Calculating Double Integral Of Xcos(2x+y) Over Region R

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Hey everyone! Today, we're diving into the fascinating world of double integrals and tackling a specific problem. We're going to calculate the double integral of the function xcos(2x + y) over a rectangular region R. This might sound intimidating, but don't worry, we'll break it down step by step. So, grab your calculators (or your favorite integration software), and let's get started!

Understanding the Problem

Before we jump into the calculations, let's make sure we understand what we're trying to solve. We're given the double integral:

∬Rxcos⁡(2x+y)dA\iint_{R} x \cos(2x + y) dA

Where R is the region defined by:

0≤x≤16π,0≤y≤12π0 \leq x \leq \frac{1}{6} \pi, \quad 0 \leq y \leq \frac{1}{2} \pi

What does this all mean? Well, we're essentially trying to find the "volume" under the surface z = xcos(2x + y) over the rectangular region R in the xy-plane. Think of it like finding the area under a curve, but in three dimensions!

The region R is a rectangle in the xy-plane. The x-values range from 0 to π/6, and the y-values range from 0 to π/2. This defines the boundaries of our integration.

The function x*cos(2x + y) is the surface whose volume we want to calculate. It's a function of two variables, x and y, which means its value changes depending on the location in the xy-plane.

The dA represents the infinitesimal area element, which can be expressed as dx dy or dy dx. This tells us the order in which we'll integrate with respect to x and y.

Now that we have a good grasp of the problem, let's move on to the solution.

Step-by-Step Solution

The key to solving double integrals is to treat them as iterated integrals. This means we'll integrate with respect to one variable at a time, treating the other variable as a constant. The order of integration (whether we integrate with respect to x first or y first) can sometimes make the problem easier or harder. In this case, integrating with respect to y first seems like a good strategy.

Step 1: Set up the iterated integral

Since we're integrating with respect to y first, we'll set up the integral as follows:

∬Rxcos⁡(2x+y)dA=∫0π6∫0π2xcos⁡(2x+y)dydx\iint_{R} x \cos(2x + y) dA = \int_{0}^{\frac{\pi}{6}} \int_{0}^{\frac{\pi}{2}} x \cos(2x + y) dy dx

Notice how the outer integral has the limits for x (0 to π/6) and the inner integral has the limits for y (0 to π/2). This is crucial for setting up the problem correctly.

Step 2: Evaluate the inner integral

Now, let's focus on the inner integral:

∫0π2xcos⁡(2x+y)dy\int_{0}^{\frac{\pi}{2}} x \cos(2x + y) dy

Remember, we're treating x as a constant here. The integral of cos(2x + y) with respect to y is sin(2x + y). We also need to remember to multiply by the constant x.

So, the inner integral becomes:

x∫0π2cos⁡(2x+y)dy=x[sin⁡(2x+y)]0π2x \int_{0}^{\frac{\pi}{2}} \cos(2x + y) dy = x [\sin(2x + y)]_{0}^{\frac{\pi}{2}}

Now, we need to evaluate the sine function at the upper and lower limits of integration and subtract:

x[sin⁡(2x+π2)−sin⁡(2x)]x [\sin(2x + \frac{\pi}{2}) - \sin(2x)]

Using the trigonometric identity sin(a + π/2) = cos(a), we can simplify this to:

x[cos⁡(2x)−sin⁡(2x)]x [\cos(2x) - \sin(2x)]

Step 3: Evaluate the outer integral

Now we have the result of the inner integral, we can plug it into the outer integral:

∫0π6x[cos⁡(2x)−sin⁡(2x)]dx\int_{0}^{\frac{\pi}{6}} x [\cos(2x) - \sin(2x)] dx

This integral is a bit more challenging. We'll need to use integration by parts. Let's break it down:

We have two terms to integrate: xcos(2x) and xsin(2x).

For the integral of x*cos(2x), let:

  • u = x (so du = dx)
  • dv = cos(2x) dx (so v = (1/2)sin(2x))

Using the integration by parts formula, ∫u dv = uv - ∫v du, we get:

∫xcos⁡(2x)dx=12xsin⁡(2x)−∫12sin⁡(2x)dx=12xsin⁡(2x)+14cos⁡(2x)\int x \cos(2x) dx = \frac{1}{2}x\sin(2x) - \int \frac{1}{2}\sin(2x) dx = \frac{1}{2}x\sin(2x) + \frac{1}{4}\cos(2x)

For the integral of x*sin(2x), let:

  • u = x (so du = dx)
  • dv = sin(2x) dx (so v = -(1/2)cos(2x))

Using the integration by parts formula, ∫u dv = uv - ∫v du, we get:

∫xsin⁡(2x)dx=−12xcos⁡(2x)+∫12cos⁡(2x)dx=−12xcos⁡(2x)+14sin⁡(2x)\int x \sin(2x) dx = -\frac{1}{2}x\cos(2x) + \int \frac{1}{2}\cos(2x) dx = -\frac{1}{2}x\cos(2x) + \frac{1}{4}\sin(2x)

Now we can put these results back into our outer integral:

∫0π6x[cos⁡(2x)−sin⁡(2x)]dx=[12xsin⁡(2x)+14cos⁡(2x)+12xcos⁡(2x)−14sin⁡(2x)]0π6\int_{0}^{\frac{\pi}{6}} x [\cos(2x) - \sin(2x)] dx = \left[\frac{1}{2}x\sin(2x) + \frac{1}{4}\cos(2x) + \frac{1}{2}x\cos(2x) - \frac{1}{4}\sin(2x)\right]_{0}^{\frac{\pi}{6}}

Step 4: Evaluate the definite integral

Now, we need to plug in the limits of integration (0 and π/6) and subtract. This is where things get a bit messy with the arithmetic, but we can do it!

Plugging in x = π/6, we get:

12(π6)sin⁡(π3)+14cos⁡(π3)+12(π6)cos⁡(π3)−14sin⁡(π3)=π12(32)+14(12)+π12(12)−14(32)\frac{1}{2}(\frac{\pi}{6})\sin(\frac{\pi}{3}) + \frac{1}{4}\cos(\frac{\pi}{3}) + \frac{1}{2}(\frac{\pi}{6})\cos(\frac{\pi}{3}) - \frac{1}{4}\sin(\frac{\pi}{3}) = \frac{\pi}{12}(\frac{\sqrt{3}}{2}) + \frac{1}{4}(\frac{1}{2}) + \frac{\pi}{12}(\frac{1}{2}) - \frac{1}{4}(\frac{\sqrt{3}}{2})

Plugging in x = 0, we get:

12(0)sin⁡(0)+14cos⁡(0)+12(0)cos⁡(0)−14sin⁡(0)=0+14+0−0=14\frac{1}{2}(0)\sin(0) + \frac{1}{4}\cos(0) + \frac{1}{2}(0)\cos(0) - \frac{1}{4}\sin(0) = 0 + \frac{1}{4} + 0 - 0 = \frac{1}{4}

Subtracting the result at x = 0 from the result at x = π/6, we get:

π12(32)+14(12)+π12(12)−14(32)−14=π324+18+π24−38−14\frac{\pi}{12}(\frac{\sqrt{3}}{2}) + \frac{1}{4}(\frac{1}{2}) + \frac{\pi}{12}(\frac{1}{2}) - \frac{1}{4}(\frac{\sqrt{3}}{2}) - \frac{1}{4} = \frac{\pi\sqrt{3}}{24} + \frac{1}{8} + \frac{\pi}{24} - \frac{\sqrt{3}}{8} - \frac{1}{4}

Simplifying this expression, we get:

π(3+1)24−38−18\frac{\pi(\sqrt{3} + 1)}{24} - \frac{\sqrt{3}}{8} - \frac{1}{8}

Step 5: The Final Answer

So, after all that work, the final answer to the double integral is:

∬Rxcos⁡(2x+y)dA=π(3+1)24−38−18\iint_{R} x \cos(2x + y) dA = \frac{\pi(\sqrt{3} + 1)}{24} - \frac{\sqrt{3}}{8} - \frac{1}{8}

This is the volume under the surface z = xcos(2x + y) over the region R. Whew! That was a journey, but we made it!

Key Takeaways

  • Double integrals are used to calculate volumes under surfaces.
  • We can solve double integrals by treating them as iterated integrals.
  • Integration by parts is a powerful technique for integrating products of functions.
  • Careful algebra and trigonometry are essential for getting the correct answer.

Why This Matters

You might be wondering, "Okay, that's cool, but why should I care about double integrals?" Well, double integrals have tons of applications in various fields, including:

  • Physics: Calculating mass, center of mass, and moments of inertia of objects.
  • Engineering: Designing structures, analyzing fluid flow, and solving heat transfer problems.
  • Probability and Statistics: Finding probabilities and expected values in two-dimensional distributions.
  • Computer Graphics: Rendering 3D images and creating realistic lighting effects.

So, understanding double integrals is a valuable skill for anyone pursuing a career in these areas.

Practice Makes Perfect

The best way to master double integrals is to practice, practice, practice! Try working through similar problems with different functions and regions. Don't be afraid to make mistakes – that's how we learn! And if you get stuck, there are plenty of resources available online and in textbooks.

Keep practicing, and you'll become a double integral pro in no time! Good luck, guys, and happy integrating!