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202401041119 Exercise 8.2.541

Evaluate the integral

\displaystyle{\underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x)},

where (a) \varGamma is an arc of the parabola y=\sqrt{x}, (b) \varGamma is a segment of a straight line, (c) \varGamma is the arc of the parabola y=x^2 connecting the points (0,0) and (1,1) in the direction indicated by the arrows (see Figure below).

Extracted from Yakov Stepanovič Bugrov. (1984). A collection of problems.

Answer. (a) 1; (b) 1; (c) 1.

Roughwork.

(a)

When a point moves along the curve \varGamma :y=\sqrt{x} in the indicated direction,

\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y^2\,\mathrm{d}y+\sqrt{x}\,\mathrm{d}x) \\ & = \bigg[\frac{y^3}{3} + \frac{2x^{3/2}}{3}\bigg]\bigg|_{(0,0)}^{(1,1)}\\ & = \bigg(\frac{(1)^3}{3} + \frac{2(1)^{3/2}}{3}\bigg) - \bigg(\frac{(0)^3}{3} + \frac{2(0)^{3/2}}{3}\bigg) \\ & = \bigg(\frac{1}{3}+\frac{2}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}

(b)

When a point moves along the curve \varGamma :y=x in the indicated direction,

\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(y\,\mathrm{d}y+x\,\mathrm{d}x) \\ & = \bigg[\frac{y^2}{2} + \frac{x^2}{2}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{(1)^2}{2} + \frac{(1)^2}{2}\bigg) - \bigg(\frac{(0)^2}{2} + \frac{(0)^2}{2}\bigg)\\ & = \bigg( \frac{1}{2} + \frac{1}{2}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}

(c)

When a point moves along the curve \varGamma :y=x^2 in the indicated direction,

\begin{aligned} &\quad \underset{\varGamma}{\int}(x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int_{(0,0)}^{(1,1)}(\sqrt{y}\,\mathrm{d}y+x^2\,\mathrm{d}x) \\ & = \bigg[\frac{2y^{3/2}}{3} + \frac{x^3}{3}\bigg]\bigg|_{(0,0)}^{(1,1)} \\ & = \bigg(\frac{2(1)^{3/2}}{3} + \frac{(1)^3}{3}\bigg) - \bigg(\frac{2(0)^{3/2}}{3} + \frac{(0)^3}{3}\bigg) \\ & = \bigg(\frac{2}{3}+\frac{1}{3}\bigg) - (0+0) \\ & = 1 \\ \end{aligned}

Afterword.

What am I doing?

\displaystyle{\int (x\,\mathrm{d}y+y\,\mathrm{d}x) = \int x\,\mathrm{d}y + \int y\,\mathrm{d}x}

Note the first term on RHS:

and the second term on RHS:

Many simple formulae in physics, such as the definition of work as W=\mathbf{F}\cdot\mathbf{s}, have natural continuous analogues in terms of line integrals, in this case \displaystyle{W=\underset{L}{\int}\mathbf{F}(\mathbf{s})\,\mathrm{d}\mathbf{s}}, which computes the work done on an object moving through an electric or gravitational field \mathbf{F} along a path L.

Cited from Wikipedia on Line integral

\begin{aligned} &\quad \int (x\,\mathrm{d}y+y\,\mathrm{d}x) \\ & = \int \begin{bmatrix} y & x \end{bmatrix} \begin{bmatrix} \mathrm{d}x \\ \mathrm{d}y \end{bmatrix} \\ & = \int \mathbf{F}(x,y)\,\mathrm{d}\mathbf{s} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ & \mathbf{F}(x,y) = y\,\hat{\mathbf{i}} + x\,\hat{\mathbf{j}} \\ \cdots & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ \end{aligned}

As an aside, work done W is the scalar (/dot) product of force \mathbf{F} and displacement \mathbf{s}. A conservative vector field in physics is analogous to a path-independent vector field in mathematics. That is, over any paths, the line integral depends only upon the starting point and the end point.

This problem is not to be attempted.