Problem 40 - Calculus: Early Transcendental Functions

Question

Show that the line integral is not independent of path by finding two paths that give different values of the integral. \(\int_{C} y^{2} d x+x^{2} d y,\) where \(C\) goes from \((0,0)\) to \((1,1)\)

Step-by-Step Solution

Answer

The two calculated integration values using different paths from (0,0) to (1,1) are \(\frac{2}{3}\) for a straight line and 0 for the semicircular path proving that the given line integral is not independent of the path.

Step 1: Path Parametrization

Parametrize two different paths that both start from (0,0) and end at (1,1). For path 1, an option is a straight line denoted by \(r_1(t) = (t, t)\) where \(0 \leq t \leq 1\). For path 2, choose a semicircular path from a point on the x-axis to a point on the y-axis in the unit plane, noted as \(r_2(t) = (cos(\pi t/2), sin(\pi t/2))\) where \(0 \leq t \leq 1\).

Step 2: Integral Calculation for Path 1

Substitute the parametrization \(r_1(t)\) into the integral expression and solve it: \(I_1 = \int_{0}^{1}(t^2 dt + t^2 dt) = \int_{0}^{1}2t^2 dt = \left[\frac{2}{3} t^3 \right]_0^1 = \frac{2}{3}\)

Step 3: Integral Calculation for Path 2

Likewise, substitute the parametrization \(r_2(t)\) into the integral expression: \(I_2 = \int_{0}^{1}(sin^2(\pi t/2) (-\pi/2 sin(\pi t/2) dt) + cos^2(\pi t/2) (\pi/2 cos(\pi t/2) dt)) = 0\) because the integral of a full period of both \(cos^2(x)\) and \(sin^2(x)\) over [0,1] will be zero.

Step 4: Compare the Two Integral Values

Comparing the two calculated values \(I_1 \neq I_2\). This shows the given line integral is not path independent since different paths yield different results.

Key Concepts

Path Independence

Path independence in the context of line integrals means that the value of the integral is the same no matter which path is taken between two points. If a line integral is path independent, it signifies there exists a potential function, and the vector field is conservative.
In simple terms, for a function to have path independence, the result of integrating over any two paths between the same points must be identical. However, in this exercise, the calculated integral values differed for two chosen paths, demonstrating that the integral is path dependent. This distinction is crucial in fields such as physics, where it can affect the work done by a force field. Path independence implies saving computational effort, as it requires knowledge of endpoint values only.

Parametrization

Parametrization is the process of defining a path in a more manageable mathematical form. In vector calculus, it's crucial as it simplifies calculating line integrals. You accomplish this by expressing each point on the path using a parameter, usually denoted by \( t \).
In the given problem, two different parametrizations were explored:

• For Path 1: A straight line given by \( r_1(t) = (t, t) \) captures a linear route from (0,0) to (1,1).
• For Path 2: A semicircular path defined using trigonometric functions \( r_2(t) = (\cos(\pi t/2), \sin(\pi t/2)) \) which adds complexity by curving between the same endpoints.

These parametrizations are used to represent different ways of traversing space and calculating respective integrals.

Integral Calculation

Integral calculation involves evaluating the integral of a function along a given path. In this context, we compute line integrals, which consider how a vector field interacts with a curve.
Breaking it down using two parametrized paths:

• For Path 1: Substitution of \( (t, t) \) into the integral simplifies to a polynomial, making it straightforward to solve. The calculation reveals that the integral evaluates to \( \frac{2}{3} \).
• For Path 2: This uses trigonometric identities, and size differences mean more complex integration. Remarkably, the complexity simplifies to zero due to the properties of the sine and cosine over a full cycle in the parameter space.

Differences in results reinforce the lack of path independence.

Vector Calculus

Vector calculus deals with vector fields, vectors, and their transformations, including integration. This broader branch of mathematics studies how to apply calculus to vector fields and is fundamental in understanding physical phenomena.
The given problem can be further understood through the following aspects:

• **Line integrals**: They measure the 'accumulation' over a path in a vector field, akin to how regular integrals 'accumulate' area under a curve.
• **Vector fields**: Assign vectors to each point in a space. Understanding how they can be manipulated through line integrals is key to solving such exercises.

This branch provides theoretical foundations with practical implications, streamlining problem-solving through systematic techniques as seen in this exercise.