. Applied calculus; principles and applications . Example. — To find the volimie of the right circular conewhose altitude is h and the radius of whose base is a. Divid-ing the volume into parts, each A 7, by passing planes Axapart parallel to the base Ah, and denoting a section at adistance x from the vertex at the origin by Ax, then, sinceAx/Ah = x^/h^, V is given approximately by XA7 = T AxAx = Xand exactly by Ah ^ Ax 7 = hm X ^k Ax=0 0 ^ Ah x^>/i2 3 h^ Ax ^Ah ph Jo x^dx I ,. (1) (2) (3) While AF is a frustum of the cone, dV may be representedby the cyhnder PMMi = Ax Ax = iry^dx. It is
. Applied calculus; principles and applications . Example. — To find the volimie of the right circular conewhose altitude is h and the radius of whose base is a. Divid-ing the volume into parts, each A 7, by passing planes Axapart parallel to the base Ah, and denoting a section at adistance x from the vertex at the origin by Ax, then, sinceAx/Ah = x^/h^, V is given approximately by XA7 = T AxAx = Xand exactly by Ah ^ Ax 7 = hm X ^k Ax=0 0 ^ Ah x^>/i2 3 h^ Ax ^Ah ph Jo x^dx I ,. (1) (2) (3) While AF is a frustum of the cone, dV may be representedby the cyhnder PMMi = Ax Ax = iry^dx. It is to be noted that the equations all apply to a pyramidwith any plane base Ah^^ well as to the cone. 268 INTEGRAL CALCULUS For another example: to find the volume of a sphere withradius a, divide by planes perpendicular to OX; then, since V = lim X A,^-^^x = ^ r {a - x^) dx Ax=0 —a Ci (X *J — a Aof . x^l Ao 4 „ 4 „ - . „ = -r a^a; — 17 = —? * o ^ = o 7^o^^ where Ao = wa^,c? \_ 3J_a a^ 3 3. Otherwise; 2) AF = 2) Ax Aaj = 2) ^?/^ ^^ where Ax = tti/^; x-\ r« 4 .*. F = Hm X TTi/^ Ax = TT I (a^ — x^) dx = -^ iroF. 156. Representation of a Volume by an Area. — In on the significance of an area as an integral it was statedthat the integrals represented by areas might be functionsof various kinds. To show an example of a volume as anintegral represented by an area under a curve, let the volumeof the paraboloid of revolution, between x = 0 and x = 4,be first found as the limit of the sum of the parts between REPRESENTATION OF A VOLUME BY AN AREA 269 the parallel planes Ax apart, as Ax = 0 and the numberof the parts increases without limit. The equation of thegenerating parabola being y^ = I x, V = Umit y\ rf Ax = -T I xdx = -r-pr\—2Tr cubic ^0 - 4 Jo 4 2 Jo
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