Evaluate ∫C X³z Ds: Line Integral Example
Hey guys! Today, we are going to dive into the exciting world of line integrals and tackle a specific problem. We'll be evaluating the line integral ∫C x³z ds, where C is the line segment connecting the points (0, 6, 6) and (8, 1, 1). This problem combines concepts from calculus and vector geometry, so buckle up and let's get started! Understanding line integrals is crucial in various fields, including physics, engineering, and computer graphics. They allow us to calculate quantities along a curve, such as work done by a force or the mass of a wire. This article will provide a step-by-step guide to solving this problem, ensuring you grasp every concept along the way. So, whether you're a student grappling with multivariable calculus or just someone curious about the beauty of mathematics, this article is for you. Let's break down this problem into manageable steps and conquer it together. We'll start by understanding the basics of line integrals and parameterizing the curve. Then, we'll set up the integral and evaluate it. By the end of this article, you'll have a solid understanding of how to evaluate line integrals along line segments. So, grab your favorite beverage, get comfortable, and let's embark on this mathematical journey!
Understanding Line Integrals
Before we jump into the specific problem, let's take a moment to understand what line integrals are all about. Simply put, a line integral is an integral where the function to be integrated is evaluated along a curve. Instead of integrating over an interval on the x-axis, like in single-variable calculus, we're integrating along a path in space. This path can be a curve in two dimensions or three dimensions, making line integrals a powerful tool for dealing with problems in multivariable calculus. In our case, we're dealing with a line integral of the form ∫C f(x, y, z) ds, where f(x, y, z) is a scalar function, and 'ds' represents an infinitesimal arc length along the curve C. The curve C is defined by the line segment connecting two points in space. To evaluate this integral, we need to parameterize the curve C, meaning we need to express x, y, and z as functions of a single parameter, usually denoted as 't'. This parameterization allows us to transform the line integral into a regular integral with respect to 't'. The function f(x, y, z) then becomes a function of 't', and 'ds' is expressed in terms of 'dt' using the arc length formula. The arc length formula is derived from the Pythagorean theorem and represents the infinitesimal length along the curve. It involves the derivatives of the parametric equations with respect to 't'. Once we have the parameterization and the expression for 'ds', we can set up the integral and evaluate it using standard integration techniques. This process might seem a bit abstract now, but don't worry, we'll break it down step by step as we solve our specific problem. The key takeaway here is that line integrals allow us to integrate functions along curves, and parameterization is the key to making this integration possible. This technique is not only useful in mathematics but also in various applications, such as calculating work done by a force along a path or finding the mass of a curved wire with varying density. So, understanding line integrals opens up a whole new world of problem-solving possibilities!
Parameterizing the Line Segment
The first crucial step in evaluating our line integral is to parameterize the line segment C that connects the points (0, 6, 6) and (8, 1, 1). Parameterization essentially means expressing the coordinates x, y, and z as functions of a single parameter, which we'll call 't'. Think of 't' as a time variable that moves us along the line segment from the starting point to the ending point. When t = 0, we should be at the starting point (0, 6, 6), and when t = 1, we should be at the ending point (8, 1, 1). To find the parametric equations, we can use the formula for a line segment in three dimensions. If we have two points, P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), the line segment connecting them can be parameterized as follows:
r(t) = (1 - t)P + tQ
where r(t) = (x(t), y(t), z(t)) represents a point on the line segment for any value of t between 0 and 1. Applying this formula to our points (0, 6, 6) and (8, 1, 1), we get:
r(t) = (1 - t)(0, 6, 6) + t(8, 1, 1)
Now, let's break this down component-wise:
x(t) = (1 - t)(0) + t(8) = 8t y(t) = (1 - t)(6) + t(1) = 6 - 6t + t = 6 - 5t z(t) = (1 - t)(6) + t(1) = 6 - 6t + t = 6 - 5t
So, our parametric equations for the line segment are:
x(t) = 8t y(t) = 6 - 5t z(t) = 6 - 5t
with 0 ≤ t ≤ 1. We've successfully parameterized the line segment! This means we've found a way to describe every point on the line segment using a single parameter 't'. This is a major step forward because it allows us to transform our line integral into an integral with respect to 't'. Remember, parameterization is a key technique in multivariable calculus, and it's not just limited to line segments. We can also parameterize curves, surfaces, and even higher-dimensional objects. The ability to parameterize geometric objects opens up a world of possibilities for solving complex problems. Now that we have our parameterization, we're ready to move on to the next step: finding 'ds'.
Finding ds
Now that we have the parametric equations for our line segment, the next step is to find an expression for 'ds', which represents the infinitesimal arc length along the curve. Remember, 'ds' is a crucial component of the line integral, as it tells us how much weight to give to the function we're integrating at each point along the curve. To find 'ds', we'll use the following formula, which is derived from the arc length formula in calculus:
ds = ||r'(t)|| dt
where r'(t) is the derivative of the vector function r(t) with respect to 't', and ||r'(t)|| represents the magnitude (or length) of the vector r'(t). First, let's find r'(t). We have r(t) = (x(t), y(t), z(t)) = (8t, 6 - 5t, 6 - 5t). So, we need to differentiate each component with respect to 't':
x'(t) = d/dt (8t) = 8 y'(t) = d/dt (6 - 5t) = -5 z'(t) = d/dt (6 - 5t) = -5
Therefore, r'(t) = (8, -5, -5). Now, we need to find the magnitude of r'(t), which is given by:
||r'(t)|| = √(x'(t)² + y'(t)² + z'(t)²)
Plugging in our values, we get:
||r'(t)|| = √(8² + (-5)² + (-5)²) = √(64 + 25 + 25) = √114
So, the magnitude of r'(t) is √114. Now we can express 'ds' as:
ds = ||r'(t)|| dt = √114 dt
We've found 'ds'! This expression tells us how the arc length along the line segment changes as we vary the parameter 't'. Notice that in this case, 'ds' is a constant multiple of 'dt', which makes sense because we're dealing with a straight line segment. For more complex curves, 'ds' might be a more complicated function of 't'. Finding 'ds' is a crucial step in evaluating line integrals. It allows us to convert the integral along the curve into a standard integral with respect to the parameter 't'. Without 'ds', we wouldn't be able to properly account for the length of the curve in our calculation. Now that we have 'ds', we're just one step away from setting up our line integral. We have the parameterization of the curve, and we have 'ds'. All that's left is to substitute these into the integral and evaluate it. Let's move on to that final step!
Setting up and Evaluating the Line Integral
Alright, guys, we've reached the final stage! We have all the pieces of the puzzle, and now it's time to put them together and evaluate our line integral. Recall that we want to evaluate:
∫C x³z ds
where C is the line segment from (0, 6, 6) to (8, 1, 1). We've already found the parameterization of C:
x(t) = 8t y(t) = 6 - 5t z(t) = 6 - 5t
and we've found 'ds':
ds = √114 dt
Now, we need to substitute these into the line integral. We replace x and z in the integrand with their parametric expressions, and we replace 'ds' with √114 dt. The limits of integration will be from t = 0 to t = 1, since those are the values of 't' that correspond to the endpoints of our line segment. So, our integral becomes:
∫C x³z ds = ∫[0 to 1] (8t)³(6 - 5t) √114 dt
Let's simplify the integrand:
(8t)³(6 - 5t) √114 = 512t³(6 - 5t) √114 = √114 (3072t³ - 2560t⁴)
Now we have a standard definite integral with respect to 't':
∫[0 to 1] √114 (3072t³ - 2560t⁴) dt
We can pull the constant √114 out of the integral:
√114 ∫[0 to 1] (3072t³ - 2560t⁴) dt
Now, let's find the antiderivative of the integrand:
∫ (3072t³ - 2560t⁴) dt = 3072(t⁴/4) - 2560(t⁵/5) + C = 768t⁴ - 512t⁵ + C
Now we can evaluate the definite integral using the Fundamental Theorem of Calculus:
√114 [768t⁴ - 512t⁵] evaluated from 0 to 1
= √114 [(768(1)⁴ - 512(1)⁵) - (768(0)⁴ - 512(0)⁵)]
= √114 (768 - 512)
= √114 (256)
So, the value of the line integral is:
256√114
We did it! We successfully evaluated the line integral ∫C x³z ds along the line segment C. This problem involved several steps, but we broke it down into manageable pieces. We parameterized the curve, found 'ds', set up the integral, and evaluated it. This process demonstrates the power of line integrals and how they can be used to solve problems in multivariable calculus. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a pro in no time!
Conclusion
Great job, everyone! We've successfully navigated the world of line integrals and evaluated a challenging problem. We started by understanding the basics of line integrals, then moved on to parameterizing the line segment, finding 'ds', and finally setting up and evaluating the integral. The result, 256√114, represents the value of the integral along the specified path. This exercise highlights the importance of line integrals in various fields, as they allow us to calculate quantities along curves in space. Whether it's finding the work done by a force, the mass of a wire, or other physical quantities, line integrals provide a powerful tool for problem-solving. The key takeaways from this exercise are the steps involved in evaluating a line integral: parameterization, finding 'ds', and setting up and evaluating the integral. Each of these steps is crucial, and mastering them will allow you to tackle a wide range of line integral problems. Remember, the journey of learning mathematics is a continuous one. Keep practicing, keep exploring, and keep challenging yourself. The more you practice, the more comfortable you'll become with these concepts. And who knows, maybe you'll even discover new and exciting applications of line integrals in your own field of study. So, keep up the great work, and I'll see you in the next mathematical adventure!
