数学竞赛积分学讲义(10月19日)

2013cÀunóõ�È©Æ¿m�Ô ·K<:�[U 1!�u = f(x)´ëY¼ê…؏",…F (t) = ∫ t 1 dy ∫ t y f(x)dx,¦‡©§ ( F ′(x) x− 1 − x )2 + 2 = du dx �Ï)u = f(x) 2!�f(x)ëY¼ê…؏0,�D = {(x, y)|x2+y2 ≤ R2},¦ ∫ ∫ D af(x) + bf(y) + (a+ b)f(xy) f(x) + +2f(xy) + f(y) dxdy 3!�D´d­‚y = x3†x = −1, y = 1¤Œ¤�k.4«,¦∫ ∫ D (y2 + sin (xy))dxdy, 4!OŽn­È©∈ ∫ ∫ Σ x2 √ x2 + y2dxdydz,Ù¥Σ´­¡z = √ x2 + y2†z = x2 + y2Œ¤�k .«. 5!¦ ∫ ∫ ∫ V xyzdxdydz,Ù¥V : 1 ≤ yz x ≤ 2, y ≤ zx ≤ 2xy, z ≤ xy ≤ 2z 6!�¼êf(x)këY,…÷vlim x→1 f(x− 1) (x− 1) = 2014¦ lim t→0 ∫ ∫ ∫ x2+y2+z2≤t2 f( √ x2 + y2 + z2)dxdydz t4 7!�Σ = {(x, y, z)| x 2 A2 + y2 B2 + z2 C2 ≤ 1},,¦∫ ∫ ∫ (x a + y b + z c )2 dv 8!�l:A(−1, 0)�:B(3, 0)�þŒ�±(x− 1)2 + y2 = 4(y ≥ 0),K∫ l (4x− y)dx+ (x+ y)dy 4x2 + y2 9!�f(u)äkëY�˜��ê,lAB±AB†»�†þŒ‡�l,lA�B,Ù¥:A(1, 1), B(3, 3),¦ 1�.­‚È© ∫ lAB ( 1 x f ( x y ) + 2y ) dx− ( 1 y f ( x y ) + x ) dy 10!�l´­‚x2 + y2 = R2�˜±,nl� {‚•þ,u(x, y)äk��ëY �ê,… ∂2u ∂x2 + ∂2u ∂y2 = x2 + y2,¦ ∮ l ∂u ∂n dl 11!�l´ x2 4 + y2 = 1�±A,K­‚ ∮ l (x+ 2x2y + x2 + 4y2)ds 1 12!�f(x)ëY¼ê,XJ3Œ7�:�?¿˜^Åã1w��•{üµ4­‚lþ,­‚È © ∮ l f(y)dx+ 2xydy 2x2 + y4 = kيäNlÃ',Ó˜~ê,k. (1):y²:éu?¿˜^Åã1w�{üµ4­‚L,§ØŒ7�:Ø²L�:, K7k∮ L f(y)dx+ 2xydy 2x2 + y4 = 0,…Ù_¤á, (2):Áy²:3?¿˜‡Ø¹�:3ÙS�üëÏ«D0þ,­‚È© ∮ cAB f(y)dx+ 2xydy 2x2 + y4 †äN �cÃ' =†A,Bk'.ef(y)äkëY��ê,¦f(y)�Lˆª 13!�S´ý¥¡ x2 9 + y2 4 +z2 = 1,®S�¡ÈA,¦1˜.­¡È© ∫ ∫ S [(2x+3y)2+(6z−1)2]dS 14!OŽ­¡È© I = ∫ ∫ S 2dydz x cos2 x + dzdx cos2 y − dxdy z cos2 z , Ù¥S´¥¡x2 + y2 + z2 = 1,{•þ• . 15!�S²¡x − y + z = 10un‹I²¡m�kÜ©,{•þ†z¶��b�,f(x, y, z)ë Y.OŽ I = ∫ ∫ [f(x, y, z) + x]dydz + [2f(x, y, z) + y]dydx+ [f(x, y, z) + z]dxdy 16!�­¡Σ´d˜m­‚C : x = t, y = 2t, z = t2(0 ≤ t ≤ 1),7z¶^=˜± ¤�^=­¡,Ù {•þ†z¶�•¤ð�,®ëY¼êf(x, y, z)÷v f(x, y, z) = (x+ y + z)2 + ∫ ∫ Σ f(x, y, z)dydz + x2dxdy ¦f(x, y, z) 17!¦­‚AB�§(A3y¶þ,B31˜–,3x¶��KC,¦�ã/OABC7x¶^=¤/ ¤�^=N�/%�î‹I�uB :�î‹I� 4 5 . 18!¦d­‚y2 = x††‚x = 1¤Œ�þ!�¡(¡—ݏ1)7L�:��?˜†‚�=Ä. þ,¿?ØT=Ä.þ�ŒŠ†�Š. 19!¦y: lim R→+∞ ∫ L ydx− xdy (x2 + xy + y2) 1 2 = 0, L : x2 + y2 = R2 20!�D : x2 + y2 ≤ a2(a > 0), f(x, y)3DþkëY �ê,…df(x, y) = 0(x, y ∈ ∂D),¦y: | ∫ ∫ D f(x, y)dxdy| ≤ pia 3 3 max (x,y)∈D √( ∂f ∂x )2 + ( ∂f ∂x )2 gK:�f(x, y)3²¡«D : ε2 ≤ x2 + y2 ≤ 1(x ≥ 0, y ≥ 0),këY� �ê,…÷v f(x, y) =  0 x2 + y2 = 1 sin ( ex 2+y2 − 1 ) − ( esin (x 2+y2) − 1 ) sin4 (3(x2 + y2)) ln2 ( x2 y2 + 1 ) x2 + y2 = ε2 ¦ lim ε→0+ ∫ ∫ D xf ′x + yf ′ y x2 + y2 dxdy 2