Line Integral Calculation: Step-by-Step Guide

Hey guys! Today, we're diving into a fun problem from calculus involving line integrals. We'll be calculating the line integral of a vector field over a curve made up of two line segments. Don't worry, it's not as scary as it sounds! We'll break it down step-by-step so you can follow along easily. So, let's jump right in!

Problem Statement

The problem we're tackling involves computing a line integral over a piecewise linear curve. Specifically, we're given a curve $C$ that is formed by the union of two line segments. The first segment goes from the origin $(0,0)$ to the point $(-1,-1)$, and the second segment goes from $(-1,-1)$ to the point $(-2,0)$. Our mission, should we choose to accept it (and we do!), is to compute the line integral $\int_C -1 dy + 1 dx$.

Breaking Down the Problem

To solve this problem, we'll need to remember a few key concepts about line integrals and how to handle curves that are defined piecewise. First, let's recall what a line integral is. In simple terms, a line integral calculates the integral of a function along a curve. In our case, the function is given in the form of a differential, $-1 dy + 1 dx$, which represents a vector field. We want to find the "work" done by this vector field along the curve $C$.

Since our curve $C$ is made up of two line segments, we'll need to compute the line integral over each segment separately and then add the results. This is because the parameterization of each segment will be different, and we need to account for that. So, let's break down the curve $C$ into its two segments: $C_1$ and $C_2$.

• $C_1$: The line segment from $(0,0)$ to $(-1,-1)$.
• $C_2$: The line segment from $(-1,-1)$ to $(-2,0)$.

Now, we'll find a parameterization for each of these segments. Remember, a parameterization is a way to describe the points on the curve using a single parameter, usually denoted by $t$. This will allow us to convert the line integral into a regular integral that we can evaluate.

Parameterizing the Curve Segments

Parameterizing $C_1$

The first segment, $C_1$, goes from $(0,0)$ to $(-1,-1)$. To parameterize this, we can use the following approach:

Let $r_1(t) = (x(t), y(t))$ be the parameterization of $C_1$. We want $r_1(0) = (0,0)$ and $r_1(1) = (-1,-1)$. A simple linear parameterization that achieves this is:

$x(t) = -t$ $y(t) = -t$

So, $r_1(t) = (-t, -t)$ for $0 \leq t \leq 1$.

Let's check if this makes sense. When $t=0$, we have $r_1(0) = (0,0)$, which is our starting point. When $t=1$, we have $r_1(1) = (-1,-1)$, which is our ending point. Perfect!

Now, we need to find the derivatives of $x(t)$ and $y(t)$ with respect to $t$, as these will be needed in the line integral formula:

$\frac{dx}{dt} = -1$ $\frac{dy}{dt} = -1$

Parameterizing $C_2$

The second segment, $C_2$, goes from $(-1,-1)$ to $(-2,0)$. We'll use a similar approach to parameterize this segment. Let $r_2(t) = (x(t), y(t))$ be the parameterization of $C_2$. We want $r_2(0) = (-1,-1)$ and $r_2(1) = (-2,0)$.

A linear parameterization for this segment can be written as:

$x(t) = -1 + (-2 - (-1))t = -1 - t$ $y(t) = -1 + (0 - (-1))t = -1 + t$

So, $r_2(t) = (-1-t, -1+t)$ for $0 \leq t \leq 1$.

Again, let's verify this. When $t=0$, we have $r_2(0) = (-1, -1)$, which is our starting point for $C_2$. When $t=1$, we have $r_2(1) = (-2, 0)$, which is our ending point. This looks good!

Now, let's find the derivatives of $x(t)$ and $y(t)$ with respect to $t$:

$\frac{dx}{dt} = -1$ $\frac{dy}{dt} = 1$

Computing the Line Integrals

Line Integral Over $C_1$

Now that we have the parameterization for $C_1$, we can compute the line integral over this segment. The line integral $\int_{C_1} -1 dy + 1 dx$ can be written in terms of the parameter $t$ as:

$\int_{C_1} -1 dy + 1 dx = \int_0^1 ${-1 \frac{dy}{dt} + 1 \frac{dx}{dt}}$ dt$

We already found the derivatives $\frac{dx}{dt} = -1$ and $\frac{dy}{dt} = -1$. Plugging these into the integral, we get:

$\int_0^1 [-1(-1) + 1(-1)] dt = \int_0^1 [1 - 1] dt = \int_0^1 0 dt = 0$

So, the line integral over $C_1$ is 0. That was a nice and easy one!

Line Integral Over $C_2$

Next, we'll compute the line integral over $C_2$. Using the parameterization $r_2(t) = (-1-t, -1+t)$ and the derivatives $\frac{dx}{dt} = -1$ and $\frac{dy}{dt} = 1$, we can write the line integral $\int_{C_2} -1 dy + 1 dx$ as:

$\int_{C_2} -1 dy + 1 dx = \int_0^1 ${-1 \frac{dy}{dt} + 1 \frac{dx}{dt}}$ dt$

Plugging in the derivatives, we have:

$\int_0^1 [-1(1) + 1(-1)] dt = \int_0^1 [-1 - 1] dt = \int_0^1 -2 dt$

Now, we can evaluate this integral:

$\int_0^1 -2 dt = -2 \int_0^1 dt = -2[t]_0^1 = -2(1 - 0) = -2$

So, the line integral over $C_2$ is -2.

Final Calculation

To find the line integral over the entire curve $C$, we simply add the line integrals over $C_1$ and $C_2$:

$\int_C -1 dy + 1 dx = \int_{C_1} -1 dy + 1 dx + \int_{C_2} -1 dy + 1 dx = 0 + (-2) = -2$

Therefore, the line integral $\int_C -1 dy + 1 dx$ is equal to -2.

Conclusion

And there you have it, guys! We've successfully computed the line integral over the piecewise linear curve $C$. We broke down the problem into smaller, manageable steps, parameterized each segment of the curve, and then calculated the line integrals over each segment separately. Finally, we added the results to get the line integral over the entire curve.

This problem highlights the importance of understanding parameterization and how it allows us to convert line integrals into regular integrals. It also shows how we can handle piecewise curves by breaking them down into individual segments and working with each segment separately.

I hope this explanation was helpful and clear. Keep practicing these types of problems, and you'll become a line integral master in no time! If you have any questions, feel free to ask. Until next time, happy calculating!