Question

Evaluate the line integral \(\int\left(x^{2}-y^{2}\right) d x-2 x y d y\) along each of the following paths from, \((0,0)\) to \((1,2)\) (a) \(y=2 x^{2}\), (b) \(x=t^{2}, y=2 t\) (c) \(y=0\) from \(x=0\) to \(x=2 ;\) then along the straight line joining \((2,0)\) to \((1,2)\).

Short answer

-29/15, -3/35, -2/1.

Step-by-step solution

Parameterize the first curve

For path (a), the parameterization is given by the equation of the curve: - Use the expression: \(y=2x^2\). - The limits for x are from 0 to 1.Express the integral in terms of x: \[ \int_{0}^{1} (x^2 - (2x^2)^2) \, dx - 2x (2x^2) \, dy \].

Simplify the integral for path (a)

Simplifying the expressions: - Transform the integrals: \[ \int_{0}^{1} (x^2 - 4x^4 ) \, dx - (4x^3) \, dy \].- Since \(dy = 4x \, dx\), substitute \(dy\): \[ \int_{0}^{1} (x^2 - 4x^4 ) \, dx - 8 x^4 \, dx \].Combine like terms: \[ \int_{0}^{1} (x^2 - 12x^4) \, dx \].

Evaluate the simplified integral for path (a)

Perform the integration: \[ \int_{0}^{1} x^2 \, dx - \int_{0}^{1} 12x^4 \, dx \]. - Solve each term individually: \[ \left[ \frac{x^3}{3} \right]_{0}^{1} - 12 \left[ \frac{x^5}{5} \right]_{0}^{1} \]. - Compute the values: \[ \left( \frac{1}{3} - 12 \cdot \frac{1}{5} \right) = \frac{1}{3} - \frac{12}{5} = \frac{-29}{15} \].

Parameterize the second curve

For path (b), parameterize using the given:- \(x = t^2\) and \(y = 2t\).- The limits for t are from 0 to 1.Express the integral in terms of t: \[ \int_{0}^{1} [(t^2)^2-(2t)^2] (2t) \, dt - 2(t^2)(2t) \frac{d (2t)}{dt} \, dt \].

Simplify the integral for path (b)

Simplify and substitute: - Transform the integrals: \[ \int_{0}^{1} t^2 [t^4 - 4t^2] \, dt \].- Combine to get: \[ \int_{0}^{1} (t^4 - 4t^2 ) \, dt \].

Evaluate the simplified integral for path (b)

Perform the integral computation: \[ \int_{0}^{1} t^4 \, dt - 4 \int_{0}^{1} t^6 \, dt \].- Solve each term: \[ \left[ \frac{t^5}{5} \right]_{0}^{1} - 4 \left[ \frac{t^7}{7} \right]_{0}^{1} \].- Evaluate: \[ \left( \frac{1}{5} - \frac{4}{7} \right) = \frac{1}{5} - \frac{4}{7} = \frac{-3}{35} \].

Parameterize the third curve: (First segment)

For path (c), - Firstly consider \(y=0\) from \((0,0)\) to \((2,0)\).- Express the integral: \[ \int_{0}^{2} (x^2-0) \, dx - 2x0 \, dy \].

Evaluate the first segment for path (c)

The integral simplifies to: - \(\int_{0}^{2} x^2 \, dx\) - Solved as \[ \left[ \frac{x^3}{3} \right]_{0}^{2}= \frac{8}{3} \].

Parameterize the third curve: (Second segment)

Next consider along the line joining \((2,0)\) to \((1,2)\).-The line can be parameterized as: \(x = 2-t\) and \(y = 2t\) for t ranging from 0 to 1.- Express the integral: \[ \int_{0}^{1} [(2-t)^2 - (2t)^2](-1) dt - 2(2-t) \cdot 2t dt \].

Simplify and evaluate the second segment for path (c)

Combine and simplify: - \(\int_0^1 [(4-4t+t^2-4t^2)(-1) -8t(2-t)] dt \).Perform the integration: - \[ \int-4+5t^2 - 12t dt = -4 - 0 - 6t |^1_0 =-2 \].

Key concepts

Parameterization of Curves

Parameterization of curves is a crucial step in solving line integrals. By parameterizing, we translate the curve into a set of equations that depend on a single variable, typically denoted as \( t \). This simplifies the evaluation of the integral over a given curve. For instance, if a curve is given by \(x = t^2\) and \(y = 2t\) for \( t \) ranging from 0 to 1, we substitute these expressions directly into our integral. By doing this, we turn a complex curve into a problem that can be handled with simpler limits and simpler forms of integration.

Integration Techniques

Understanding and applying proper integration techniques is essential for calculating line integrals efficiently. To begin, once the curve is parameterized, we represent the integral in terms of our parameter \( t \), \( x \), or \( y \), whichever makes the integral straightforward. For example, when given the integral \( \int_{0}^{1} (x^2 - 4x^4) \, dx \), we need to recognize that it can be split into separate easily integrable parts: \( \int_{0}^{1} x^2 \, dx \) and \( \int_{0}^{1} - 4x^4 \, dx \). By solving each separately, it simplifies our computations. The importance of substituting correctly using derivatives like \( dy \) cannot be overstated, as accurately converting these terms maintains the integrity of the integral's representation.

Evaluation of Line Integrals

Evaluating line integrals involves integrating the parameterized equation over its given limits. Let's break down one example. Suppose we are evaluating an integral along a curve parameterized by \( y = 2t \) and \( x = t^2 \). First, express the integral: \( \int_{0}^{1} (y^2 - x^2) \, dt \). Next, substitute the parameters, which makes this: \( \int_{0}^{1} [(2t)^2 - (t^2)^2] \, dt \), simplifying to \( \int_{0}^{1} [4t^2 - t^4] \, dt \). Finally, integrating each part separately, \(4 \int_{0}^{1} t^2 \, dt \) and \( - \int_{0}^{1} t^4 \, dt \), we find the value of the line integral. It's essential to follow each step diligently to ensure the accuracy of the final result.