. Differential and integral calculus. e) and when this inte-gration has been performed (0 and itsdifferential being constant) we will havethe area of the sector OPS which forms the differential area of 342 Integral Calculus OTP. Hence, the limits of 6 are from o to —. And the com- 2 plete integration would give the area of the upper half. For theentire area we have n/* 2 /*2a cos 0 A = 2 ) I dBrdr J? 4 a2 cos2 0™ vC1 las* 3 dd = 4 a2 I cos2 0 #00 2 */ 7T = 4«2j -[ See Ex. 2, p. 322. val 3. Show that 7r^^ measures the area of the ellipse. 4. Find the area of the cardioid, r = a (1 — cos 0). Ans
. Differential and integral calculus. e) and when this inte-gration has been performed (0 and itsdifferential being constant) we will havethe area of the sector OPS which forms the differential area of 342 Integral Calculus OTP. Hence, the limits of 6 are from o to —. And the com- 2 plete integration would give the area of the upper half. For theentire area we have n/* 2 /*2a cos 0 A = 2 ) I dBrdr J? 4 a2 cos2 0™ vC1 las* 3 dd = 4 a2 I cos2 0 #00 2 */ 7T = 4«2j -[ See Ex. 2, p. 322. val 3. Show that 7r^^ measures the area of the ellipse. 4. Find the area of the cardioid, r = a (1 — cos 0). Ans. f na2. 5. Find the area between the line, ay — bx, and the parabola, ay2 = bx. ^«j. I j a dxdy = a 216. Surfaces and Volumes in General by Double and TripleIntegration. I. Surfaces. From § 210 (a), we have •dS = Pds. Differentiating with respect to P, we have d*S = dPds. Let SpS be the position of the generating curve at any instant—/ being any point of that curve. Suppose the surface to be Geometric Applications 343. Fig- 59- generated by the curve moving in the direction of X and sochanging that its coordi-nates always satisfy theequation of the surface,y=zf(x, z). Then, atthe instant of reachingthe position SpS anypoint such as p has amotion (represented bypA) in a direction per-pendicular to the tangentpB of the curve SpS byvirtue of the motion ofthe plane of the curve in the direction of X, and a motion(represented by pB) in the direction of the tangent pB by virtueof the change in the curve as it conforms to the configuration ofthe surface, j =/(x, z), it generates ; and this whatever may bethe absolute or resultant motion of p. ButpB = dP and pA = ds. Hence, d2S = dPds = area, rectangle pA CB. The projection of the rectangle pACB on the coordinateplanes XY, XZ, YZ, are obviously dxdy, dxdz, dydz, we now let , 6, \j/ represent the angles which the plane of therectangle pACB makes with XY, XZ, and YZ, respectively, wehave d2S cos <£ = d
Size: 1872px × 1334px
Photo credit: © Reading Room 2020 / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1900, bookdecade1910, booksubjectcalculu, bookyear1918
