Q.No: 1 (a) King Saud University Department of Mathematics M-203[Final Examination] (Differential and Integral Calculus) (First-Semester 1428/29) Max.Marks:50 Marking Scheme: Q.No:1[3+4+5+4], Q.No2:[4+4+4+4], Q.No:3[4+4+4+6] Discuss the convergence of the sequence { ln( 2n) ln( 2n 1)}. 2n Lim[ln( 2n) ln( 2n 1)] Lim ln Limln( 1) 0 . n n 2 n 1 n Solution: (b) 1 Determine whether the series n 1 n n 2n 1 2 is absolutely convergent, Conditional ly convergent, or divergent. 1 1 and is divergent 2n 2n 1 2n n n is divergent. 1 2 2n 1 n 1 Solution: Since n 2 Let us now try Alternating series test: n Hence (c) n 0, 2n 1 x (ii) f ( x) , f / ( x) 0 for x 1. 2 2x 1 is conditionally convergent. 1n 2n 2n 1 n 1 (i) Lim 2 Find the interval of convergence and radius of convergence of the power Series 1 n 1 n 1 ( x 1) n . n n3 ( x 1) n1 n3n 1 1 Lim x 1 x 1 n 1 n n (n 1)3 n 3 3 ( x 1) 1 x 1 1 x 1 3 For convergence 3 3 x 1 3 2 x 4 . Check convergence at x 2 , 1 1 n 1 2n n n 1 ( 2 1 ) 1 ( 3 ) = (Divergent) n n n3 n3 n 1 n 1 n 1 n Check convergence at x 4 n n 3 n 1 n 1 n 1 ( 4 1 ) 1 1 (Convergent) n n n n3 n3 n 1 Therefore Interval of convergence is (2,4]. Radius of Convergence =3. Solution: (d) Lim Find Maclaurin’s series for the function f ( x) 1 1 x 2 . Time:3hrs f ( x) Solution: 1 f (0) 1 1 x 2 2 f ( x) 1 x f / ( x) (2)(1 x ) 3 (1) f / (0) 2 f // ( x) (2)(3)(1 x ) 4 f // (0) 6 f /// ( x) (2)(3)(4)(1 x ) 5 f /// (0) 24 f iv ( x) (2)(3)(4)( 5)(1 x ) 6 f /// (0) 120 and so on…………. 6 2 24 3 120 5 x x x 2! 3! 4! f ( x) 1 2 x Therefore Q.No: 2 (a) 2 0 Solution: 2 2 Cos( y 0 2 2 y )dydx x 0 Cos( y )dxdy 2 2 Cos( y Sketch the region of integration and evaluate the integral )dydx . x 2 yCos( y 0 2 2 )dy 0 1 1 [ Sin Sin (0)] . 2 2 2 (b) Find the area of the surface of the portion of paraboloid z x 2 y 2 that Is cut off by the plane z 1. Solution: Surface Area= 1 ( f x ) 2 ( f y ) 2 dA. R Here z x y f ( x, y) f x ( x, y ) 2 x & f y ( x, y ) 2 y . 2 2 1 (2 x) 2 (2 y ) 2 dA. 1 4( x 2 y 2 )dA . R R 2 0 (c ) 1 6 0 5 1 1 1 4r rdrd 2 1 4r 2 12 0 2 5 1. A lamina has the shape of the region bounded by y a 2 x 2 and the density is directly proportional to the distance from x-axis. Find the mass of the lamina. Solution: Mass of a lamina= a a2 x2 a 0 ( x, y)dA R a k kydydx a 2 x 2 dx 2 a a k a 2 x3 2 a x 2 dx k a 2 x 3 0 2 0 2 ka3 . 3 4 (d) Evaluate the integral 0 Solution: 2 2 16 x 2 32 x y 0 x 2 y 2 z 2 dzdydx . x2 y2 On converting this integral into spherical we get: 2 4 32 2 4 32 Sinddd = Sinddd 0 0 0 0 0 2 2 0 1 4 4 2 Sindd 2 0 4 4 0 2 1 2 256 Cos 0 4 d 256 1 d 37.49 2 0 0 On converting the given integral into Cylindrical we get: 4 0 Q.No: 3 (a) 2 2 16 x 2 32 x y 0 x y z dzdydx 2 2 2 x2 y2 0 Show that the integral ye xy 32 r2 2 4 0 r 2 z 2 rdzdrd r dx ( xe xy 2 y )dy is independent of path C by finding the potential function. Solution: We need to show F ( x, y ) f x ( x, y) i f y ( x, y) j F ( x, y) ye xy i ( xe xy 2 y) j f x ( x, y) i f y ( x, y) j f x ( x, y ) ye xy ………………………..(1) f y ( x, y) xe 2 y …………………….(2) Integrating (1) with respect to x and (2) with respect to y , we get 1 f ( x, y ) ye xy g ( y )......................(3) y 1 f ( x, y ) xe xy y 2 h( y )...............(4) x xy Comparing (3) and (4), we get g ( y ) y 2 h( y ) Therefore f ( x, y) e xy y 2 c which is a potential function. (b) Evaluate the integral y 2 dx 3xydy , where C is the boundary of the region C that lies inside the circle x 2 y 2 4 and outside the circle x 2 y 2 1 in upper half plane. Solution: We apply Green’s theorem and we have (3xy) ( y 2 ) 2 y dx 3 xydy R dx dy dA R (3 y 2 y)dA R ydA C 2 r3 7 14 rSin ( )rdrd Sin ( )d Sin ( )d 3 1 30 3 0 1 0 (c ) Use Stokes’s theorem to evaluate 2 F dr , where C is the boundary of the C Portion of z 4 x y above the xy-plane oriented upward and 2 2 F ( x, y , z ) ( x 2 e x y ) i y2 1 j z3 k . Solution: i curl F x 2 x x e y We find k z z 3 j y y2 1 i (0) j (0) k (1) ( M 1 i N1 j P1 k ) 2 2 z 4 x y g ( x, y) g x 2 x, g y 2 y curl F n dS ( M 1 g x N1 g x P1 )dA (0 0 1)dA dA R R S 2 2 2 2 0 0 0 0 (d) dA rdrd 4 Verify the divergence theorem by evaluating both the surface integral and the Triple integral for the function F ( x, y, z ) x i y j z k , and the surface S which is the solid bounded by the graphs of z x 2 y 2 and z 4. Solution: M N z F n dS dV .......(1) y z x Q Here we have to prove S L.H.S=R.H.S…………………………………………(1) First we calculate L.H.S. of (1) i.e. Q M N z dA. (1 1 1)dV y z x Q dV Q Now we evaluate the L.H.S. of (1) i.e. 2 2 4 0 0 r2 rdzdrd 8 F n dS S The surface S is divided into two parts namely: S1: Plane z 4 g1 ( x, y) this portion has upper normal vector S2: Paraboloid z x 2 y 2 g 2 ( x, y) this portion has lower normal. Hence, F n dS = F n dS1 F n dS 2 S1 S S2 ( Mg 1x Ng1 y P)dA Mg 2 x Ng 2 y P) dA Rxy Rxy ( Mg 1x Ng1 y P)dA Mg 2 x Ng 2 y P) dA Rxy Rxy ( Mg 1x Ng1 y P)dA Mg 2 x Ng 2 y P) dA Rxy Rxy ( x(0) ( y)(0) z )dA 2 x 2 2 y 2 z ) dA Rxy Rxy zdA (2 x 2 2 y 2 x 2 y 2 )dA Rxy Rxy 2 2 0 16 0 rdrd ( x 2 3 y 2 ) dA Rxy 2 2 0 0 (r 2 Cos 2 3r 2 Sin 2 )rdrd 16 8 8
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