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Spring 2021 MTH301: Calculus II Assignment No. 1 (Lectures # 11 to 16) Total Marks: 15 Due Date: 11-06- 2021 Question No.1 Marks: 5 Find volume bounded by the planes 𝑥 = 0, 𝑦 = 0, 𝑧 = 0 and 2𝑥 + 𝑦 + 𝑧 = 6 0 ≤ 𝑥 ≤ 2,0 ≤ 𝑦 ≤ 6 − 2𝑥 Solution The Equation of the plane is 2𝑥 + 𝑦 + 𝑧 = 6 can be rechanged as Z = 6-y2x Representing the integral as integrated integral (see next page Please) v 2 62 x 0 0 2 62 x v dx v dx 0 62 x 0 0 2 v 0 2 v 0 2 v 0 6 y 2 x dy dz 0 dy (6 y 2 x ) 0 2 v dx dxdydz (6 y 2 x)dy (6 2 x) 2 2 x(6 2 x) dx 6(6 2 x) 2 (36 24 x 4 x 2 ) (36 12 x ) (12 x 4 x) dx 2 (36 24 x 4 x 2 ) 2 (36 24 x 4 x ) dx 2 (72 48 x 8 x 2 ) (36 24 x 4 x 2 ) v dx 2 0 2 2 v 1 72 48 x 8 x 2 36 24 x 4 x 2 ) dx 20 2 1 v 36 24 4 x 2 ) dx 20 1 24 x 2 4 x3 v |02 36 x 2 2 3 1 2 24 x 2 4 x 3 v |0 36 x 2 2 3 1 2 4 x3 2 v |0 36 x 12 x 2 3 1 4(2)3 v 36(2) 12(2) 2 2 3 1 4(8) v 72 12(4) 2 3 1 32 v 24 2 3 1 72 32 v 2 3 1 104 v 2 3 52 v 3 Question No.1 Marks: 10 Let 𝑓(𝑥, 𝑦, 𝑧) = 𝑥𝑦𝑒 𝑥 2 +𝑧 2 −5 . Calculate the gradient of 𝑓 at the point (1,3,−2) and calculate the directional derivative 𝐷𝒖 𝑓 at the point (1,3,−2) in the direction of the vector v=(3,−1,4). Solution f ( x, y, z) xye x 2 z 2 5 Given points (1,3,-2) f f f 2 2 f ( x, y , z ) , , e x z 5 x x x 2 2 f ( y 2 x 2 y )e x z 5 x 2 2 f (3 2(1) 2 3)e(1) ( 2) 5 x f (3 2(3))e1 45 x f (9)e0 x f 9 x Value put in 1 f ( x, y, z ) (9,1, 12) Du f ( x, y, z ) f ( x, y, z ) u......(a ) Unit vector v .........(b) |v| v (3, 1, 4) u | v | (3) 2 (1) 2 (4) 2 | v | 9 12 16 | v | 26 Putting in b u( 3 1 4 ) 26 26 26 Putting in a 3 1 4 ) 26 26 26 27 1 48 Du f ( x, y, z ) ( ) 26 26 26 27 1 48 Du f ( x, y, z ) ( ) 26 22 Du f ( x, y, z ) ( ) 26 Du f ( x, y, z ) (9,1, 12) (