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) \(\quad y=2 x^{2}\) (b) \(\quad x=t^{2}, y=2 t\) (c) \(\quad y=0\) from \(x=0\) to \(x=2 ;\) then along the straight line joining (2,0) to (1,2)

Short answer

For (a) and (b) the integrals evaluate to zero. For (c), split the path and compute for each segment.

Step-by-step solution

Path (a) - Setup

Consider the path defined by the equation y = 2x^2. Parameterize the curve by setting y = 2x^2. We need to compute dydx and substitute them in the integral.

Path (a) - Compute dydx

Differentiate y = 2x^2 to find dy: dy = 4x dx. Substitute these expressions into the integral.

Path (a) - Integral Setup

The integral becomes: dx - 2xy dydy = dx (x^2 - (2x^2)^2) dx - 2x (2x^2) * 4x dx = dx.x ^ 2 - 4x^4 - 8x^4 dx. Calculate this integral from y 0 1dy.

Path (a) - Evaluate the Integral

Integrate dx.x ^ 2 - 4x^4 - 8x^4 dx from dx 0 1 to get the result: 0.

Path (b) - Parameterize the Curve

Consider the parameterized path dx=t^2dt, dy=2tdt. Compute the integral with dx=t^2 - (2t)^2dx = -4t^2 dt^2.

Path (b) - Substitute and Integrate

Integral becomes 0^1dxt^2 dt=t^2 1dt= -4t^2 dt^2 dx= -4dt.

Path (c) - Split the Path

The path is split into two segments: from (0,0) to (2,0), and from (2,0) to (1,2). A straight line then connects the two points.

Path (c) - First Segment

Evaluate the integral dx^2 dx^2 -2x(0)^2dx by splitting 0^ 2^1. The first segment dt becomes: dx = dx,x ^ 2.

Path (c) - Second Segment

For the second segment, parameterize the line from (2,0) to (1,2)t =1 - t, y = 2t^x+2: dt(.1,t=1.^dxdtdx) t.

Key concepts

path parameterization

Path parameterization is an essential concept when calculating line integrals. It involves expressing the coordinates of points on a path as functions of a single parameter, typically denoted as t. This allows us to convert a multivariable integral into a single-variable integral, making it easier to evaluate. For instance, if a curve is given by the equation \(y = 2x^2\), you can parameterize the path by setting \(x = t\) and \(y = 2t^2\). By substituting these expressions into the integral, you can simplify the computation. Path parameterization is a powerful tool because it transforms complex geometrical paths into simpler parametric forms, enabling the application of standard calculus techniques.

differential calculus

Differential calculus is crucial when dealing with parameterized curves for line integrals. It involves the computation of derivatives to understand how functions change. For a parameterized path like \(x = t\) and \(y = 2t^2\), you need to compute the derivatives of x and y with respect to t. The derivative \(dy/dt\) tells you how y changes as t changes, and similarly, \(dx/dt\) tells you how x changes with t. These derivatives are used to transform the line integral into a form that can be evaluated using single-variable calculus techniques. For example, if \(y = 2x^2\), differentiating gives \(dy = 4x \, dx\). These differential expressions are then substituted back into the integral for easier computation.

integral calculus

Integral calculus comes into play after path parameterization and differentiation. It focuses on the accumulation of quantities and helps in evaluating line integrals. Once you have parameterized the curve and computed the necessary derivatives, you set up the integral. For the line integral \(\int (x^2 - y^2) dx - 2xy dy\) along a path, parameterizing and substituting derivatives simplifies the evaluation. For instance, if parameterized by \(x = t\) and \(y = 2t^2\), the integral can be transformed into a single-variable form \(\int_t (t^2 - (2t^2)^2) dt - 2t(2t^2) 4t dt\). You then compute the integral over the relevant range of t values. Integral calculus allows us to find the total 'effect' along the path specified.

multivariable calculus

Multivariable calculus deals with functions of multiple variables and is essential for understanding line integrals in higher dimensions. It involves partial derivatives and multiple integrals, expanding the concepts of single-variable calculus to more dimensions. When given a path described by equations like \(y = 2x^2\) or \(x = t^2, y = 2t\), multivariable calculus enables us to compute line integrals by transforming them into manageable single-variable integrals through parameterization and differentiation. For example, to evaluate \(\int (x^2 - y^2) dx - 2xy dy\), you use properties of multivariable functions, transform using parameterization, and compute partial derivatives. Understanding these concepts is crucial for correctly evaluating the integral and interpreting its physical meaning in the context of fields such as physics and engineering.