Question

Q1. Evaluate the line integral f (x2 + y2)dx + 2xydy by two methods a) directly, b) using Green's Theorem, where C consists of the arc of the parabola y = x2 from (0,0) to (2,4) and the line segments from (2,4) to (0,4) and from (0,4) to (0,0). [Answer: 0] Q2. Use Green's Theorem to evaluate the line integral $. F. dr or the work done by the force field F(x, y) = (3y - 4x)i +(4x - y)j when an object moves once counterclockwise around the ellipse 4x2 + y2 = 4. [Answer: 27) Q3. Evaluate the surface integral S), xy ds where o is part of the plane 2x+y +z = 2 that lies in the first octant. [Answer: Q4. Evaluate the surface integral So x ds where o is the surface y = x2 + 4z, 0SX S2,0 SzS2. [Answer: 33/33-17/17 Q5. Let o be the portion of the surface z = 1 – x2 - y2 that lies above the xy-plane. Suppose that o is oriented outward. Evaluate the surface integral SS, Finds where F = xi + yj + zk. [Answer: Зл 2

Answer 1

(1).Solution: Given integral is f(x+y?)dx + 2xydy C:The arc of the parabola y=x- from (0,0) to (2,4) and the line segment from (2,4) to (0,4) and from (0,4) to (0,0). yn (0,4) y=4 (2,4) y=x2 0(0,0) >x (a). Directly (i). Along y = x2 from (0,0) to (2,4) dy = 2xdx (2;9(x2 + y2)dx + 2xydy = }(x2 + x4)dx + 2x(x2)(2xdx ) (0,0) x=0 }(x2 + 5x4)dx x=0 :: [x"dx (= +C n+1 + 5x5 5 Jx=0 3 x=0 8 +32 104 3 (ii). Along the line segment from (2,4) to (0,4) i.e., y = 43 dy = 0 (0,4) *** (x2 + y2)dx + 2xydy = $(x2+1 +16) dx + 0 (2,4) x=2 j(x2 +16)dx x=2 10 +16x 3 Jx=2 8 =[CO: = (0+0) - +32 104 3 (iii). Along the line segment from (0,4) to (0,0). i.e., x = 0 dx = 0 (0,0) (x2 + y2)dx + 2xydy = fo+0=0 (0,4) y=4 Adding (1),(ii) and (iii),we get 104 104 = [(x² + y^)dx + 2xydy +0=0 w . - S(x2 + y2)dx + 2xydy = 0

(b). Using Green's Theorem : Given integral is $(x2 + y2)dx + 2xydy с M=x2 + y2, N = 2xy ОМ ду ON = 0 +2y = 2y, ax = = 2y Green's theorem fMdx +Ndy = SIR aN OM Ox ду dA с [(x2 + y2)dx + 2xydy = SIR (2y – 2y) dA C = IIRO DA = 0 [(x2 + y2)dx + 2xydy = 0 с