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# Plane Curves Expressed in $$x-y$$ coordinates

Figure I.7 shows how an elemental length $$\delta s$$ is related to the corresponding increments in $$x$$ and $$y$$:

$\delta s = \sqrt{\delta x^{2} + \delta y^{2}} = \sqrt{1+\left(\dfrac{dy}{dx}\right)^{2}} \delta x = \sqrt{ \left(\dfrac{dx}{dy}\right)^{2} + 1} \, dy \label{eq:1.4.1}$

Consider a wire of mass per unit length (linear density) $$\lambda$$ bent into the shape $$y = y(x)$$ between $$x = a$$ and $$x=b$$. The mass of an element $$ds$$ is $$\lambda \delta s$$, so the total mass is

$\int \lambda \,ds = \int_a^b \lambda \sqrt{1 + \left(\dfrac{dy}{dx}\right)^{2}} \, dx \label{eq:1.4.2}$

The first moments of mass about the $$y$$ - and $$x$$ -axes are respectively

$\int_a^b \lambda x \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} \, dx \label{eq:1.4.3A}$

and

$\int_a^b \lambda y \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} \, dx \label{eq:1.4.3B}$

If the wire is uniform and $$\lambda$$ is therefore not a function of $$x$$ or $$y$$ , $$\lambda$$ can come outside the integral signs in Equations $$\ref{eq:1.4.2}$$ - $$\ref{eq:1.4.3B}$$, and we hence obtain

$\overline{x} = \dfrac{\displaystyle \int_a^b x \sqrt{ 1+\left( \dfrac{dy}{dx} \right)^2} dx } { \displaystyle \int_a^b \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} dx} \label{eq:1.4.4A}$

and

$\overline{y} = \dfrac{\displaystyle \int_a^b y \sqrt{ 1+\left( \dfrac{dy}{dx} \right)^2} dx } { \displaystyle \int_a^b \sqrt{ 1+ \left(\dfrac{dy}{dx}\right)^2} dx} \label{eq:1.4.4B}$

the denominator in each of these expressions merely being the total length of the wire.

Example $$\PageIndex{1}$$

Consider a uniform wire bent into the shape of the semicircle $$x^{2} + y^{2} = a^{2}$$ , $$x >0$$.

First, it might be noted that one would expect $$\overline{x} > 0.4244a$$ (the value for a plane semicircular lamina).

The length (i.e. the denominators in Equations $$\ref{eq:1.4.4A}$$ and $$\ref{eq:1.4.4B}$$) is just $$\pi a$$. Since there are, between $$x$$ and $$x + \delta x$$, two elemental lengths to account for, one above and one below the $$x$$ axis, the numerator of Equation $$\ref{eq:1.4.4A}$$ must be

$2 \int_0^a x \sqrt{1+ \left(\dfrac{dy}{dx}\right)^{2}} dx$

In this case

$y = \sqrt{a^2 - x^2}$

and

$\dfrac{dy}{dx} = \dfrac{-x}{ \sqrt{a^2 - x^2} }$

The first moment of length of the entire semicircle is

$\overline{x} = 2 \int_0^a x \sqrt{ 1 + \dfrac{x^2}{a^2-x^2} } dx = 2a \int_0^a \dfrac{x\,dx}{ \sqrt{ a^2-x^2 }}$

From this point the student is left to his or her own devices to solve this integral and derive $$\overline{x} = \dfrac{2a}{\pi} = 0.6366a$$.

# Plane Curves Expressed in Polar Coordinates

Figure I.8 shows how an elemental length $$\delta s$$ is related to the corresponding increments in $$r$$ and $$\theta$$:

$\delta s = \sqrt{( \delta r)^{2} + (r \delta \theta )^{2}} = \sqrt{\left( \dfrac{dr}{d \theta }\right)^{2} + r^{2}} \, \delta \theta = \sqrt{1+\left(r \dfrac{d \theta }{dr}\right)^{2}} \,\delta r. \label{eq:1.4.5}$

The mass of the curve (between $$\theta = a$$ and $$\theta = b$$ ) is

$\int_ \alpha ^ \beta \lambda \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta.$

The first moments about the $$y$$ - and $$x$$ -axes are (recalling that $$x = r \cos \theta$$ and $$y = r \sin \theta$$ )

$\int_ \alpha ^ \beta \lambda r \cos \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta$

and

$\int_ \alpha ^ \beta \lambda r \sin \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta.$

If $$\lambda$$ is not a function of $$r$$ or $$\theta$$ , we obtain

$\overline{x} = \dfrac{1}{L} \int_ \alpha ^ \beta r \cos \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} \, d \theta \label{eq:1.4.6A}$

and

$\overline{y} = \dfrac{1}{L} \int_ \alpha ^ \beta r \sin \theta \sqrt{ \left(\dfrac{dr}{d \theta }\right) ^{2} + r^{2}} d \theta \label{eq:1.4.6B}$

where $$L$$ is the length of the wire.

Example $$\PageIndex{2}$$

Again consider the uniform wire of Figure I.8 bent into the shape of a semicircle. The equation in polar coordinates is simply $$r = a$$, and the integration limits are $$\theta = \dfrac{- \pi}{2}$$ to $$\theta = \dfrac{+ \pi}{2}$$ and the length is $$\pi a$$.

Thus

$\overline{x} = \dfrac{1}{ \pi a} \int_{- \pi/2} ^{+ \pi/2} acos \theta [ 0 - a^{2}]^ \frac{1}{2} d \theta = \dfrac{2a}{ \pi } .$

The reader should now find the position of the center of mass of a wire bent into the arc of a circle of angle $$2 \alpha$$. The expression obtained should go to $$\dfrac{2 a }{\pi}$$ as $$\alpha$$ goes to $$\dfrac{\pi}{2}$$ , and to $$a$$ as $$\alpha$$ goes to zero.