Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/381

287.] the current normal to its plane is $$u,$$ so that the quantity which enters through this triangle is $$ \frac r^2 \frac.$$

The quantities which enter through the triangles $$OCA$$ and $$OAB$$ respectively are

If $$ \gamma $$ is the component of the velocity in the direction $$ OR, $$ then the quantity which leaves the tetrahedron through $$ ABC $$ is Since this is equal to the quantity which enters through the three other triangles, multiplying by $$ \frac, $$ we get

and make $$ l', m', n' $$ such that

Hence, if we define the resultant current as a vector whose magnitude is $$ \Gamma ,$$ and whose direction-cosines are $$ l', m', n'$$ and if $$ \gamma $$ denotes the current resolved in a direction making an angle $$ \theta $$ with that of the resultant current, then shewing that the law of resolution of currents is the same as that of velocities, forces, and all other vectors.

287.] To determine the condition that a given surface may be a surface of flow. be the equation of a family of surfaces any one of which is given by making $$ \lambda $$ constant, then, if we make the direction-cosines of the normal, reckoned in the direction in which $$ \lambda $$ increases, are