# Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/67

X is constant. In this case $\frac{dX}{d\beta}=0$, and the expression becomes by integration with respect to a,

 $\int \int \frac{dX}{da}\frac{dx}{d\beta} d\beta da = \int X \frac{dx}{d\beta} d\beta$; (9)

where the integration is now to be performed round the closed curve. Since all the quantities are now expressed in terms of one variable $\beta$ we may make $s$, the length of the bounding curve, the independent variable, and the expression may then be written

 $\int X \frac {dx}{ds}\, ds$; (10)

where the integration is to be performed round the curve s. We may treat in the same way the parts of the surface-integral which depend upon Y and Z, so that we get finally,

 $\iint (l\xi +m\eta+n\zeta)dS= \int (X \frac {dx}{ds}+ X \frac {dy}{ds}+X \frac {dz}{ds}) ds$; (11)

where the first integral is extended over the surface, and the second round the bounding curve s [1].

### On the effect of the operator $\nabla$ on a vector function.

25.] We have seen that the operation denoted by $\nabla$ is that by which a vector quantity is deduced from its potential. The same operation, however, when applied to a vector function, produces results which enter into the two theorems we have just proved (III and IV). The extension of this operator to vector displacements, and most of its further development, is due to Professor Tait[2].

Let $\sigma$ be a vector function of $\rho$, the vector of a variable point. Let us suppose, as usual, that

$\rho = ix+jy + kz,$
and
$\sigma = iX+jY+kZ;$
where Xy Y, Z are the components of $\sigma$ in the directions of the axes.

We have to perform on $\sigma$ the operation

$\nabla = i\frac{d}{dx}+j\frac{d}{dy}+k\frac{d}{dz}$
Performing this operation, and remembering the rules for the -

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