. Applied calculus; principles and applications . 324 INTEGRAL CALCULUS Example 2. — Find the volume generated by revolving thecardioid, p = 2 a (1 — cos 6) about the initial line. y = 27r167ra J^TT /2 a (1-0 Jo ■cos 6) p2 sin 6 dd dp 64 cos Oy sin Odd = -^-Tra^. Example 3. — Find the volume made by revolving thelemniscate p^ = a^ cos 2 d about the initial line. Jo Jo a V cos 2 d p^ sin 6 do dp 4:Td ~3~ TraM J/ ^ ^x 3 • ^ 7o 4 7ra^(cos 2^)2 sm0(i^ = ^r—0 o log (V2 + 1) 1\ r (2cos2(9-l)^sin(9( 2\/2 6; 170. Volumes by Double Integration — Cylindrical Co-in finding the volume of some solids thein
. Applied calculus; principles and applications . 324 INTEGRAL CALCULUS Example 2. — Find the volume generated by revolving thecardioid, p = 2 a (1 — cos 6) about the initial line. y = 27r167ra J^TT /2 a (1-0 Jo ■cos 6) p2 sin 6 dd dp 64 cos Oy sin Odd = -^-Tra^. Example 3. — Find the volume made by revolving thelemniscate p^ = a^ cos 2 d about the initial line. Jo Jo a V cos 2 d p^ sin 6 do dp 4:Td ~3~ TraM J/ ^ ^x 3 • ^ 7o 4 7ra^(cos 2^)2 sm0(i^ = ^r—0 o log (V2 + 1) 1\ r (2cos2(9-l)^sin(9( 2\/2 6; 170. Volumes by Double Integration — Cylindrical Co-in finding the volume of some solids theintegration is performed morereadily with the use of cylin-drical coordinates. In this system of coordi-nates the position of a pointis given by the cylindrical co-ordinates {r, </), z), where (r, 0)are the polar coordinates ofthe projection (x, y, 0), on theXY plane, of the point {x, y, z). It is evident that the equa-tions of transformation fromrectangular to cylindrical co-ordinates are:. x = r cos 0, 2/ = ^ sin 0, z = z) (1) VOLUMES BY DOUBLE INTEGRATION 325 and those from cylindrical to rectangular, -iV r = Vx^ + 2/2, 0 = cos-i - = sin-i ^ = tan-i ^, z = z. (2) To derive a formula for volume the differential element ofarea in the XY plane may be taken as the rectangular baseof an elementary right prism with altitude z, the base of theactual prisms into which the solid may be divided beingbounded by lines two only of which are right lines, the othertwo being circular arcs, and the altitude of possibly only oneedge being z, since the surface of the solid may be curved,or not parallel to the XY plane, even when expression for the volume of the solid is V = JJzrd(j>dr, (1) where z must be expressed in terms of r or (^ in order to effectthe integration, and where the hmits are to be such as willgive the volume sought. Corollary. — 5r/7 = zr dcp dr is a right prism with rectangu-lar base. The double integral in (1) is the Umit of the sum ofthe eleme
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