# Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/224

The denominator of this fraction is the product of the squares of the semi-axes of the surface $\lambda_1$.

If we put

 $D_{1}^{2}=\lambda_{3}^{2}-\lambda_{2}^{2},\ D_{2}^{2}=\lambda_{3}^{2}-\lambda_{1}^{2},\ \mathrm{and}\ D_{3}^{2}=\lambda_{2}^{2}-\lambda_{1}^{2},$ (5)

and if we make $a=0$, then

 $\frac{ds_{1}}{d\lambda_{1}}=\frac{D_{2}D_{3}}{\sqrt{b^{2}-\lambda_{1}^{2}}\sqrt{c^{2}-\lambda_{1}^{2}}}$ (6)

It is easy to see that $D_2$ and $D_3$ are the semi-axes of the central section of $\lambda_1$ which is conjugate to the diameter passing through the given point, and that $D_2$ is parallel to $ds_2$, and $D_3$ to $Ds_3$.

If we also substitute for the three parameters $\lambda_{1},\lambda_{1},\lambda_{1}$ their values in terms of three functions $\alpha,\ \beta,\ \gamma$ defined by the equations

 $\begin{array}{lll} \frac{da}{d\lambda_{1}}=\frac{c}{\sqrt{b^{2}-\lambda_{1}^{2}}\sqrt{c^{2}-\lambda_{1}^{2}}} & & \lambda_{1}=0\ \mathrm{when}\ \alpha=0,\\ \\\frac{d\beta}{d\lambda_{2}}=\frac{c}{\sqrt{\lambda_{2}^{2}-b^{2}}\sqrt{c^{2}-\lambda_{2}^{2}}} & & \lambda_{2}=b\ \mathrm{when}\ \beta=0,\\ \\\frac{d\gamma}{d\lambda_{3}}=\frac{c}{\sqrt{\lambda_{3}^{2}-b^{2}}\sqrt{\lambda_{3}^{2}-c^{2}}} & & \lambda_{3}=c\ \mathrm{when}\ \gamma=0;\end{array}$ (7)

then

 $ds_{1}=\frac{1}{c}D_{2}D_{3}d\alpha,\ ds_{2}=\frac{1}{c}D_{3}D_{1}d\beta,\ ds_{3}=\frac{1}{c}D_{1}D_{2}d\gamma.$ (8)

148.] Now let $V$ be the potential at any point $\alpha,\ \beta,\ \gamma$, then the resultant force in the direction of $ds_1$ is

 $R_{1}=-\frac{dV}{ds_{1}}=-\frac{dV}{d\alpha}\frac{d\alpha}{ds_{1}}=-\frac{dV}{d\alpha}\frac{c}{D_{2}D_{3}}.$ (9)

Since $ds_{1},\ ds_{2}$, and $ds_3$ are at right angles to each other, the surface-integral over the element of area $ds_{2}\ ds_{3}$ is

 $\begin{array}{ll} R_{1}ds_{2}ds_{3} & =\frac{dV}{d\alpha}\frac{c}{D_{2}D_{3}}\cdot\frac{D_{3}D_{1}}{c}\cdot\frac{D_{1}D_{2}}{c}\cdot d\beta\ d\gamma\\ \\ & =\frac{dV}{d\alpha}\frac{D_{1}^{2}}{c}d\beta\ d\gamma.\end{array}$ (10)

Now consider the element of volume intercepted between the surfaces $\alpha,\ \beta,\ \gamma$, and $\alpha+d\alpha,\ \beta+d\beta,\ \gamma+d\gamma.$. There will be eight such elements, one in each octant of space.

We have found the surface-integral for the element of surface intercepted from the surface $\alpha$ by the surfaces $\beta$ and $\beta+d\beta$, $\gamma$ and $\gamma+d\gamma$.