TEAM PROJECT. First Fundamental Form of a Surface. Given a

Chapter 10, Problem 10.6

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TEAM PROJECT. First Fundamental Form of a Surface. Given a surface S: r(lI, v), the corresponding quadratic differential fonn (13) ds2 = E du2 + 2F du dv + G dv2with coefficientsis called the first fundamental form of S. (E, F, G arestandard notations that have nothing to do with F andG that occur at some other places in this chapter.) Thefirst fundamental form is basic in the theory of surfaces,since with its help we can determine lengths, angles,and areas on S. To show this, prove the following.(a) For a curve C: u = u(t), v = vet), a ~ t ~ b, onS, formulas (10), Sec. 9.5, and (14) give the length(15)bI = I v'r'(t).r'(t) dt ab= I YEu'2 + 2Fu'v' + Gv'2dt.a(b) The angle 'Y between two intersecting curvesCr: u = gUY, v = h(t) and C2 : u = p(t), v = q(t) onS: r(u, v) is obtained from(16) cos 'Y =where a = rug' + rvh' and b = rup' + rvq' aretangent vectors of Cr and C2 .(c) The square of the length of the normal vector Ncan be writtenso that formula (8) for the area A(S) of S becomesA(S) = I I dA = I I INI du dv(18) S R= I I Y EG - F2 du dv.R(d) For polar coordinates u (= r) and v (= 8) definedby x = u cos v, y = u sin v we have E = 1, F = O.G = u2 , so thatds 2 = du2 + u2 dv2 = dr2 + r2 d82.Calculate from this and (18) the area of a disk ofradius a.(e) Find the tirst fundamental fOlm of the torus inExample 5. Use it to calculate the area A of the torus.Show that A call also be obtained by the theorem of pos,7 which states that the area of a surface ofrevolution equals the product of the length of ameridian C and the length of the path of the center ofgravity of C when C is rotated through the angle In.(I) Calculate the first fundamental form for the usualrepresentations of important surfaces of your ownchoice (cylinder, cone, etc.) and apply them to thecalculation of length~ and areas on these ~urfaces.