Tiêu đề Tổng hợp đề thi cuối kỳ Giải tích 2 - Nhóm ngành CTTT.pdf ✅

FINAL EXAM OF CALCULUS 2 – Semester 20212 Course: 66 CTTT. Time: 90 minutes Question 1 (1 point). Find the curvature of elip elip x ty t = = 3cos , 3 sin at the point corresponding to t = 0. Question 2 (1 point). Find 2 2 1 D dxdy + + x y ∫∫ with 2 2 Dx y x y : 9, 0, 0 +≤ ≥ ≥ . Question 3 (1 point). Find ( ) 2 2 4 V − − x y dxdydz ∫∫∫ , with 2 22 Vx y z x y : ++≤ ≤ ≤ 4, 0, 0 . Question 4 (1 point). Find the parametric equation for the curve of intersection of surfaces: 2 2 9 18 4 8 23 x xy y − + += , 32 1 x yz + −= . Question 5 (1 point). Find (2 3 ) C x y dx + ∫ , with C is the arc defined by 2 cos , sin , 0 2 x ty t t π = + = ≤≤ . Question 6 (1 point). Find ( ) ( ) ( ) (3,1) 2 1, 1 2 B A x y dx ydy − x y + + + ∫ . Question 7 (1 point). Find S ydS ∫∫ , with S is the half-sphere 2 22 xyz x ++= ≥ 9, 0 Question 8 (1 point). Find ( ) S x z dxdy + ∫∫ , with S is the above of surface 2 2 z x yz =+ ≤ 9, 9 . Question 9 (1 point). Find the volume of V defined by x y x y xyz − + + + + + +++ ≤ 4 2 3 24 22 . Question 10 (1 point). Find (6 sin cos 2 6 3 2 5 ) ( ) ( ) C y y x z dx x y z dy y x dz − − + +− + ++ ∫ with C is the intersection of sphere 2 22 xyz ++= 3 and plane x y + − 0 ,...
Q1: Find the curvature of elip x ty t = = 3cos , 3 sin at the point corresponding to t = 0 . ( ) ( ) ( ) ( ) 3sin 3cos 3 cos 3 sin xt t xt t yt t y t t  =−  =− ′ ′′     →     ′ = ′′ = − ( ) ( ) ( ) ( ) ( ) ( ( ) ) ( ( ) ) ( ) 0 0 0 0 0 3 3 2 2 2 2 2 0 0 0 3 3 0 3 9 0 3 t t t t t t t x y x y C x y = = = = = = = ′ ′ ′′ ′′ − = = =   ′ ′ + +   . Vậy độ cong tại điểm t=0 là 3 9 C = Q2: Find 2 2 1 dxdy + + x y ∫∫ with 2 2 Dx y x y : 9, 0, 0 +≤ ≥ ≥ . 0 3 cos , sin 0 2 r x r J r y r  ≤ ≤  = φ    ⇒ = π  = φ ≤φ≤   ( ) 3 3 2 2 2 0 0 0 1 . ln 1 ln10 1 22 4 r I d dr r r π π π =φ = + = + ∫ ∫ . Vậy 2 2 ln10 1 4 D dxdy x y π = + + ∫∫ . Q3: Find ( ) 2 2 4 V − − x y dxdydz ∫∫∫ , with 2 22 Vx y z x y : ++≤ ≤ ≤ 4, 0, 0. 2 3 cos sin 2 sin sin 0 cos 0 2 sin x r y r r z r J r  π π≤φ≤   = φθ    = φ θ→ ≤θ≤π     = θ ≤ ≤    = θ ( ) 3 2 2 22 2 0 0 I d d r r dr 4 sin sin π π π =θφ − θ θ ∫∫∫ 3 0 32 32 sin sin 23 5 d π π = θ− θ θ ∫ Vậy ( ) 2 2 32 4 5 V −− = π x y dxdydz ∫∫∫
Q4: Find the parametric equation for the curve of intersection of surfaces 2 2 9 18 4 8 23 x xy y − + += , 32 1 x yz + −= . ( ) ( ) 2 2 ⇔ −+ + = 3 3 2 2 36 x y Đặt: 6cos 3 2cos 1 3 3 6cos 3 2 2 6sin 6sin 2 3sin 1 2 t x t x t y t t y t  + = = +  − =    ⇒  + = −  = = −  ⇒= ++ −− zt t 3(2cos 1) 2(3sin 1) 1 = + 6cos 6sin t t ⇒ Parametric equation: 2cos 1 3sin 1 6cos 6sin x t y t z tt  = +   = −   = + . Q5: Find (2 3 ) C x y dS + ∫ , with C is the arc defined by 2 cos , sin , 0 2 x ty t t π = + = ≤≤ . ( ) 2 2 2 0 I 4 2cos 3sin sin cos t t t tdt π =+ + + ∫ 2 0 (4 2cos 3sin ) 2 5 t t dt π = + + = π+ ∫ . Vậy (23 25 ) C x y dS + = π+ ∫ Q6: Find ( ) ( ) ( ) (3;1) 2 3; 1 2 B A x y dx ydy − x y + + + ∫ ( ) 1 2 1 3 ln 2 3 4 ydy I − y = = − + ∫ . Vậy ( ) ( ) ( ) (3;1) 2 3; 1 2 3 ln 2 4 B A x y dx ydy − x y + + = − + ∫
Q7: Find S ydS ∫∫ , with S is the half-sphere 2 22 xyz x ++= ≥ 9, 0 . 12 1 2 2 2 2 2 SS S 9 9 S y y I ydS ydS dxdy dxdy x y x y =+= + − − − − ∫∫ ∫∫ ∫∫ ∫∫ 2 3 2 2 0 2 9 2 cos 2.2. 9 9 4 r d dr r π π− π = φφ = = π − ∫ ∫ . Q8: Find ( ) S x z dxdy + ∫∫ with S is the above of surface 2 2 z x yz =+ ≤ 9, 9 . ( ) 2 2 2 2 :99 9 Dx y I x x y dxdy + ≤ = ++ ∫∫ Set 0 2 3 cos 0 1 sin 3 x r r y r J r  ≤φ≤ π  = φ    → ≤≤  = φ   = ( ) 2 1 2 0 0 I d r r rdr 3 cos 3 π = φ φ+ ∫ ∫ 2 0 3 3 3cos 4 2 d π π = φ+ φ= ∫ . Vậy ( ) 3 2 S x z dxdy π + = ∫∫ Q9: Find the volume of V defined by: x y x y xyz − + + + + + +++ ≤ 4 2 3 24 22 . 1 4 2 1 11 3 2 1 40 7 4 2 130 411 ux y vx y J J t xyz −  =− +   =+ + ⇒ = = = −   = +++ ( , , ) 1 64 1 .8. .2.2.2 3 21 V xyz V V dxdydz J dudvdt J ′ = = = = ∫∫∫ ∫∫∫ . Vậy 64 21 V =