Evaluate ∫_C X³z Ds Along A Line Segment
Hey everyone! Today, we're diving into the fascinating world of line integrals, specifically focusing on evaluating the line integral ∫_C x³z ds, where C is the line segment connecting the points (0, 6, 6) and (8, 7, 1). This might sound intimidating at first, but trust me, we'll break it down step by step and make it super clear. So, buckle up, and let's get started!
Understanding Line Integrals
Before we jump into the nitty-gritty details of this specific problem, let's take a moment to understand what line integrals are all about. Think of a line integral as a way to integrate a function along a curve. Instead of integrating over an interval on the x-axis (like in regular calculus), we're integrating along a path in space. This path, which we denote as C, could be a straight line, a curve, or any other trajectory. Line integrals have numerous applications in physics and engineering, such as calculating the work done by a force along a path or determining the mass of a wire with varying density.
In our case, we're dealing with the line integral ∫_C x³z ds. This means we're integrating the function f(x, y, z) = x³z along the line segment C. The 'ds' represents an infinitesimal arc length element along the curve. To actually compute this integral, we'll need to parameterize the curve C and express everything in terms of a single parameter, usually denoted as 't'. This parameterization allows us to transform the line integral into a regular definite integral that we can evaluate using standard calculus techniques. The parameterization process is a crucial step in solving line integrals, and it's something we'll delve into shortly. Without a proper understanding of line integrals, tackling problems like this can feel like navigating a maze blindfolded. So, let's keep this foundational knowledge in mind as we move forward. Grasping the concept of integrating along a curve, rather than just a straight line, opens up a whole new dimension of calculus applications.
Parameterizing the Line Segment
The first crucial step in evaluating our line integral is to parameterize the line segment C. Remember, C connects the points (0, 6, 6) and (8, 7, 1). To parameterize a line segment, we can use the following general formula:
r(t) = (1 - t)r₀ + tr₁, 0 ≤ t ≤ 1
Where r₀ and r₁ are the position vectors of the starting and ending points, respectively, and t is a parameter that varies from 0 to 1. In our case, r₀ = <0, 6, 6> and r₁ = <8, 7, 1>. Plugging these values into the formula, we get:
r(t) = (1 - t)<0, 6, 6> + t<8, 7, 1>
r(t) = <0, 6(1 - t), 6(1 - t)> + <8t, 7t, t>
r(t) = <8t, 6 - 6t + 7t, 6 - 6t + t>
r(t) = <8t, 6 + t, 6 - 5t>
So, our parameterization is x(t) = 8t, y(t) = 6 + t, and z(t) = 6 - 5t, with 0 ≤ t ≤ 1. This parameterization essentially describes the position of a point moving along the line segment as t varies from 0 to 1. When t = 0, we are at the starting point (0, 6, 6), and when t = 1, we are at the ending point (8, 7, 1). Parameterizing the line segment is like creating a roadmap that allows us to navigate the curve and perform the integration. A common mistake is not correctly applying the formula or making errors in the vector arithmetic. So, double-check your calculations and ensure you understand the underlying concept. This parameterized form is essential because it allows us to express the line integral in terms of a single variable, 't', making it much easier to evaluate. Think of it as translating the problem from a 3D space into a 1D world, where we can apply our familiar integration techniques. The key here is to remember that each component of the vector function r(t) represents the x, y, and z coordinates as a function of t.
Finding ds
Now that we have our parameterization r(t) = <8t, 6 + t, 6 - 5t>, the next step is to find ds, which represents the infinitesimal arc length element along the curve. Recall that ds is given by the formula:
ds = ||r'(t)|| dt
Where r'(t) is the derivative of the parameterization vector function and ||r'(t)|| is its magnitude. First, let's find r'(t):
r'(t) = <d/dt(8t), d/dt(6 + t), d/dt(6 - 5t)>
r'(t) = <8, 1, -5>
Next, we need to find the magnitude of r'(t):
||r'(t)|| = √(8² + 1² + (-5)²)
||r'(t)|| = √(64 + 1 + 25)
||r'(t)|| = √90
So, ds = √90 dt. This ds element is crucial because it connects the infinitesimal change in the parameter 't' to the infinitesimal arc length along the curve. It's like a conversion factor that allows us to integrate with respect to 't' instead of the arc length. A common mistake here is forgetting to take the magnitude of the derivative vector. Remember, we need the length of the tangent vector, not just the vector itself. This magnitude calculation ensures we're accounting for the actual distance traveled along the curve as 't' changes. Think of it as calculating the speed of a particle moving along the curve; we need the magnitude of the velocity vector, not just its components. The square root often trips people up, so take your time and double-check your arithmetic. With ds in hand, we're one step closer to transforming our line integral into a standard definite integral. We've essentially found the infinitesimal piece of the curve's length, which will be crucial in the next stage of our evaluation.
Setting Up the Integral
With the parameterization r(t) = <8t, 6 + t, 6 - 5t> and ds = √90 dt in hand, we're now ready to set up the integral. Remember, our original line integral is ∫_C x³z ds. We need to express x and z in terms of t using our parameterization. We have x(t) = 8t and z(t) = 6 - 5t. Plugging these into our integral, we get:
∫_C x³z ds = ∫₀¹ (8t)³(6 - 5t)√90 dt
Notice that the limits of integration are from 0 to 1, which corresponds to the range of our parameter t. This is because our parameterization covers the line segment C as t varies from 0 to 1. The setup of the integral is the culmination of all our previous work. We've successfully transformed the line integral, which was defined over a curve in space, into a regular definite integral with respect to the parameter 't'. A common error at this stage is forgetting to substitute both x and z in terms of t, or messing up the limits of integration. Double-check that you've correctly replaced all the variables and that your limits of integration correspond to the parameter range for your curve. Think of this step as translating a sentence from one language to another; we're expressing the same mathematical idea, but in a form that's easier to work with. This definite integral is now in a form that we can tackle using standard calculus techniques. We've essentially distilled the complexity of integrating along a curve into a straightforward integration problem.
Evaluating the Integral
Now comes the moment we've been working towards: evaluating the integral. We have the integral set up as:
∫₀¹ (8t)³(6 - 5t)√90 dt
Let's simplify this a bit. First, (8t)³ = 512t³, so our integral becomes:
∫₀¹ 512t³(6 - 5t)√90 dt
We can pull the constant √90 out of the integral:
512√90 ∫₀¹ t³(6 - 5t) dt
Now, let's distribute the t³:
512√90 ∫₀¹ (6t³ - 5t⁴) dt
Now we can integrate term by term:
512√90 [∫₀¹ 6t³ dt - ∫₀¹ 5t⁴ dt]
512√90 [6∫₀¹ t³ dt - 5∫₀¹ t⁴ dt]
Using the power rule for integration, we get:
512√90 [6(t⁴/4)|₀¹ - 5(t⁵/5)|₀¹]
512√90 [6(1/4 - 0) - 5(1/5 - 0)]
512√90 [6/4 - 5/5]
512√90 [3/2 - 1]
512√90 [1/2]
256√90
So, the value of the line integral is 256√90. Evaluating this integral required us to combine our knowledge of algebra, calculus, and the parameterization we derived earlier. Each step, from simplifying the expression to applying the power rule for integration, was crucial. A common mistake during this phase is making arithmetic errors, especially when dealing with fractions and exponents. Double-check your work at each step to ensure accuracy. Think of this process as solving a puzzle; each piece (each step) needs to fit perfectly to reveal the final answer. The result, 256√90, represents the value of the line integral along the specified curve. It's a single number that encapsulates the integration of the function f(x, y, z) = x³z along the line segment C. And that, my friends, is how we evaluate a line integral! We've taken a seemingly complex problem and broken it down into manageable steps, from parameterizing the curve to performing the final integration. Pat yourselves on the back for making it this far!
Conclusion
And there you have it! We've successfully _evaluated the line integral ∫C x³z ds, where C is the line segment from (0, 6, 6) to (8, 7, 1). We walked through the process step by step, from understanding the concept of line integrals to parameterizing the curve, finding ds, setting up the integral, and finally, evaluating it. Remember, the key to mastering line integrals is to break them down into smaller, manageable steps. Parameterization is crucial, and understanding how ds relates to the parameter is essential. Don't be afraid to practice and work through different examples. Line integrals might seem daunting at first, but with a solid understanding of the underlying concepts and a bit of practice, you'll be solving them like a pro in no time!
I hope this explanation was clear and helpful. If you have any questions or want to explore more examples, feel free to ask. Keep practicing, and you'll conquer those line integrals in no time! Remember, the world of calculus is vast and fascinating, and every problem you solve adds another tool to your mathematical toolkit. So, keep exploring, keep learning, and most importantly, keep having fun with math! Whether you're a student grappling with multivariable calculus or just someone curious about the beauty of mathematical concepts, line integrals offer a glimpse into the power and elegance of calculus in higher dimensions. They are not just abstract equations; they are tools that can be used to model and understand real-world phenomena, from the flow of fluids to the forces acting on objects in motion. So, embrace the challenge, and let the journey of mathematical discovery continue!
