. Differential and integral calculus, an introductory course for colleges and engineering schools. ng upon the contour C or the number of faces of P has become very great, each polyg-onal face, AP, coincides very nearly with the surface S itself, andmay be regarded as a small portion of that surface. That is, Vp Vp -frjrdS , or -—rdxdy, dz dz is looked upon as the area of a small portion of the surface S, andis termed the element of surface. If, as in Art. 199, we use as the direction cosines — V —o 1 Vl+p2 + g2 Vl+p2 + g2 Vl+pt + q*forniula (C) becomes x=5 v=p(x) (CO S=f fVl+p* + q* x=
. Differential and integral calculus, an introductory course for colleges and engineering schools. ng upon the contour C or the number of faces of P has become very great, each polyg-onal face, AP, coincides very nearly with the surface S itself, andmay be regarded as a small portion of that surface. That is, Vp Vp -frjrdS , or -—rdxdy, dz dz is looked upon as the area of a small portion of the surface S, andis termed the element of surface. If, as in Art. 199, we use as the direction cosines — V —o 1 Vl+p2 + g2 Vl+p2 + g2 Vl+pt + q*forniula (C) becomes x=5 v=p(x) (CO S=f fVl+p* + q* x=S y=P(x) „ dx. x=y y=a(x) In this case the element of surface is V1 + p2 + g2 cfo cfo/. By projecting $ upon the ?/2-plane or the £2-plane, we get S inthe form \^dS= f f—dydz, J J dx J J dx S 2=7 V=a(z) §217and (C2) s = MULTIPLE INTEGRALS 2=5 z=/9(z) Vr 331 firii dxdz. S z=y x=a(.z) S denotes a different region, of course, in each of the three for-mulae (C), (Ci), (C2). Example. Let us find by thismethod the surface of the spherex2 + y2 + z2 = a2. We have dx dz dy2 0, # = 4a2. Then. VR=a= a df z Va2 — x2 — y2dz If S be the area of that part of the surface which lies in the first octant,we have x=a y=Vfl2-x2 x=a y^^at-z* dx x=0 waz S = af f—=JL dx = a r[sin-i -=¥=1 J J Va2-x2-y2 J [ Va2 - x2\ z=0 y=0 = I (sin-1 l)dx — ~o I dx = 2 = 0 Hence the total surface = 8 # = 4 ira2. 217. Exercises. 1. Find the area of each portion of the surface of the spherex2 + y2 + z2 = a2which is cut out by the cylinder x2 + y2 — ax = 0. Hint. S a ( sin-1 V _ + * cfc. To integrate, set x = a tan20. 2. Find the area of the surface of the cone x2 + y2 — 22 tan2a = 0which is cut out by the cylinder (x — c)2 -\-y2 = a2, (a is constant.) 332 INTEGRAL CALCULUS §218 3. By the method of Art. 216, find the area of that portion of thesurface of the cylinder ya -j- z* = a* which is contained in the cylinder 2 X3 +2/3 = a*. 4. Use the method of this article to find the surface of t
Size: 1584px × 1577px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1912
