. A new treatise on the elements of the differential and integral calculus . huno 64 426 INTEGRAL CALCULUS, Ex. 6. The areas of surfaces and volumes of solids havethus far been found by single integration. As an example ofdouble integration, let it be required to find the volumebounded by the surface determined by the equation xy ziz az,and by the four planes having for their equations x = x^, x=x,^, y — Vi, y — expression for this volume is =/:;/;;f**=(:-<) = J {^2 - ^1) {y-i — ^1) (2i + 2., + ^3 + 24). in which z^, 29; ^3? ^4? ^^^ the ordinates of the points in which the late
. A new treatise on the elements of the differential and integral calculus . huno 64 426 INTEGRAL CALCULUS, Ex. 6. The areas of surfaces and volumes of solids havethus far been found by single integration. As an example ofdouble integration, let it be required to find the volumebounded by the surface determined by the equation xy ziz az,and by the four planes having for their equations x = x^, x=x,^, y — Vi, y — expression for this volume is =/:;/;;f**=(:-<) = J {^2 - ^1) {y-i — ^1) (2i + 2., + ^3 + 24). in which z^, 29; ^3? ^4? ^^^ the ordinates of the points in which the lateral edges of the volume considered pierce the surface xy =z az. Ex. 7. To illustrate triple integration geometrically, in the figure suppose planes to bepassed perpendicular to theaxis of z. Let two of theseplanes be at the distances zand z -\- AZ res23ectively fromthe origin of co-ordinates,cutting from the elementarycolumn PQ^ a rectangularparellelopij^edon ah measuredby AX Ay A z. This parallel-opipedon may be considered as an element of the whole volume V: hence. V=J J Jdxdydz. EXAMPLES. 427 Required the portion of the volume of the right cylinderthat is intercepted by the planes z =: , z ^^; theequation of the base of the cylinder being x^ + 2/^ — 2aa:; =z 0-Here the limits of the integral are z := , z ^,2/ ^= — V2ax — x^, ?/ z= -|- \^2ax — x\ x=zO, x^:z2a: there-fore, denoting the values of y by — y^ + 2/u / dxdydz 0 *y —y^Jx tan. ^ / _ (tan. d — tan. 6) xdxdy /2« xs/2ax-^xdx0 = (tan. ^ — );ta^.The base of this cylinder is a circle in the plane {x, y) tan-gent to the axis of y at the origin of co-ordinates; and thesecant planes pass through theorigin, and are perpendicularto the plane (2, x). The re-quired volume is therefore theportion of the cylinder includedbetween the sections OP, OF,It can be seen from this exam-ple why, as was observed inArt. 244, Avhen there is a rela-tion between the variables atthe limi
Size: 1507px × 1658px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1800, bookdecade1860, bookpublish, bookpublisheretcetc
