MIT18_02SC_SupProbSol3
8
�� �� 3. Double Integrals 3A. Double integrals in rectangular coordinates 3A-1 � 1 � 2 a) Inner: 6x 2 y + y 2 = 12x 2 ; Outer: 4x 3 = 32 . y=−1 0 � π b) Inner: −u cos t + 1 t 2 cos u =2u + 1 π 2 cos u 2 2 t=0 � π/2 Outer: u 2 + 2 1 π 2 sin u =( 2 1 π) 2 + 2 1 π 2 = 4 3 π 2 . 0 2 � x � 1 7 1 4 1 1 3 c) Inner: x 2 y 2 √ x = x 6 − x 3 ; Outer: 7 1 x − 4 x 0 = 7 − 4 = − 28 d) Inner: v √ u 2 +4 � u = u √ u 2 + 4; Outer: 3 1 (u 2 + 4) 3/2 � 1 = 3 1 (5 √ 5 − 8) 0 0 3A-2 �� � 0 � 2 �� � 2 � 0 a) (i) dy dx = dy dx (ii) dx dy = dx dy R −2 −x R 0 −y b) i) The ends of R are at 0 and 2, since 2x − x 2 = 0 has 0 and 2 as roots. 2 �� � 2 � 2x−x dydx = dydx R 0 0 ii) We solve y =2x − x 2 for x in terms of y: write the equation as x 2 − 2x + y = 0 and solve for x by the quadratic formula, getting x =1 ± √ 1 − y. Note also that the maximum point of the graph is (1, 1) (it lies midway between the two roots 0 and 2). We get �� � 1 � 1+ √ 1−y dxdy = dxdy, R 0 1− √ 1−y y=-x x y 2 1 2 y=2x-x 2 �� � √ 2 � x � 2 � √ 4−x 2 c) (i) dy dx = dy dx + dy dx 2 y=x √ R 0 0 2 0 �� � √ 2 � √ 4−y 2 (ii) dx dy = dx dy 2 2 R 0 y d) Hint: First you have to find the points where the two curves intersect, by solving simultaneously y 2 = x and y = x − 2 (eliminate x). The integral dy dx requires two pieces; dx dy only one. R R �� � 2 � 1−x/2 3A-3 a) x dA = x dy dx; R 0 0 � 2 Inner: x(1 − 1 2 x) Outer: 1 2 x 2 − 1 6 x 3 0 = 4 2 − 8 6 = 2 3 . 1
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Transcript of MIT18_02SC_SupProbSol3
3. Double Integrals3A. Double integrals in rectangular coordinates3A-112a) Inner:6x2y +y2= 12x2; Outer:4x3= 32.y=1 0b) Inner:ucos t+1t2cos u = 2u+12cos u2 2t=0/2 Outer:u2+212sin u = (21)2+212=432. 0 2 x17 1 4 1 1 3c) Inner:x2y2 x=x6x3; Outer:71x 4x0=74=28d) Inner:vu2+4u=uu2+4; Outer:31(u2+4)3/21=31(55 8)0 03A-2 0 2 2 0a) (i) dy dx= dy dx (ii) dxdy= dxdyR 2 x R 0 yb) i)TheendsofRareat0and2,since2xx2= 0has0and2asroots.2 2 2xxdydx= dydxR 0 0ii) Wesolvey= 2x x2forxintermsofy: writetheequationasx22x+y= 0 andsolve forxby the quadratic formula, gettingx= 11 y.Notealso that themaximum pointofthe graph is (1, 1) (it liesmidwaybetweenthetworoots0and2).Weget 1 1+1ydxdy= dxdy,R 0 11yy=-x x y 2 12 y=2x-x 2 2 x 2 4x2c) (i) dy dx= dy dx + dy dx2 y=x R 0 0 2 0 2 4y2(ii) dxdy= dxdy2 2 R 0 yd)Hint: Firstyouhavetondthepointswherethetwocurvesintersect,bysolvingsimultaneouslyy2=xandy=x2 (eliminatex).The integral dy dxrequires two pieces; dxdyonlyone.R R 2 1x/23A-3 a) xdA= xdy dx;R 0 02 Inner:x(1 12x) Outer:12x2 16x30=42 86=23 .1 2 E. 18.02 EXERCISES2 1 1yb) (2x+y2) dA= (2x+y2) dxdyR 0 01y2 1 Inner:x2+y2x = 1 y2; Outer:y 31 332 y = .0 0 1 1yc) y dA = y dxdyR 0 y1 1y1 Inner:xyy1=y[(1 y) (y 1)]= 2y 2y2Outer:y2 2y30=1 .3 3 /2 cosx3A-4 a) sin2xdA= sin2xdy dxR /2 0cosx Inner:ysin2x = cos xsin2x Outer:1sin3x/2/2=1 2(1 (1))= .3 3 30 1 xb) xy dA= (xy) dy dx.R 0 x2x 4 611 x x 1 1 1Inner:1xy2=1(x3x5) Outer: .2x222 4 6=2 12=240c) Thefunctionx2 y2iszeroonthelinesy=xandy=x,and positiveon theregionRshown, lyingbetweenx= 0andx= 1.y=xTherefore 1 x Volume= (x2y2) dA= (x2y2) dy dx. 1 2R 0 x x1y=-x y=xInner:x2y 13y3x=43x3; Outer:13x40=13.2 5a 2 2 2 y 22 2 2 2 1213A-5 a) eydy dx= eydxdy= eyy dy=2ey0=2(1 e4)0 x 0 0 01 1 1 2 11 u u u4 2e2e2 2 11e b) dudt= dt du= ueudu= (u1)eu= 1 30 tu0 0u00 2x=u3 1 1 1 u 1 3 111 1 u 1 ln 2.5cc) dudx= dxdu= du= ln(1 +u4) =0 x1/3 1 +u40 01 +u401 +u44 0 43A-6 0; 2 exdA, S=upperhalfofR; 4 x2dA, Q =rstquadrantS Q0; 4 x2dA; 0Q13A-7 a) x4+y4 0x4+y4 1 1 +xdA1 1x 11ln 2 .7b)R1 +x2+y2 0 01 +x2dxdy=2ln(1 +x2)0=2
