not to cross at the knot A but first to trace out the loop AC and then to trace out the loop A B, in both cases going round these two loops in the direction

B

C

B

FIG. 45..

indicated by the arrow-heads. We are thus able in all cases to convert one line cutting itself in a knot into two lines, each bounding a separate loop, which just touch at the point indicated by the former knot. This dissolution of knots may be suggested to the reader by leaving a vacant space where the boundaries of the loops really meet. The two knots in the following figure are shown dissolved in this fashion :

FIG. 46.

The reader will now find no difficulty in separating the most complex tangle into simple loops. The positive or negative character of the areas of these loops

will be sufficiently indicated by the arrow-heads on their perimeters. We append an example :

FIG. 47.

In this case the tangle reduces to a negative loop a, and to a large positive loop b, within which are two other positive loops c and d, the former of which con

tains a fifth small positive loop e. The area of the entire tangle then equals b+c+d+e a. The space marked s in the first figure will be seen from the second to be no part of the area of the tangle at all.

§ 13. On the Volumes of Space-Figures. Let us consider first the space-figure bounded by three pairs of parallel planes mutually at right angles. Such a space-figure is technically termed a rectangular parallelepiped,' but might perhaps be more shortly described as a 'right six-face.' We may first observe that when one edge of such a right six-face is lengthened or shortened in any ratio, the other nonparallel edges being kept of a fixed length, the volume

B'

C

A

A

FIG. 49.

will be increased in precisely the same ratio. Hence, in order to make any right six-face out of a cube we have only to give the cube three stretches (or it may be squeezes), parallel respectively to its three sets of parallel edges. Let o A, 0 B, o c be the three edges of the cube which meet in a corner o. Let o A be stretched to o A', so that the ratio of OA' to OA is represented by a; then if the figure is to remain right all lines parallel to o A will be stretched in the same ratio. The figure has now become a six-face whose section perpendicular to o A' only is a square. Now stretch o B to O B', so that the ratio o B' to oв be represented by b, and let all lines parallel to oв be

increased in the same ratio; the figure is now a right six-face, only one set of edges of which are equal to the edge of the criginal square. Finally stretch oc to oc', so that oc and all lines parallel to it are increased in the ratio of oo' to oc, which we will represent by c. By a process consisting of three stretches we have thus converted our original cube into a right six-face. If the cube had been of unit-volume, the volume of our six-edge would obviously be abc, and we may show as in the case of a rectangle (see p. 115) that abccba = bac, &c.; or the order of multiplying together three ratios is indifferent. If we term the face A'c' of our

right six-face its base and O B' its height, ac will represent the area of its base, and b its height, or the volume of a right six-face is equal to the product of its base into its height.

Let us now suppose a right six-face o A D C EBF G to receive a shear, or the face B E F G to be moved in its own plane in such fashion that its sides remain parallel to their old positions, and B and E move respectively along B F and E G. If B' E' G'F' be the new position of the face BEG F, it is easy to see that the two wedgeshaped figures BEE' B'OC and F G G'F'A D are exactly equal; this follows from the equality of their corresponding faces. Hence the volume of the sheared

figure must be equal to the volume of the right six-face. Now let us suppose in addition that the face B' E'G' F' is again moved in its own plane into the position B′′ E′′ G"F", so that B' and E' move along B' E' and F'G' respectively. Then the slant wedge-shaped figures B' B" F" FAO and EE"G" F'DC will again be equal, and the volume of the six-face B" E" G" F" AD CO obtained by this second shear will be equal to the volume of the figure obtained by the first shear, and therefore to the volume of the right six-face. But by m. as of two shears we can move the face BE G F to any position in its plane, B" E" G" F", in which its sides. remain parallel to their former position. Hence the volume of a six-face will remain unchanged if, one of its faces, O C D A, remaining fixed, the opposite face, B E G F, be moved anywhere parallel to itself in its own plane. We thus find that the volume of a six-face formed by three pairs of parallel planes is equal to the product of the area of one of its faces and the perpendicular distance between that face and its parallel. For this is the volume of the right six-face into which it may be sheared; and, as we have seen, shear does not alter volume.

The knowledge thus gained of the volume of a sixface bounded by three pairs of parallel faces, or of a so-called parallelepiped, enables us to find the volume of an oblique cylinder. A right cylinder is the figure generated by any area moving parallel to itself in such wise that any point P moves along a line P P' at right angles to the area. The volume of a right cylinder is the product of its height P P' and the generating area. For we may suppose that volume to be the sum of a number of elementary right six-faces whose bases, as at P, may be taken so small that they will ultimately