(88)-高数专项练题9_08.2026考研数学高途王喆全程班_赠送2025课程_25考研数学（三）全年智达班_{2}--资料

2025 专项刷题班 -第六章 二重积分n n n 例6.1 lim    ( 1 7 ).   n  n  i  n2  j2 i1 j1 1 1 1 x 1 x (A)  dx dy (B)  dx dy 0 0  1  x   1  y2  0 0  1  x  1  y  1 1 1 1 1 1 (C)  dx dy (D)  dx dy 0 0  1  x  1  y  0 0  1  x   1  y2  N I Im = (H)(H)n M = vexity to (ax) Is vx yay ay = ax =1 例6.2 设D由 x  0, y  0, x  y  , x  y  1围成，且 2 I   ln3( x  y)dxdy,I   ( x  y)3dxdy,I  [1  cos(x  y)]3dxdy， 1 2 3 D D D 其中，则I , I , I 之间的大小顺序为( C ) 1 2 3 (A)I  I  I (B)I  I  I (C)I  I  I (D)I  I  I 1 2 3 3 2 1 1 3 2 3 1 2 ↓ Y Estel B ** Int Ed " t X+ t 1-cost = , , , , , X+y = / : t: 1 : In t = 0 EO Most O . D , X+ y z = Int S : ↑ fitt (1 flol = +- - cost) = t - I + cost = o , fitl 1-Sm+> 0 : fit) : f(t) fio) O = > = : to Most , 1 < Is < 12 .例6.3 设l : x2  y2  1,l : x2  y2  2,l : x2  2 y2  2,l : 2x2  y2  2,为 1 2 3 4 y2 四条封闭曲线，记I   (1  x2  )dxdy(i  1,2,3,4)，则 i 2 D i max{I , I , I , I } （ ）. 1 2 3 4 （A）I （B）I （C）I （D） 1 2 3 I 4 X 1) ) 1 : = . Bil : iR fo ED C D2 , , , ↓ fixyl I fixy no = as 1- X - 30X # = (D) +DulA fIxy)30 ED CD+ : 2 <14 : , . fiya I fixy) not In = 10 far 11 fia # Is I + = , H D4 - F : DE fixy) TE* fix d Eit 0 * : , 000 1) Adxay : D  例6.4 设 f x, y 为连续函数，则使 2 2 4x  f (x, y)dxdy  4 dx f  x, y  dy成立的一个充分条件是( D ). 0 0 . 2 2 x  y 4 ( X A) f   x, y    f  x, y  D = g ↑ X         (B) f  x, y   f x, y 且 f  x, y  f x, y =8 'I/1 X (C) f   x, y   f  x, y  , &         2 (D) f  x, y  f x, y 且 f x, y  f x, y - 2例6.5 设区域D  {(x, y) || x |  | y | 1}，则二重积分 I   (1  x)(1  y)(1  x  y )dxdy =_______. ↑ D X 2 = () (1 - x - y +xy) (1 - 1) - 191)dxay 11 Y = 1 1= X !x( ! (1 ( 11) axay 1x) MI)axay - - - - - = + -11 y(-M-Bildxay xy (l-x-1d 4()ux) 5 4x5 4b(1 -X - y)axay = ( -x - y(xy = = =例6.6 极坐标系(,)中的累次积分  1 I   2 d f (cos,sin)d可化为直角坐标系(x, y)中的累次 1 0 cossin 积分 ( B ). 2 2 f (x, y) 1 1x 1 1x (A)  dx f (x, y)dy (B)  dx dy 0 1x 0 1x x2  y2 2 2 f (x, y) 1 1x 1 1x (C)  dx f (x, y)dy (D) dx dy 0 x 0 x x2  y2 ootsm o xy ↑ X+ Y = 1 P= = , ⑪ = ' fro psmo 73 , = - xay a1 例6.7 设函数 f (x)在区间[0,1]上连续,并设 f (x)dx  A，求 0 1 1  dx f (x) f ( y)dy. Y =1 N 0 x Day 1, 11 fix-fly) 10 fillay 75 fina axy x = . = & " Cay) fix figiux /fig) my ? fixax (X) X= · = = ! = fix flyoxay 11 f(x) = faxay ff TB yi = fillay 1 fixux = ↓ ! = · ·1 例6.7 设函数 f (x)在区间[0,1]上连续,并设 f (x)dx  A，求 0 1 1  dx f (x) f ( y)dy. 0 x ( 1. ( f()my xfay 75 ) fax = 35 = = = ) . fitldt Fix =t t 例6.8 设 f (x)为连续函数，F(t)   dy f ( x)dx，则F(2)等于（ B ） 1 y (A)2 f (2) (B) f (2) (C) f (2) (D) 0 N Ct E) FR 1) fix axay = (ax) fix my , · Y=2 = XXD y = y =1 (1) ( , fix) = (x+ ) ax S F(l f(t) f(2) (t 1) = + = . ==2例6.9 设区域D  {(x, y) | x2  y2  R2}，则二重积分 x2 y2 I   (  )dxdy ______. a2 b2 Y =X xx D # 4/1 (x 1 ? 1 laxa : = FP) : 33 - EJBXJFF 13 , =: xaxayaxy 4/1 =) * · I + axay 4(a axay = = +例6.9 设区域D  {(x, y) | x2  y2  R2}，则二重积分 x2 y2 I   (  )dxdy ______. a2 b2 D 2(at + b Ex y's axay ) + = , )n01 * + p2pdp Cat = - = R +    例6.10 设I   max x, y dxdy, I   min x, y dxdy, 1 2 D D   I  [x  y]dxdy其中D  {(x, y) | 0  x  1,0  y  1}， a 表示不超过 3 D a 的最大整数， 则有( D ). . (A)I  I  I (B) I  I  I (C)I  I  I (D)I  I  I 1 2 3 1 2 3 1 3 2 1 3 2 y= X ↑ I axay yaxay 2 2 /x xaya = = uxx Dr yax , 2/jax/xay 3 3 = = 8 I !yaxay + 2)'ax) 22 x axay 21 y axay yay - = = = = ! ,DA0 : x+ Y = 2 x , Ds Dx 0xty < / 0, I [x y] + = Ds Do LEX+y c2 1 3 , (E) 1y ven (11) 2 = 2 =1 X , blEx+y] axay- oaxa llaxay = : + = x 3 : 2 ] 22. i x = = ,↑ 例6.11 设平面区域D  {(x, y) |1  x2  y2  4, x  0. y  0}．计算 xsin( x2  y2 ) X  dxdy. x  y EN D EFE(ER) 15- : EFEFE FJ 33 =: x Sman +yaxd % = =// . Smay 1 i Y x+ 2221 mx y : 10 P Esma = · I xityaxey = =2 2 1 a a a x 例6.12 计算I   dx dy (a  0). 0 x x2  y2 4a2  x2  y2 Y = a+ a Y > y +a = ax ) X + (4 + ai = a 1 X=a af a) pE x y 2ay zaPSmO = + + + = - S I ⑪ axay To = ** xty2 - 4 · S 8 zusmO id - 1 (a -a) . pap = a · a s (iarism P (cod = ao =e(x y), x  0, y  0 例6.13 设 f (x, y)   ，求I   f (x, y)dxdy，其中 0, 其它  D D 是由 x  y  1, x  y  2, y  0和 y  3所围成的闭区域. & D2 : 2 11 fixylaxay fixylaxiy x + = "Gile Py) , (x+ - · leavy X+y 2 11/an = = y x+ =1 > 12 leaxayl eay P 35- 2 0 : = S eaxay -11 1 = Eaxy 35 i =: YA 16= Pax/eay -)ux) ze e ay = = -=P cosotsmo Du 35 = = x+ y = 1 < po + PSmO = 1 = x 7 x+y 2 Ep (x+ y)) = = e O cat mo X+y 2 = y /00/o x+ =1 = e-ploso + sholp > ap ( 2 ① O S Co2O+O = cosonsmo I do e-Plosotsmosp ⑮otsma (cosothold Cosataq 100+Sma 7) # Z I I ~e-t cosotsmo - no + at . , · N I I d no = o = Su