. Differential and integral calculus, an introductory course for colleges and engineering schools. heorem is an analytic one, but we shall give a geometricproof of it. On R as a base we construct a truncated right cylinder like thatof the preceding article, whose upper base lies in the surface z =f(x, y). On each of the small poly-gons AR we construct small trun-cated right cylinders.* Plainly, thelarge cylinder on base R is the sumof the small cylinders. Consider oneof these small truncated cylindersABC DM, and the corresponding pro-duct f(xk,yk)ARk. This product isthe volume of a cylinder no
. Differential and integral calculus, an introductory course for colleges and engineering schools. heorem is an analytic one, but we shall give a geometricproof of it. On R as a base we construct a truncated right cylinder like thatof the preceding article, whose upper base lies in the surface z =f(x, y). On each of the small poly-gons AR we construct small trun-cated right cylinders.* Plainly, thelarge cylinder on base R is the sumof the small cylinders. Consider oneof these small truncated cylindersABC DM, and the corresponding pro-duct f(xk,yk)ARk. This product isthe volume of a cylinder not trun-cated, ABCDM, whose base is ARkand altitude f(xk, yk), and which co-incides with the truncated cylinderexcept in the small upper it is fairly obvious that the limitof the sum of such cylinders asABCDM is V, that is, that \im^f(x,y)AR = V. R As soon as this has been proved our theorem is proved. * We term such a small solid a cylinder, although its lateral surface con-sists in general of parts of several cylindrical surfaces, rather than of a singleclosed cylindrical 324 INTEGRAL CALCULUS §214 The proof is as follows: Let vh v2, . . be the volumes of the small truncated V = 2) v and V = lim V v. R R Let z and Z be the smallest and greatest 2-coordinate in the trun-cated cylinder ABCDM and let vk be its volume. Then zARk = vk^ ZARk. Let zk=f(xk,yk), and divide the inequality by zkARk. There results z_ ^ vk ^ Z2a zkARk zk Since by hypothesis /(#, ?/) is continuous, 2, ^, and Z all have thesame limit, which is not 0; and therefore 1 ^ lim—^- ^ Hence lim —t-=- = 1, and by Duhamels Theorem, Art. 173, inZk&Kk the equation V = lim x ^, the ys may be replaced by the prod- R ucts Z&AZ4 or f(xk, yk)ARk, and therefore F = lim ]£/(*, &)A#. Moreover, it is plain that the above proof does not depend in theleast upon the way in which R is subdivided. Hence our theoremis proved. The limit, lim X/fe 2/)Ai2, ls defined to be the double int
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