Triple Integral Calculation: Step-by-Step Solution

Hey guys! Today, we're diving into the world of triple integrals! Specifically, we're going to tackle a problem where we need to calculate the value of a triple integral over a defined region. Don't worry, we'll break it down step-by-step so it's super easy to follow. So, let's jump right into it and learn how to solve this kind of problem. Stick with me, and you'll be a pro in no time!

Understanding the Triple Integral Problem

So, the question we're tackling is: What is the value of the triple integral ∫∫∫_R 2xy³z² dV, where R is the region bounded by 1 ≤ x ≤ 2, 0 ≤ y ≤ 3, and 0 ≤ z ≤ 1? Essentially, we need to find the volume under the function 2xy³z² within the specified boundaries. This might sound intimidating, but trust me, it's just a matter of breaking it down into smaller, manageable integrals.

Before we even start crunching numbers, it's vital to understand what a triple integral represents. Think of it as an extension of the familiar single and double integrals. Single integrals help us find the area under a curve, and double integrals help us find the volume under a surface. Now, a triple integral takes it a step further, allowing us to integrate a function over a three-dimensional region. This is super useful in fields like physics, engineering, and, of course, mathematics for calculating things like mass, moments of inertia, and other properties of 3D objects. The function we're integrating (2xy³z² in our case) is like a density function, and the triple integral essentially sums up the values of this function over the entire region R.

Now, let's talk about the region R. It's described by the inequalities 1 ≤ x ≤ 2, 0 ≤ y ≤ 3, and 0 ≤ z ≤ 1. What does this mean? Well, it defines a rectangular box in 3D space. The x-coordinate ranges from 1 to 2, the y-coordinate ranges from 0 to 3, and the z-coordinate ranges from 0 to 1. This rectangular shape makes our lives a little easier because it simplifies the setup of our triple integral. If the region were more complex, we'd have to deal with variable limits of integration, which can get a bit hairy. But for this problem, we have constant limits, which is fantastic news!

Understanding the integrand, 2xy³z², is just as crucial. This function tells us how the value we're integrating changes within the region R. Notice how each variable (x, y, and z) plays a role. The x-term contributes linearly, the y-term contributes cubically (y³), and the z-term contributes quadratically (z²). This means that the value of the function will increase more rapidly as y increases compared to x or z. Keeping this in mind can sometimes help you anticipate the result of the integration or catch errors along the way. So, with a solid grasp of what the triple integral represents and a clear picture of the region R and the integrand, we're well-prepared to dive into the actual calculation. Let's get to it!

Setting Up the Triple Integral

Okay, so now we know what we're trying to calculate, let's actually set up the triple integral. This is a crucial step because the order in which we integrate with respect to x, y, and z can affect the complexity of the calculations. For our problem, since the region R is a simple rectangular box and our limits of integration are constants, the order doesn't really matter that much. But it's still good practice to think about the best order to make things as smooth as possible.

In this case, we can choose any order we like. Let's go with integrating with respect to x first, then y, and finally z. This means our triple integral will look like this:

∫[0 to 1] ∫[0 to 3] ∫[1 to 2] 2xy³z² dx dy dz

Notice how the limits of integration correspond to the bounds of our region R. The innermost integral, ∫[1 to 2] 2xy³z² dx, represents integrating with respect to x while treating y and z as constants. The limits 1 and 2 are the bounds for x. Then, the next integral, ∫[0 to 3] (result from the x-integration) dy, integrates with respect to y, using the bounds 0 and 3. And finally, the outermost integral, ∫[0 to 1] (result from the y-integration) dz, integrates with respect to z, using the bounds 0 and 1. This nested structure is what makes triple integrals work. We're essentially performing a series of single integrations, one variable at a time.

One thing that's super important to remember is that when you're integrating with respect to one variable, you treat the other variables as constants. This is a key concept that makes triple integrals (and multiple integrals in general) manageable. So, when we integrate 2xy³z² with respect to x, we'll treat y³ and z² as constants and just focus on the x term. This simplifies the integration process significantly. It's like doing a regular single-variable integral, but with some extra constants hanging around. Once we've done the x-integration, we'll plug in the limits of integration (1 and 2) and simplify. Then, we'll move on to the y-integration, and so on. The key is to take it one step at a time and keep track of which variables you're treating as constants at each stage.

So, now that we've got the triple integral set up and we understand the order of integration, we're ready to actually start calculating. The next step is to evaluate the innermost integral, which is the integration with respect to x. Let's dive into that and see how it's done!

Evaluating the Inner Integral (with respect to x)

Alright, let's get our hands dirty and start evaluating this triple integral! We're going to start with the innermost integral, which is the integration with respect to x. Remember, our integral looks like this:

∫[0 to 1] ∫[0 to 3] ∫[1 to 2] 2xy³z² dx dy dz

So, we're focusing on the ∫[1 to 2] 2xy³z² dx part first. As we discussed earlier, when we integrate with respect to x, we treat y³ and z² as constants. This makes the integration much simpler. We can think of 2y³z² as a constant coefficient multiplying the x term. The integral of x with respect to x is simply x²/2. So, the integral of 2xy³z² with respect to x is (2y³z²)(x²/2), which simplifies to x²y³z².

Now, we need to evaluate this expression at the limits of integration, which are x = 1 and x = 2. This means we'll plug in 2 for x and then subtract the result of plugging in 1 for x. So, we have:

(2²y³z²) - (1²y³z²) = 4y³z² - y³z² = 3y³z²

This is the result of our first integration! Notice that we've eliminated x from the expression, and we're left with a function that depends only on y and z. This is exactly what we want. We've effectively "integrated out" the x variable. This 3y³z² will now become the integrand for our next integral, the one with respect to y.

It's super important to be careful with your algebra and keep track of your constants and variables throughout this process. A small mistake in the integration or evaluation can throw off the whole calculation. So, double-check your work as you go along. Now, before we move on to the next integral, let's take a moment to appreciate what we've accomplished. We've successfully integrated with respect to x and obtained a simpler expression. This is a big step forward in solving the triple integral. The key takeaway here is to treat the other variables as constants when integrating with respect to a specific variable. This technique makes the whole process much more manageable. So, with this first integration under our belts, we're feeling confident and ready to tackle the next one. Let's move on to integrating with respect to y!

Integrating with Respect to y

Fantastic! We've conquered the inner integral and now we're ready to move on to the next step: integrating with respect to y. Remember, after integrating with respect to x and evaluating the limits, we were left with the expression 3y³z². This now becomes the integrand for our y-integral. So, we're looking at this integral:

∫[0 to 3] 3y³z² dy

Just like before, when we integrate with respect to y, we treat the remaining variable, z, as a constant. This means we can think of 3z² as a constant coefficient. The integral of y³ with respect to y is y⁴/4. So, the integral of 3y³z² with respect to y is (3z²)(y⁴/4), which we can write as (3/4)y⁴z².

Now comes the evaluation part. We need to plug in the limits of integration for y, which are 0 and 3. So, we have:

[(3/4)(3⁴)z²] - [(3/4)(0⁴)z²] = (3/4)(81)z² - 0 = (243/4)z²

There we go! We've successfully integrated with respect to y. Notice that we've eliminated y from the expression, and we're now left with a function that depends only on z. This is exactly what we want as we move through the layers of the triple integral. This expression, (243/4)z², will be the integrand for our final integration, the one with respect to z.

It's worth pausing for a moment to appreciate how each integration simplifies the problem. We started with a function of three variables (x, y, and z) and, by integrating with respect to x, we reduced it to a function of two variables (y and z). Then, by integrating with respect to y, we further reduced it to a function of just one variable (z). This step-by-step reduction is the key to solving triple integrals. And it really highlights the beauty and power of calculus. By breaking down a complex problem into smaller, manageable steps, we can arrive at the solution. So, we're now in the home stretch! We've got just one more integration to go. Let's finish strong and integrate with respect to z!

Final Integration: Integrating with Respect to z

Okay, guys, we've made it to the final leg of our journey! We're now ready to integrate with respect to z. After our previous integrations, we're left with the expression (243/4)z². This is the integrand for our final integral:

∫[0 to 1] (243/4)z² dz

This integral looks much simpler than our original triple integral, doesn't it? That's the magic of iterated integration! We've systematically reduced the complexity of the problem step by step. Now, let's tackle this last integral.

(243/4) is a constant coefficient, so we can just focus on integrating z² with respect to z. The integral of z² is z³/3. Therefore, the integral of (243/4)z² with respect to z is (243/4)(z³/3), which simplifies to (81/4)z³.

Now, we need to evaluate this expression at the limits of integration for z, which are 0 and 1. Plugging in these values, we get:

[(81/4)(1³)] - [(81/4)(0³)] = (81/4) - 0 = 81/4

And there we have it! The value of our triple integral is 81/4. This is the final answer to our problem. We started with a seemingly complex triple integral, and by carefully breaking it down into a series of single integrations, we arrived at a simple numerical result. This demonstrates the power and elegance of calculus in solving multi-dimensional problems.

So, to recap, we integrated with respect to x first, then y, and finally z. We treated the other variables as constants during each integration, and we carefully evaluated the limits of integration at each step. This systematic approach allowed us to solve the triple integral without getting lost in the details. It's a good practice to double-check your work and make sure your answer makes sense in the context of the problem. In this case, 81/4 is a positive value, which is reasonable given that our integrand (2xy³z²) is positive over the region R.

Final Answer and Conclusion

So, guys, after all that meticulous calculation, we've arrived at our final answer! The value of the triple integral ∫∫∫_R 2xy³z² dV, where R is the region bounded by 1 ≤ x ≤ 2, 0 ≤ y ≤ 3, and 0 ≤ z ≤ 1, is 81/4 which equals to 20.25.

To recap the entire process, we:

1. Understood the problem: We identified that we needed to calculate a triple integral over a rectangular region.
2. Set up the integral: We chose an order of integration (dx dy dz) and set up the integral with the correct limits.
3. Integrated with respect to x: We treated y and z as constants and integrated 2xy³z² with respect to x.
4. Integrated with respect to y: We treated z as a constant and integrated the result from the previous step with respect to y.
5. Integrated with respect to z: We integrated the result from the previous step with respect to z.
6. Evaluated the limits: We plugged in the limits of integration at each step to obtain our final answer.

This step-by-step approach is key to solving triple integrals. It might seem like a lot of work at first, but with practice, it becomes much more manageable. The important thing is to stay organized, keep track of your variables and constants, and double-check your work along the way.

Triple integrals are a powerful tool in mathematics and have applications in various fields, including physics, engineering, and computer graphics. They allow us to calculate volumes, masses, moments of inertia, and other important quantities in three-dimensional space. Mastering the techniques for solving triple integrals will open up a whole new world of problem-solving possibilities.

I hope this step-by-step explanation has helped you understand how to calculate triple integrals. Remember, practice makes perfect! The more problems you solve, the more comfortable you'll become with the process. So, keep practicing, and you'll be a triple integral pro in no time! Keep up the great work, and I'll catch you in the next mathematical adventure!