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

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

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