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

and if we make $$a=0$$, then

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

then

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

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

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 $$\gamma+d\gamma$$.