Evaluating The Line Integral ∫(x²+y) Dx + (y²+x) Dy Along Different Paths

This article delves into the evaluation of the line integral ∫(x²+y) dx + (y²+x) dy along three distinct paths connecting the points (0,1) and (1,2). This exploration will demonstrate how the path of integration significantly influences the value of the line integral, highlighting a key concept in multivariable calculus. We will consider a straight line path, a piecewise linear path, and a parabolic path, showcasing the different techniques required for each.

i. Along a Straight Line from (0,1) to (1,2)

In this section, we will evaluate the line integral along the straight line path connecting the points (0,1) and (1,2). To begin, we need to parameterize this line segment. The equation of the line passing through these two points can be found using the point-slope form. The slope of the line is (2-1)/(1-0) = 1. Using the point-slope form with the point (0,1), we get y - 1 = 1(x - 0), which simplifies to y = x + 1. Now, we can parameterize the line segment by letting x = t, where t varies from 0 to 1. Then, y = t + 1. Thus, our parameterization is r(t) = (t, t+1), with 0 ≤ t ≤ 1. To evaluate the line integral, we need to express dx and dy in terms of dt. Since x = t, dx = dt. Similarly, since y = t + 1, dy = dt. Now we can substitute these expressions into the line integral:

∫(C) (x² + y) dx + (y² + x) dy = ∫(0 to 1) [(t² + (t+1)) dt + ((t+1)² + t) dt]. Next, we simplify the integrand: ∫(0 to 1) [t² + t + 1 + t² + 2t + 1 + t] dt = ∫(0 to 1) [2t² + 4t + 2] dt. Now, we can integrate term by term: [2t³/3 + 2t² + 2t] evaluated from 0 to 1. Plugging in the limits of integration, we get (2/3 + 2 + 2) - (0) = 2/3 + 4 = 14/3. Therefore, the line integral along the straight line path from (0,1) to (1,2) is 14/3. This calculation underscores the importance of parameterization in evaluating line integrals. By expressing the path in terms of a single parameter, we can transform the line integral into a standard single-variable integral, which can then be evaluated using familiar techniques. The straight line path represents the most direct route between the two points, and its parameterization is relatively straightforward. This provides a foundation for comparing the results obtained along other paths, where the parameterization might be more complex.

ii. Along a Straight Line from (0,1) to (1,1) and from (1,1) to (1,2)

In this section, we will evaluate the line integral along a piecewise linear path. This path consists of two line segments: the first segment goes from (0,1) to (1,1), and the second segment goes from (1,1) to (1,2). We need to evaluate the line integral separately for each segment and then add the results. For the first segment, from (0,1) to (1,1), we can parameterize the path by letting x = t, where t varies from 0 to 1. Since the y-coordinate remains constant at 1, we have y = 1. Thus, the parameterization is r1(t) = (t, 1), with 0 ≤ t ≤ 1. For this segment, dx = dt and dy = 0. Now, let's evaluate the line integral along this segment:

∫(C1) (x² + y) dx + (y² + x) dy = ∫(0 to 1) [(t² + 1) dt + (1² + t) (0)] = ∫(0 to 1) (t² + 1) dt. Integrating, we get [t³/3 + t] evaluated from 0 to 1, which equals (1/3 + 1) - (0) = 4/3. For the second segment, from (1,1) to (1,2), we can parameterize the path by letting y = t, where t varies from 1 to 2. Since the x-coordinate remains constant at 1, we have x = 1. Thus, the parameterization is r2(t) = (1, t), with 1 ≤ t ≤ 2. For this segment, dx = 0 and dy = dt. Now, let's evaluate the line integral along this segment: ∫(C2) (x² + y) dx + (y² + x) dy = ∫(1 to 2) [(1² + t) (0) + (t² + 1) dt] = ∫(1 to 2) (t² + 1) dt. Integrating, we get [t³/3 + t] evaluated from 1 to 2, which equals (8/3 + 2) - (1/3 + 1) = 7/3 + 1 = 10/3. Now, we add the results from the two segments to get the total value of the line integral along the piecewise linear path: ∫(C) (x² + y) dx + (y² + x) dy = ∫(C1) (x² + y) dx + (y² + x) dy + ∫(C2) (x² + y) dx + (y² + x) dy = 4/3 + 10/3 = 14/3. Interestingly, the value of the line integral along this piecewise linear path is the same as the value along the straight line path. This suggests that the line integral might be path-independent, which is a characteristic of conservative vector fields. However, to definitively conclude path-independence, we need to examine the behavior of the integral along a third path, which we will do in the next section. The piecewise linear path provides a different perspective on the integration process, highlighting how the integral can be broken down into simpler components when the path consists of multiple segments.

iii. Along the Parabolic Path x = t, y = t² + 1

In this section, we will evaluate the line integral along the parabolic path defined by x = t and y = t² + 1. The path starts at (0,1) and ends at (1,2), which corresponds to the parameter t varying from 0 to 1. This path provides a curved alternative to the straight line and piecewise linear paths considered previously. To evaluate the line integral, we first express dx and dy in terms of dt. Since x = t, we have dx = dt. Since y = t² + 1, we have dy = 2t dt. Now, we can substitute these expressions into the line integral:

∫(C) (x² + y) dx + (y² + x) dy = ∫(0 to 1) [(t² + (t² + 1)) dt + ((t² + 1)² + t) (2t dt)]. Next, we simplify the integrand: ∫(0 to 1) [2t² + 1 + 2t(t^4 + 2t² + 1 + t)] dt = ∫(0 to 1) [2t² + 1 + 2t^5 + 4t³ + 2t + 2t²] dt = ∫(0 to 1) [2t^5 + 4t³ + 4t² + 2t + 1] dt. Now, we can integrate term by term: [t^6/3 + t^4 + (4/3)t³ + t² + t] evaluated from 0 to 1. Plugging in the limits of integration, we get (1/3 + 1 + 4/3 + 1 + 1) - (0) = 1/3 + 1 + 4/3 + 2 = 5/3 + 3 = 14/3. The value of the line integral along this parabolic path is also 14/3. This result, consistent with the values obtained along the straight line and piecewise linear paths, strongly suggests that the line integral is path-independent. Path-independence is a crucial property in vector calculus, indicating that the integral's value depends only on the endpoints of the path and not on the specific path taken. This property is closely related to the concept of conservative vector fields, which we will discuss further in the conclusion.

Conclusion: Path Independence and Conservative Vector Fields

In this exploration, we evaluated the line integral ∫(x² + y) dx + (y² + x) dy along three different paths connecting the points (0,1) and (1,2): a straight line, a piecewise linear path, and a parabolic path. Remarkably, the value of the line integral was 14/3 in all three cases. This consistent result strongly suggests that the line integral is path-independent. Path-independence is a fundamental concept in vector calculus, and it has significant implications for the properties of the vector field involved. A line integral is path-independent if its value depends only on the endpoints of the path of integration and not on the specific path taken between those endpoints. This property is closely related to the concept of conservative vector fields. A vector field F is said to be conservative if it is the gradient of a scalar potential function, i.e., F = ∇f for some scalar function f. If a vector field is conservative, then the line integral of F along any path between two points is equal to the difference in the potential function evaluated at those points. In other words, ∫(C) F · dr = f(B) - f(A), where A and B are the endpoints of the path C. The fact that our line integral yielded the same value along different paths suggests that the vector field F(x, y) = (x² + y, y² + x) is conservative. To confirm this, we can check if the curl of F is zero. The curl of a two-dimensional vector field F(x, y) = (P(x, y), Q(x, y)) is given by ∂Q/∂x - ∂P/∂y. In our case, P(x, y) = x² + y and Q(x, y) = y² + x. Therefore, ∂Q/∂x = 1 and ∂P/∂y = 1. The curl of F is 1 - 1 = 0, which confirms that F is a conservative vector field. Since F is conservative, there exists a potential function f(x, y) such that ∇f = F. To find f, we need to solve the following system of partial differential equations: ∂f/∂x = x² + y and ∂f/∂y = y² + x. Integrating the first equation with respect to x, we get f(x, y) = x³/3 + xy + g(y), where g(y) is an arbitrary function of y. Now, we differentiate this expression with respect to y: ∂f/∂y = x + g'(y). Comparing this with the second partial differential equation, ∂f/∂y = y² + x, we see that g'(y) = y². Integrating g'(y) with respect to y, we get g(y) = y³/3 + C, where C is an arbitrary constant. Thus, the potential function is f(x, y) = x³/3 + xy + y³/3 + C. Now, we can evaluate the line integral using the potential function: ∫(C) F · dr = f(1, 2) - f(0, 1) = (1³/3 + 12 + 2³/3) - (0³/3 + 01 + 1³/3) = (1/3 + 2 + 8/3) - (1/3) = 1/3 + 2 + 8/3 - 1/3 = 10/3 + 2 = 16/3. However, we made an error in our earlier calculation. Let's re-evaluate the difference in the potential function: f(1, 2) - f(0, 1) = (1/3 + 2 + 8/3) - (0 + 0 + 1/3) = (9/3 + 2) - 1/3 = 3 + 2 - 1/3 = 5 - 1/3 = 14/3. This confirms our earlier result and reinforces the path-independence of the line integral. The consistency of the results obtained along different paths, the zero curl of the vector field, and the existence of a potential function all point to the conservative nature of the vector field F(x, y) = (x² + y, y² + x). This example provides a valuable illustration of the interplay between path-independence, conservative vector fields, and potential functions in multivariable calculus.