110.202. Calculus III2004 Sum m erPractice Final8/3/20041. Com p u te the following limits if they exist:(a) lim(x,y)→(0,0)exyx+1.(b) lim(x,y)→(0,0)cos x−1−x22x4+y4.[Solution](a) Sin ce lim(x,y)→(0,0)exy=1and lim(x,y)→(0,0)x +1 = 1 6=0in aneigh bourhood of (0, 0), we use quotient property of limitsto getlim(x,y)→(0,0)exyx +1=11=1.(b) If w e approaches (0, 0) along y =0, we ha ve, by l’hospitalrule,lim(x,0)→(0,0)cos x − 1 −x22x4=limx→0−sin x − x4x3=limx→0−cos x − 112x.This last limit tends to −∞ as x → 0. Thus, the lim(x,y)→(0,0)cos x−1−x22x4+y4does not exist.¥2. Find the absolute m axim um and minim um for the functionf (x, y, z)=x + y + zon the b allB =©(x, y, z) | x2+ y2+ z2≤ 1ª.[Solution]Write B = U ∪ ∂B where U = {(x, y, z) | x2+ y2+ z2< 1}and ∂B = {(x, y, z) | x2+ y2+ z2=1}.Since ∇f =(1, 1, 1) 6=(0, 0, 0),wehavenocriticalpointsonU.12To find the critical points on ∂B,letf (x, y, z)=x + y + zand g (x, y, z)=x2+ y2+ z2− 1. By the method of Lagrangemultip lie r , we have∇f = λ∇g,that is,(1, 1, 1) = λ (2x, 2y, 2z) .So, w e have the followin g system of equ a tion s:⎧⎪⎪⎨⎪⎪⎩1=2λx1=2λy1=2λzx2+ y2+ z2=1.By 2λx =1,weknowλ 6=0.Wehavex =12λ= y = z.Bythe last equation, we ha ve34λ2=1.Thisimpliesthatλ = ±√32.Moreover, we have two critical points (x, y, z)=³1√3,1√3,1√3´or (x, y, z)=³−1√3, −1√3, −1√3´.Since we hav e only two candidates of absolute maximumand minim um, one m ust be the absolute maxim um and an-other must be the absolute m inimum. It is easy to see thatf³1√3,1√3,1√3´=3√3=√3 is the absolute m aximum and f³−1√3, −1√3, −1√3´=−3√3= −√3 is the absolute minim um .¥3. EvaluateZZZD¡x2+ y2+ z2¢dxdydzo ver the sphere D = {(x, y, z) | x2+ y2+ z2≤ 1}.[Solution]Using spherical coordinate s, let x = ρ sin φ cos θ, y = ρ sin φ sin θand z = ρ cos φ where 0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.3So, x2+ y2+ z2= ρ2and the Jacobian is ρ2sin φ.Therefore,ZZZD¡x2+ y2+ z2¢dxdydz=Z10Zπ0Z2π0ρ2ρ2sin φdθdφdρ=Z10Zπ0Z2π0¡ρ4· sin φ¢dθdφdρ=µZ10ρ4dρ¶·µZπ0sin φdφ¶·µZ2π0dθ¶=Ãρ55¯¯¯¯1ρ=0!·³−cos φ|πφ=0´·¡θ|2πθ=0¢=15· (−(−1) − (−1)) · 2π=4π5.¥4. Determin e whether the in tegra lZZD1√xydxdyexists where D =[0, 1] × [0, 1]. If it exists, compute its value.[Solution]4This integra l is imp roper w h enever xy =0.LetDδ,=[δ, 1]×[ε, 1].WehaveDδ,ε→ D as δ → 0 and ε → 0 and1√xyis well-defined and integrable in Dδ,ε. Therefore,ZZD1√xydxdy=limδ,→0Z1δZ11√xydydx=limδ→0Z1δlim→0"µ2√xyx¶¯¯¯¯1y=#dx=limδ→0Z1δlim→0∙2√xx− 2√xx¸dx=limδ→0Z1δ2√xdx=limδ→0h¡4√x¢¯¯1x=δi=limδ→0h4 − 4√δi=4.Since limδ→0δ ln δ = limδ→0ln δ1δ=limδ→01δ−1δ2=limδ→0−δ =0,wehaveRRDx+yx2+2xy+y2dxdy =2ln2.¥5. EvaluateZ10Z1yex2dxdy.[Solution]5By drawing the graph of the region, w e change the order ofintegration intoZ10Z1yex2dxdy =Z10Zx0ex2dydx=Z10xex2dx=ex22¯¯¯¯¯1x=0=e2−12=e − 12.¥6. EvaluateZZSz2dSwhere S is the portion of the cylind er x2+ y2=4between theplanes z =0and z = x +3.[Solution]Use cylindrical coordinates for the cylinder. Let x =2cosθ,y =2sinθ and z = z where 0 ≤ z ≤ x +3=2cosθ +3 and0 ≤ θ ≤ 2π. So, the parametrization Φ of S isΦ (θ, z)=(2cosθ, 2sinθ, z)where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 2cosθ +3.We calculateTθ× Tz=(−2sinθi +2cosθj) × (zk)=2cosθi +2sinθj.So, we havekTθ× Tzk =2.6Therefore,ZZSz2dS =Z2π0Z2cosθ+30z2(2) dzdθ=Z2π023(2 cos θ +3)3dθ=23Z2π0¡8cos3θ +36cos2θ +54cosθ +27¢dθ=60π.¥7. Auniformfluid that flows vert ically downw ard (like, heavy rain)is described by the vector fieldF (x, y, z)=(0, 0, −1) .Find the total flux through the cone z =px2+ y2,x2+y2≤ 1.[Solution]Parametrize the cone by letting x = r cos θ, y = r sin θ andz =px2+ y2=q(r cos θ)2+(r sin θ)2= r where 0 ≤ r ≤ 1and 0 ≤ θ ≤ 2π. So, the parametrization Φ of S isΦ (r, θ)=(r cos θ, r sin θ, r)where 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π.We calculate (since our fliud flows downward, we calculateTθ× Trinstead of Tr× Tθ.)Tθ× Tr=(−r sin θi + r cos θj) × (cos θi +sinθj + k)= r cos θi + r sin θj − rk.The flux is the surface in tegral of F over S,thatis,RRSF·dS.Therefore, w e ha veZZSF·dS =ZZΦF· (Tθ× Tr) dθdrZ2π0Z10(0, 0, −1) · (r cos θ, r sin θ, −r) drdθ=Z2π0Z10rdrdθ= π.¥78. EvaluateZCx2ydx + ydywhere C is the bournary of the region between the curv e y = xand y = x3where 0 ≤ x ≤ 1.[Solution]By Green’s theorem,ZCx2ydx + ydy =ZZD∙∂ (y)∂x−∂ (x2y)∂y¸dxdywhere C is the boundry of D with counterclock w ise orientation.We can d escr ibe D as 0 ≤ x ≤ 1 and x3≤ y ≤ x. H ence,ZZD∙∂ (y)∂x−∂ (x2y)∂y¸dxdy=Z10Zxx3−x2dydx=Z10h¡−x2y¢¯¯xy=x3idx=Z10¡−x3+ x5¢dx= −112.¥9. LetF =2yzi +(−x +3y +2)j +¡x2+ z¢k.EvaluateZZS(∇ × F) · dSwhere S is the cylinder x2+ y2= a2, 0 ≤ z ≤ 1 (without thetop and bottom ).[Solution]We have∇ × F =¯¯¯¯¯¯ij k∂∂x∂∂y∂∂z2yz −x +3y +2 x2+ z¯¯¯¯¯¯= −(2x − 2y) j +(−1 − 2z) k.We can parametrize S as x = a cos θ, y = a sin θ and z = zwhere 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 1. Th u s, the parametr ization Φ8of S isΦ (θ, z)=(a cos θ, a sin θ, z) .where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ 1.We calculateTθ× Tz=(−a sin θi + a cos θj) × (zk)= a cos θi + a sin θj.Therefore,ZZS(∇ × F) · dS=ZZS(∇ × F) · (Tθ× Tz) dθdz=Z2π0Z10(0, −2(a cos θ)+2a sin θ, −1 − 2z) · (a cos θ, a sin θ, 0) dzdθ=Z2π0Z10¡−2a2sin θ cos θ +2a2sin2θ¢dθ=2a2Z2π0¡sin2θ − sin θ cos θ¢dθ=2a2π.By Stokes’ theorem,ZZS(∇ × F) · dS =Z∂SF · dswhere ∂S is the boundary of the surface S. Therefore, we ha vet wo parts of ∂S.Oneisx2+ y2= a2and z =0.Anotheris x2+ y2= a2and z =1.Weparametrizex2+ y2= a2byletting x = a cos θ and y = a sin θ where 0 ≤ θ ≤ 2π. Hence,dx = −a sin θdθ and dy = a cos θdθ. Mor eov er, in both cases,dz =0.9Th us, for the first o ne, we ha veZ∂SF · ds=Z∂S(2yz) dx +(−x +3y +2)dy +¡x2+ z¢dz=Z2π0(−a cos θ +3a sin θ +2)(a cos θdθ)=Z2π0¡−a2cos2θ +3a2sin θ cos θ +2a cos θ¢dθ= −a2π.For th e second one, we hav eZ∂SF · ds=Z∂S(2yz) dx +(−x +3y +2)dy +¡x2+ z¢dz=Z2π0(2 · a sin θ · 1) (−a sin θdθ)+(−a cos θ +3a sin θ +2)(a