. These lines divide the rectangle into four subregions, which correspond to the four “leaves” or humps observed in the Desmos visualization. This step is crucial because it allows us to replace the absolute value function with piecewise definitions, making the integration process much simpler. By identifying the lines where cos(x+y) changes sign, we can set up the integral correctly for each subregion. This approach leverages the piecewise nature of the absolute value function and the periodic behavior of the cosine function. Without this breakdown, we would struggle to handle the absolute value effectively. Furthermore, understanding the geometry of these subregions can help us choose the most convenient order of integration (dxdy or dydx). In some cases, one order might lead to simpler calculations than the other. So, remember to carefully analyze the function and the region of integration to identify the best strategy for breaking down the integral.

Now, let's define our four regions:

1. Region 1: Where x+y is between -π and -π/2. Here, cos(x+y) is negative, so |cos(x+y)| = -cos(x+y).
2. Region 2: Where x+y is between -π/2 and π/2. Here, cos(x+y) is positive, so |cos(x+y)| = cos(x+y).
3. Region 3: Where x+y is between π/2 and 3π/2. Here, cos(x+y) is negative, so |cos(x+y)| = -cos(x+y).
4. Region 4: Where x+y is between 3π/2 and π. This region doesn't actually exist within our integration domain D because the maximum value of x+y in D is π + π = 2π, but 3π/2 is already greater than that.

Therefore, we only need to consider the first three regions. We can now express our double integral as a sum of integrals over these three regions:

$$\iint_D|\cos(x+y)| \ dx \ dy = \iint_{Region \ 1} -\cos(x+y) \ dx \ dy + \iint_{Region \ 2} \cos(x+y) \ dx \ dy + \iint_{Region \ 3} -\cos(x+y) \ dx \ dy$$

Setting Up the Limits of Integration

This is where things get a little tricky, but don't worry, we'll get through it! We need to determine the limits of integration for x and y in each of the three regions. This involves finding the intersection points of the lines x+y = -π/2, x+y = π/2, and the boundaries of our rectangle D (x=0, x=π, y=-π, y=π).

• Region 1: x+y ranges from -π to -π/2. The limits are: -π ≤ y ≤ -π/2 - x and 0 ≤ x ≤ π/2.
• Region 2: x+y ranges from -π/2 to π/2. The limits are: -π/2 - x ≤ y ≤ π/2 - x and 0 ≤ x ≤ π.
• Region 3: x+y ranges from π/2 to 3π/2. The limits are: π/2 - x ≤ y ≤ π and π/2 ≤ x ≤ π.

Setting up the correct limits of integration is arguably the most challenging part of this problem. It requires a solid understanding of the geometry of the regions and how they relate to the boundaries defined by x+y = constant lines. A common mistake is to overlook the intersections of these lines with the edges of the rectangle D. Drawing a careful diagram of the region D and the lines x+y = -π/2 and x+y = π/2 is highly recommended. This visual aid will help you identify the correct bounds for x and y in each subregion. Remember, the limits of integration for the inner integral will be functions of the outer integration variable. This reflects the fact that we are integrating over a region that is not simply a rectangle aligned with the coordinate axes. Moreover, choosing the correct order of integration (dxdy or dydx) can significantly impact the complexity of the calculations. In some cases, one order might lead to simpler integrals than the other. So, take your time and carefully analyze the geometry before setting up the limits of integration. It will save you a lot of headaches down the road.

Evaluating the Integrals

Now for the fun part – actually integrating! We'll evaluate each double integral separately:

Region 1:

$$\int_{0}^{\pi/2} \int_{-\pi}^{-\pi/2 - x} -\cos(x+y) \ dy \ dx$$

First, integrate with respect to y:

$$\int_{0}^{\pi/2} [-\sin(x+y)]_{-\pi}^{-\pi/2 - x} \ dx = \int_{0}^{\pi/2} [-\sin(-\pi/2) + \sin(x-\pi)] \ dx = \int_{0}^{\pi/2} [1 - \sin(x)] \ dx$$

Now, integrate with respect to x:

$$[x + \cos(x)]_{0}^{\pi/2} = (\pi/2 + 0) - (0 + 1) = \pi/2 - 1$$

Region 2:

$$\int_{0}^{\pi} \int_{-\pi/2 - x}^{\pi/2 - x} \cos(x+y) \ dy \ dx$$

Integrate with respect to y:

$$\int_{0}^{\pi} [\sin(x+y)]_{-\pi/2 - x}^{\pi/2 - x} \ dx = \int_{0}^{\pi} [\sin(\pi/2) - \sin(-\pi/2)] \ dx = \int_{0}^{\pi} 2 \ dx$$

Integrate with respect to x:

$$[2x]_{0}^{\pi} = 2\pi$$

Region 3:

$$\int_{\pi/2}^{\pi} \int_{\pi/2 - x}^{\pi} -\cos(x+y) \ dy \ dx$$

Integrate with respect to y:

$$\int_{\pi/2}^{\pi} [-\sin(x+y)]_{\pi/2 - x}^{\pi} \ dx = \int_{\pi/2}^{\pi} [-\sin(x+\pi) + \sin(\pi/2)] \ dx = \int_{\pi/2}^{\pi} [\sin(x) + 1] \ dx$$

Integrate with respect to x:

$$[-\cos(x) + x]_{\pi/2}^{\pi} = (1 + \pi) - (0 + \pi/2) = \pi/2 + 1$$

The process of evaluating these integrals involves applying the fundamental theorem of calculus and carefully handling trigonometric functions. A common mistake is to forget the chain rule when integrating cos(x+y) or sin(x+y). Remember that the integral of cos(x+y) with respect to y is sin(x+y), and similarly, the integral of sin(x+y) with respect to y is -cos(x+y). Also, be mindful of the signs when evaluating the definite integrals at the limits of integration. It's easy to make a mistake with the negative signs, which can lead to an incorrect result. After performing the integration, we obtain three separate results for the three regions. These results represent the volumes under the surface |cos(x+y)| over each subregion. To get the final answer, we need to add these volumes together. So, double-check your calculations and make sure you have correctly evaluated the integrals before moving on to the final step.

Summing the Results

Finally, we add the results from the three regions to get the total double integral:

$$\iint_D|\cos(x+y)| \ dx \ dy = (\pi/2 - 1) + 2\pi + (\pi/2 + 1) = 3\pi$$

So, the value of the double integral is 3π. Woohoo! We made it! This final step is crucial for obtaining the overall solution to the problem. It involves simply adding the individual results from each subregion. However, it's important to double-check that you have correctly identified all the subregions and that you haven't missed any contributions. In this case, we had three regions to consider, and we carefully calculated the integral over each region. By adding the results together, we obtain the total volume under the surface |cos(x+y)| over the given rectangular region D. The final answer, 3π, is a concise and elegant result that encapsulates the entire integration process. It's a testament to the power of calculus and its ability to solve complex problems involving multivariable functions. So, congratulations on reaching the end of this journey! You've successfully tackled a challenging double integral and gained valuable experience in the process.

Conclusion

Calculating double integrals, especially those involving absolute values and trigonometric functions, can seem daunting at first. But by breaking the problem down into smaller steps – visualizing the function and region, identifying subregions, setting up the limits of integration, and carefully evaluating the integrals – we can conquer even the most challenging problems. Remember, practice makes perfect, so keep practicing, guys! This problem serves as a great example of how multivariable calculus can be used to solve real-world problems involving volumes, areas, and other quantities. The techniques we've learned here, such as breaking down the integral into subregions and setting up the limits of integration, are applicable to a wide range of problems in physics, engineering, and computer graphics. So, keep honing your skills and exploring the fascinating world of calculus! And remember, don't be afraid to ask for help when you get stuck. There are plenty of resources available, including textbooks, online tutorials, and your fellow students. The journey of learning calculus is a challenging but rewarding one, and with perseverance and practice, you'll be able to master these concepts and apply them to solve complex problems. So, keep exploring, keep learning, and keep having fun with calculus!