Dr. Z’s Math251 Handout #16.3 [The Fundamental Theorem for Line Integrals]By Doron ZeilbergerProblem Type 16.3a: Determine whether or notF = P (x, y)i + Q(x, y)jis a conservative vector field. If it is, find a function f that F = 5f.Example Problem 16.3a: Determine whether or notF = (ey+ y cos x)i + (xey+ sin x + 2y)jis a conservative vector field. If it is, find a function f that F = 5f.Steps Example1. Compute∂P∂yand∂Q∂x. If they are notthe same, then the vector field is not con-servative. End of story. If they are, thenit is a conservative vector field, and youmust go on.1.∂P∂y=∂∂y(ey+ y cos x) = ey+ cos x .∂Q∂x=∂∂x(xey+sin x+2y) = ey+cos x .Since they are the same, the vector fieldis conservative. We must go on.12. Look for a function f (x, y) such that∂f∂x= P ,∂f∂y= Q .To this end, you integrate P w.r.t. to xf =ZP dx ,getting an answer up to an arbitrary con-stant (from x’s viewpoint), that is a func-tion of y, g(y). You then plug this tenta-tive form of f into the second equationand find what g0(y) is.2. Since∂f∂x= ey+ y cos x ,we get:f =Z(ey+y cos x) dx = xey+y sin x+g(y) .Plugging this to the second equation, weget∂∂y(xey+y sin x+g(y)) = xey+sin x+2y ,which meansxey+ sin x + g0(y) = xey+ sin x + 2ywhich means g0(y) = 2y .3. Integrate the expression that you gotfor g0(y) w.r.t. y, in order to get g, andplug it into f above.3. g(y) =R2y dy = y2(now you don’thave to bother about the C). Going backto f, we getf = xey+ y sin x + y2.Ans.: F is conservative and the potentialfunction f such that F = 5f is f = xey+y sin x + y2.Problem Type 16.3b : (a) Find a function f such that F = 5f and (b) use part (a) to evaluateRCF · dr along the given curve C.F(x, y, z) = P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k ,C : x = x(t) , y = y(t) , z = z(t) ; a ≤ t ≤ b .Example Problem 16.3b: (a) Find a function f such that F = 5f and (b) use part (a) toevaluateRCF · dr along the given curve C.F(x, y, z) = (2xz + y2) i + 2xy j + (x2+ 3z2) k ,2C : x = t2, y = t + 1 , z = 2t − 1 , 0 ≤ t ≤ 1 .Steps Example1. We are looking for a function f(x, y, z)such that fx= P , fy= Q, fz= R. Firstintegrate P w.r.t. to x gettingf =ZP dx = A(x, y, z) + g(y, z) ,where A(x, y, z) is explicit but g(y, z) isyet to be determined. then plug it intofy= Q gettingAy+ gy= Q ,so gy= Q − Ay, and integrating w.r.t. ygivesg(y, z) =Z(Q−Ay) dy = B(y, z)+h(z) ,where B(y, z) is explicit but h(z) is yet tobe determined. Now we havef = A(x, y, z) + B(y, z) + h(z) ,we plug it into fz= R and find h0(z) fromwhich we get h, and plug it back into f .1. fx= 2xz + y2meansf = x2z + xy2+ g(y, z) .fy= 2xy means2xy + gy= 2xy ,so gy= 0 and g(y, z) = h(z). Sof = x2z + xy2+ h(z) .fz= x2+ 3z2meansx2+ h0(z) = x2+ 3z2,so h0(z) = 3z2andh(z) = z3.Going back to f we havef = x2z + xy2+ z3.32. Plug-in t = a, t = b to get r(a) andr(b), and computef(r(b)) − f(r(a)) .2.r(t) = ht2, t + 1, 2t − 1i ,Sor(0) = h0, 1, −1i ,r(1) = h1, 2, 1i .Since our f equals x2z + xy2+ z3, thevalue of the line integral is:f(1, 2, 1) − f(0, 1, −1) =(12·1+1·22+13)−(02·(−1)+0·12+(−1)3) = 7 .Ans.: