Page:BatemanElectrodynamical.djvu/6

228 A term such as $$\int\int E_{x}dx\ dt$$ may now be understood to mean

and a term such as $$\int\int\int\rho w_{x}dy\ dz\ dt$$ to mean

The equation (II) may now be written

Transforming the left-hand side by means of Green's theorem, we get

$$\begin{array}{ll} \int\int\int\left[\frac{d}{dx}\left(E_{x}-H_{y}\frac{\partial t}{\partial z}+H_{z}\frac{\partial t}{\partial y}\right)\right. & +\frac{d}{dy}\left(E_{y}-H_{z}\frac{\partial t}{\partial x}+H_{x}\frac{\partial t}{\partial z}\right)\\ \\ & \left.+\frac{d}{dz}\left(E_{z}-H_{x}\frac{\partial t}{\partial y}+H_{y}\frac{\partial t}{\partial x}\right)\right]dx\ dy\ dz.\end{array}$$

Now

hence the above integral may be written

$$\begin{array}{l} \int\int\int\left[\left(\frac{\partial E_{x}}{\partial t}-\frac{\partial H_{z}}{\partial y}+\frac{\partial H_{y}}{\partial z}\right)\frac{\partial t}{\partial x}+\left(\frac{\partial E_{y}}{\partial t}-\frac{\partial H_{x}}{\partial z}+\frac{\partial H_{z}}{\partial x}\right)\frac{\partial t}{\partial y}\right.\\ \qquad\left.+\left(\frac{\partial E_{z}}{\partial t}-\frac{\partial H_{y}}{\partial x}+\frac{\partial H_{x}}{\partial y}\right)\frac{\partial t}{\partial z}+\left(\frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}\right)dx\ dy\ dz,\right]\end{array}$$