cylinders and the fum of the hemisphere and the cone, is e-Book VIII. qual to the difference of two folids, which are each of them lefs than W; but this difference must also be less than W, therefore the difference between the two feries of cylinders and the fum of the hemifphere and cone is lefs than the given folid W. Q. E. D. TH 'HE fame things being fuppofed as in the laft propofition, the fum of all the cylinders inscribed in the hemifphere, and defcribed about the cone, is equal to a cylinder, having the fame base and altitude with the hemifphere. Let the figure DCB be conftructed as before, and fuppofed to revolve about CD; the cylinders infcribed in the hemifphere, that is, the cylinders defcribed by the revolution of the rectangles Hh, Gg, Ff, together with those defcribed about the cone, that is, the cylinders defcribed by the revolution of the rectangles Hs, Gr, Fq, and DN, are equal to the cylinder defcribed by the revolution of the rectangle DB. Let L be the point in which GO meets the circle DB, then, because CGL is a right angle if CL be joined, the circles defcribed with the distances CG and GL are equal to the circle defcribed with the distance CLa or GO; a 2. cor. 6. now CG is equal to GR, becaufe CD is equal to DE, 8. and therefore alfo, the circles defcribed with the distances GR and GL are together equal to the circle defcribed with the distance GO, that is, the circles defcribed by the revolution of GR and GL about the point G, are together equal to the circle defcribed by the revolution of GO about the fame point G; therefore alfo, the cylinders that stand upon the two first of these circles having the common altitude GH, are equal to the cylinder which ftands on the remaining circle, and which has the fame altitude GH. The cylinders defcribed by the revolution of the reangles Gg and Gr are therefore equal to the cylinder defcribed by the rectangle. GP. And as the fame may be fhewn of all the reft, therefore the cylinders defcribed by the rectangles Hh, Gg, Ff, and by the rectangles Hs, Gr, Fq, DN, are together equal to the cylinder defcribed by DB, that is, to the cylinder having the fame bafe and altitude with the hemifphere. Q. E. D. PROP. Book VIII. PRO P. XIV. VERY fphere is two thirds of the circum E fcribing cylinder. Let the figure be constructed as in the two last propofitions, and if the hemifphere described by BDC be not equal to two thirds of the cylinder defcribed by BD, let it be greater by the folid W. Then, as the cone defcribed by CDE is one 10. S. third of the cylinder a described by BD, the cone and the hemifphere together will exceed the cylinder by W. But b 13. 8. A that cylinder is equal to the fum of all the cylinders defcribed by the rectangles Hh, Gg, Ff, Hs, Gr, Fq, DNь; therefore the hemifphere and the cone added together exceed the fum of all these cylinders by the given folid W; c 12. 8. which is abfurd, for it has been fhewn, that the hemifphere and the cone together differ from the fum of the cylinders by a folid less than W. The hemifphere is therefore equal to two thirds of the cylinder defcribed by the rectangle BD; and therefore the whole sphere is equal to two thirds of the cylinder defcribed by twice the rectangle BD, that is, to two thirds of the circumfcribing cylinder. Q.E. D. ELEMENTS OF PLANE AND SPHERICAL TRIGONOMETRY. EDINBURGH: PRINTED FOR BELL & BRADFUTE, AND T. DUNCAN; AND G. G. & J. ROBINSON, LONDON. M,DCC,XCV. ELEMENTS OF PLANE TRIGONOMETRY. A LEMMA İ. IN angle at the centre of a circle is to four right angles as the arch on which it ftands is to the whole circumference. Let ABC be an angle at the centre of the circle ACF, ftanding on the circumference AC: the angle ABC is to four right angles as the arch AC to the whole circumference ACF. Produce AB till it meet the circle in E, and draw DBF perpendicular to AE. Then, because ABC, ABD are two angles at the centre of the circle ACF, the angle ABC is to the angle ABD as the arch AC to the arch AD, (36. 6.); and therefore alfo, the angle ABC is to four times the angle ABD as the arch AC to four times the arch AD (4.5.). But ABD is a right angle, and therefore, four times the arch AD is equal to the whole circumference ACF; there U fore, |