Mastering Line Integrals On A Line Segment

Dec 9, 2025 by Admin 43 views

Understanding Line Integrals: What Are We Even Doing Here, Guys?

Hey there, calculus explorers! Ever felt like line integrals are some kind of arcane magic? Let's bust that myth right now. A line integral, at its core, isn't as scary as it sounds. Imagine you're walking along a specific, curved path in 3D space, and at every tiny step you take, something around you is changing. Maybe it's the temperature, the density of a wire, or the strength of an electric field. What a line integral helps us do is sum up all those little changes or values along that entire path. It's like a super-powered accumulation over a curve, rather than just a flat area or a bulky volume. When we encounter an expression like ∫C xyz^2 ds, the C immediately tells us we're integrating along a particular curve (in our case, a line segment). The xyz^2 represents the function whose value we're accumulating at each point (x,y,z) along that path. And then there's the star of the show, ds. This ds is the infinitesimally small piece of arc length along our path, and it's super important because it ensures we're summing up the function's values proportionally to the actual distance covered, not just some arbitrary change in a parameter. It's the engine that converts our journey along the curve into a calculable measure.

Line integrals pop up in some seriously cool real-world scenarios! For instance, physicists lean on them to calculate the work done by a force field as an object moves along a specific trajectory. Think about pushing a box across a bumpy floor – a line integral could tell you the total work involved. Engineers might use them to find the total mass of a non-uniform wire, where xyz^2 could represent the varying density at any point (x,y,z) and ds is our tiny segment of wire. Or, perhaps, you’re designing a complex fluid system and need to determine the total flow or circulation along a curved pipe – yep, line integrals are the heroes there too! The critical takeaway here is that we're essentially transforming a multi-variable function f(x,y,z) into a single-variable integral ∫ f(r(t)) ||r'(t)|| dt. We're taking a problem that lives in 3D space and smartly converting it into a more manageable 1D problem along a parameter t. It’s a powerful mathematical tool, and once you get the hang of parameterizing curves and accurately calculating that ds, you'll unlock a whole new level of calculus mastery. This particular problem involves a scalar line integral, meaning our final answer will be a single numerical value, representing the total accumulated quantity. So, get ready, because we're about to dive deep into solving our specific problem, making sure every concept clicks into place, one step at a time. This process might seem a bit abstract initially, but with practice, you'll see how intuitive and logically structured it truly is!

The Crucial First Step: Parameterizing Your Line Segment

Alright, guys, before we can even dream about integrating anything, we need to mathematically describe our path. Our problem gives us a line segment that runs from point A = (-3,2,0) to point B = (-1,3,5). This is where the magic of parameterization comes in! The most common and incredibly useful formula for parameterizing a line segment connecting two points, say A and B, is: r(t) = A + t(B - A), where t typically ranges from 0 to 1. This formula is brilliant because it literally traces your path. When t=0, r(0) = A + 0(B - A) = A, so you start exactly at point A. And when t=1, r(1) = A + 1(B - A) = A + B - A = B, meaning you end precisely at point B. It covers our entire segment perfectly, no more, no less!

Let's meticulously break this down for our specific points. First, we need to calculate the vector B - A. This vector is essentially the displacement from A to B, telling us the direction and magnitude of our line segment.

B - A = (-1 - (-3), 3 - 2, 5 - 0) B - A = (-1 + 3, 1, 5) B - A = (2, 1, 5)

This vector (2, 1, 5) represents the direction and length of our segment. Now, let's plug our starting point A = (-3,2,0) and this direction vector (B - A) into our parameterization formula:

r(t) = (-3,2,0) + t(2,1,5)

To make things super clear for the next steps, it's extremely helpful to write this out component-wise, explicitly defining x(t), y(t), and z(t):

x(t) = -3 + 2t
y(t) = 2 + t
z(t) = 0 + 5t = 5t

And let's not forget those critical bounds for t: 0 ≤ t ≤ 1. If you mess up this parameterization, guys, the rest of your problem is basically toast, so take your sweet time here. Make sure your x(t), y(t), z(t) actually start at A when t=0 and end at B when t=1. A quick mental check is always a good idea:

At t=0: x(0) = -3, y(0) = 2, z(0) = 0. This gives us (-3,2,0). Perfect!
At t=1: x(1) = -3 + 2(1) = -1, y(1) = 2 + 1 = 3, z(1) = 5(1) = 5. This gives us (-1,3,5). Spot on!

Looks like we've nailed the parameterization! This is arguably the most important foundational step in the entire process, as it converts our geometric path into a form we can readily manipulate using standard calculus techniques. Without a correct parameterization, the entire integral will be flawed, no matter how sharp your integration skills are. So, always double-check this part with a smile, knowing you've built a strong foundation.

Calculating `ds`: The Arc Length Element Explained

Okay, now that we've got our curve beautifully parameterized by r(t), the next big piece of our integral puzzle is understanding and calculating ds. Remember, ds isn't just some random variable; it represents an infinitesimally small piece of arc length along our curve. Think of it like taking a microscopically tiny ruler and measuring each micro-segment of your path. We can't simply use dt directly, because t (our parameter) might not correspond linearly to the actual physical distance covered. For instance, if r(t) described a very fast journey, a small Δt could cover a substantial distance. Conversely, a sluggish path might cover a minimal distance for the same Δt. This is where ds steps in, acting as the bridge between the parameter change and the actual distance along the curve.

The fundamental formula that connects ds to dt for a curve r(t) = (x(t), y(t), z(t)) is: ds = ||r'(t)|| dt. This formula is incredibly crucial for line integrals. Let's break it down into digestible pieces. First, we need to find r'(t), which is simply the derivative of each component of our position vector r(t) with respect to t. This r'(t) is often called the velocity vector.

From our previous parameterization step, we derived:

x(t) = -3 + 2t
y(t) = 2 + t
z(t) = 5t

Now, let's take the derivative of each component with respect to t:

x'(t) = d/dt (-3 + 2t) = 2
y'(t) = d/dt (2 + t) = 1
z'(t) = d/dt (5t) = 5

So, our velocity vector is r'(t) = (2, 1, 5). Notice something quite interesting here? For a line segment, r'(t) is a constant vector! This makes perfect sense because a line segment, when parameterized linearly, implies a constant direction and a constant speed. If you were parameterizing a more complex curve, like a helix or a parabola, r'(t) would typically be a function of t, making the calculations a bit more involved. But for our current problem, this constancy simplifies things beautifully.

Next, we need to find the magnitude of this velocity vector, denoted as ||r'(t)||. This magnitude represents the speed at which we are traversing the curve at any given point (or, in this case, constantly). The formula for the magnitude of a 3D vector (a, b, c) is √(a^2 + b^2 + c^2).

||r'(t)|| = √( (x'(t))^2 + (y'(t))^2 + (z'(t))^2 )

Plugging in our derived values:

||r'(t)|| = √( (2)^2 + (1)^2 + (5)^2 ) ||r'(t)|| = √( 4 + 1 + 25 ) ||r'(t)|| = √(30)

Therefore, our arc length element ds is simply √(30) dt. This means for every tiny increment dt of our parameter, we are consistently covering a distance of √(30) units along the line segment. This constant value for ||r'(t)|| is a strong indicator that our parameterization and ds calculation for a line segment are spot on. Understanding why ds is ||r'(t)|| dt is just as vital as knowing the formula itself, as it truly connects the abstract parameter t to the physical geometry of our path.

Setting Up the Integral: Putting All the Pieces Together

Alright, guys, we've gathered all the essential ingredients! We have our original function f(x,y,z) = xyz^2, we've successfully parameterized our line segment with x(t) = -3 + 2t, y(t) = 2 + t, z(t) = 5t for 0 ≤ t ≤ 1, and we've meticulously calculated our arc length element, ds = √(30) dt. Now, it's time for the exciting part: assembling the master integral! The general form for our scalar line integral is ∫C f(x,y,z) ds. Our primary goal is to transform this multi-variable expression into a much more manageable single-variable integral with respect to t.

This conversion involves two critical substitutions:

Replace x, y, z with their t-expressions: We take our function f(x,y,z) = xyz^2 and systematically plug in our parameterized components: x(t), y(t), and z(t).

f(r(t)) = (-3 + 2t)(2 + t)(5t)^2

Let's expand the (5t)^2 term first to simplify things: (5t)^2 = 25t^2

So, the function becomes: f(r(t)) = (-3 + 2t)(2 + t)(25t^2)

This is where careful algebra is your best friend, folks! Don't rush this step. We need to fully expand and simplify this polynomial expression before we attempt to integrate it.
Replace ds with ||r'(t)|| dt: We found earlier that ds = √(30) dt. Simple and direct!

Now, let's substitute everything into our integral, making sure to use our correct limits for t, which are from 0 to 1:

∫C xyz^2 ds = ∫ from 0 to 1 [ (-3 + 2t)(2 + t)(25t^2) ] √(30) dt

Before we jump into integration, we absolutely must simplify that integrand. It's a polynomial, so we'll multiply it out step by step. First, let's multiply the two binomials:

(-3 + 2t)(2 + t) = (-3 * 2) + (-3 * t) + (2t * 2) + (2t * t) = -6 - 3t + 4t + 2t^2 = 2t^2 + t - 6

Now, we take this result and multiply it by 25t^2:

(2t^2 + t - 6)(25t^2) = (2t^2 * 25t^2) + (t * 25t^2) - (6 * 25t^2) = 50t^4 + 25t^3 - 150t^2

Voilà! Our simplified integrand is (50t^4 + 25t^3 - 150t^2). Our full integral now looks much cleaner and far less intimidating:

∫ from 0 to 1 (50t^4 + 25t^3 - 150t^2) √(30) dt

And here’s a neat trick: √(30) is a constant. We can pull it right outside the integral to make our lives even easier!

√(30) ∫ from 0 to 1 (50t^4 + 25t^3 - 150t^2) dt

This setup is absolutely critical. Any slip-up in parameterization, ds calculation, or algebraic expansion here will lead to a completely incorrect answer. Double-check your signs, your exponents, and your basic arithmetic. Trust me, most errors pop up at this stage, not necessarily in the final integration. A well-set-up integral, my friends, is already half the battle won!

Tackling the Integration: Step-by-Step Calculation

Okay, guys, we've done all the heavy lifting in meticulously setting up the integral. Now comes the satisfying part: actually performing the integration! We're looking at a straightforward polynomial, which is fantastic because it means we can confidently use the good old power rule for integration: ∫ t^n dt = (t^(n+1))/(n+1). Since this is a definite integral, we don't need to worry about adding the + C constant. Remember, we wisely pulled √(30) outside the integral, so we'll just integrate the polynomial and then multiply by √(30) at the very end.

Our beautifully simplified integral is:

√(30) ∫ from 0 to 1 (50t^4 + 25t^3 - 150t^2) dt

Let's integrate each term separately using the power rule:

For the first term: ∫ 50t^4 dt = 50 * (t^(4+1))/(4+1) = 50 * (t^5)/5 = 10t^5
For the second term: ∫ 25t^3 dt = 25 * (t^(3+1))/(3+1) = 25 * (t^4)/4
For the third term: ∫ -150t^2 dt = -150 * (t^(2+1))/(2+1) = -150 * (t^3)/3 = -50t^3

Now, we combine these integrated terms to form our antiderivative and prepare to evaluate it from t=0 to t=1.

So, we have:

√(30) [ 10t^5 + (25/4)t^4 - 50t^3 ] evaluated from t=0 to t=1

Next, we apply the Fundamental Theorem of Calculus, which states that we evaluate the antiderivative at the upper limit and subtract its value at the lower limit: F(upper_limit) - F(lower_limit).

Let's first evaluate the expression at the upper limit, t=1:

F(1) = 10(1)^5 + (25/4)(1)^4 - 50(1)^3 F(1) = 10 + 25/4 - 50

To combine these terms, we need a common denominator, which is 4:

F(1) = (40/4) + (25/4) - (200/4) F(1) = (40 + 25 - 200) / 4 F(1) = (65 - 200) / 4 F(1) = -135 / 4

Now, let's evaluate the expression at the lower limit, t=0:

F(0) = 10(0)^5 + (25/4)(0)^4 - 50(0)^3 F(0) = 0 + 0 - 0 = 0

This is one of those nice parts about integrating a polynomial from 0 to 1 – the lower limit often simplifies directly to zero, saving us a bit of calculation!

Finally, we bring it all together by subtracting F(0) from F(1) and multiplying by our √(30) constant:

√(30) [ F(1) - F(0) ] = √(30) [ -135/4 - 0 ] = -135√(30) / 4

And there you have it, guys! The final numerical value of our line integral. It's often quite satisfying when you arrive at a clean fraction or a number that isn't excessively convoluted, though sometimes complex decimals are unavoidable. This final value represents the total