Question

2. Determine whether there is a potential function for the vector field V= <yz, xz, xy>. You may use any legitimate method but you must justify your claim. If it there is a potential function, then find it and use it to evaluate the line integral ſ v.dr along the curve r(t) = <V7,4-4,6+1>ifor Osts 4. [10]

Answer 1

2. e o (42) - V = < y2, 2x, XY). VXV = curl ve sopim (my) - (2x), 2) com cry), (2x) - dy = <0,0,0) ::vis conservative vector field. lut f(X,Y,Z) = V (*14,2). = y2 =) F(XYZ) = xyz + g(Y,Z) Fe dx he 07 dy of the = x2 + ogl 7,2) = x2 =) 3g-o gluie) = g(2). atz = xy+g(2) = xy=) 9122)=0 912) = 0 +(X,Y, 2) = XYZ. v.dr = Sot dr se = f(olt)) 110 : +12,0, 5)- +10,-4,1) こ

4. Incitors z = x2+442 ; Z31, V = <y, - * 2,*22) S : 2 = x2+4 y2 ; O%231. 5'(x,y) = < x, y, x2 + 4y2> = <1,0, 2x) <0,1, 8X). Sxx5y =<-2x,-8y, I> ds <-2*, -84, dydx. vids =<4,-xz, xz?>. <-22,- 8y, 1) dydx E [-2xy + 8xy2 + x22]dy dx. [-2xy +8y(x2+442) + x(x2+4y27 x5 + 8x3y 3 + 8x3y + bxy4 + 32xy3 - 2xy] dydx. - dydx 2 at boundary : x2 + 4y2 = 1 VT-x2 Eye VT-x2 2 -15 X 51. Vi [ x5 + 8x3y3 + 8x3y VI-X² 4 + 6X Y + 32xy3 - 2xy] dydx. as y is symmetric, all odd powers of y will be zero. Ilv. 25 - 2

VI-2 2 [ * XS t 6xy] ayax -vi-x2 2 VT-x2/2 dx xsyt GxYS oxys 111-x2/2 ✓ [ *S 11=x2 + 3 x (1-x2)512 7d* xs Vi-y2 + 3x (1-x2) 9/2] ax 7 518) - „'[ * xs andx are oad, (1-x2) are even function.