Page:Scientific Papers of Josiah Willard Gibbs - Volume 2.djvu/49

57. Integration.—If $dv$ represents an element of any space, and $d\sigma$ an element of the bounding surface,

$\iiint \nabla .\omega dv=\iint \omega .d\sigma .$

For the first member of this equation represents the sum of the surface-integrals of all the elements of the given space. We may regard this principle as affording a means of integration, since we may use it to reduce a triple integral (of a certain form) to a double integral.

The principle may also be expressed as follows:

The surface-integral of any vector function of position in space for a closed surface is equal to the volume-integral of the divergence of that function for the space enclosed.

58. Line-integrals.—The integral $\int \omega .d\rho ,$ in which $d\rho$ denotes the element of a line, is called the line-integral of $\omega$ for that line. It is implied that one of the directions of the line is distinguished as positive. When the line is regarded as bounding a surface, that side of the surface will always be regarded as positive, on which the surface appears to be circumscribed counter-clockwise.

59. Integration.—From No. 51 we obtain directly

$\int \nabla u.d\rho =u''-u',$

where the single and double accents distinguish the values relating to the beginning and end of the line.

In other words,—The line-integral of the derivative of any (continuous and single-valued) scalar function of position in space is equal to the difference of the values of the function at the extremities of the line. For a closed line the integral vanishes.

60. Integration.—The following principle may be used to reduce double integrals of a certain form to simple integrals.

If $d\sigma$ represents an element of any surface, and dp an element of the bounding line,

$\iint \nabla \omega \times \omega .d\sigma =\int \omega .d\rho .$

In other words,—The line-integral of any vector function of position in space for a closed line is equal to the surface-integral of the curl of that function for any surface bounded by the line.

To prove this principle, we will consider the variation of the line-integral which is due to a variation in the closed line for which the integral is taken. We have, in the first place,

	$\delta \int \omega .d\rho =$	$\int \delta \omega .d\rho +\int \omega .\delta d\rho .$
But⁠	$\omega .\delta \,d\rho =$	$d(\omega .\delta \rho )-d\omega .\delta \rho .$

Therefore, since $\int d(\omega .\delta \rho )=0$ for a closed line,

$\delta \int \omega .d\rho =\int \delta \omega .d\rho -\int d\omega .\delta \rho .$