CHANGES DURING EVAPORATION OF EMULSION V = 1.275TI0.0.1952/3 = 0.051 The volume of the truncated pyramid with the larger base at h: V = [ 1r(0.1529 h + 0.1506)2(h + 0.985 )/3] - 0.051 = 0.245 h3 + 0.0726 h2 - 0.0715 h - 0.0746 The volume of the cylinder for heights in excess of 1.99 cm: V = 0.6132 + 1r0.4552(h = 1.99) 31 The volumes are added to obtain a curve of the volume versus the height above the bottom of the test tube (Figure App. I(b)). APPENDIX II CORRECT LOCATION OF DIVIDING PLANE IN A TRUNCATED CIRCULAR PYRAMID (FIGURE APP. II(a)) +R d -R Figure App. II(a).

32 JOURNAL OF COSMETIC SCIENCE In the present article the tilted plane has been replaced by a horizontal plane at height (h 1 + h2)/2. (See Appendix I.) This value is obviously too low. The correct value is obtained by integration of the volume from h2 to the h value for the correct plane and by putting the value of that integral equal to the value of the integral from the plane in question to the plane at h 1 . The integrals are in principle rather simple: V = h 2J h Adh, in which the A is the area of the truncated circle. A is obtained from an integral A = 2 -dJ R (R 2 - d2)112dd, which per se has an explicit solution: J(R 2 - d2)112dd = (1/2)[d(R2 - d2) 112 + R 2 arcsin(d/R)}. The expression is simplified by putting R = 1. J(1 - d2)112dd = (1/2)[d(l - d2)112 + arcsin(d)} in which d is measured in radians. For the present case, both R and d are functions of h, a fact which complicates the inte­ gration process. Fortunately, the dependence of both the d and the R on h is linear and a numeric evaluation is possible. With d = ah + b and R = ch + d, dd = adh and the integral to calculate the surface of the truncated circle becomes A = [a / (c2 - a2)112} J [(h2 + 2h{dx - ab) / (c2-a2)} + (d2-b2) / (c2 -a2)}112dh. The correct value of h for the dividing plane is obtained from the equation h 2J h Adh = h J h 1 Adh.