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# NCSU MA 242 - C4PracticeTest4Sol

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MA 242 Test 4 Solutions1. [8 points] Given f(x, y) = xy − 2x, find the gradient vector field of f.Solution:∇f = fxˆi + fyˆj = (y − 2)ˆi + xˆj2. [15 points] Evaluate the line integral,ZCx2y√zdz, where the curve C is given by the parametric equa-tions, x = t3, y = t, z = t2over the interval 0 ≤ t ≤ 1.Solution:ZCx2y√zdz =Z10(t3)2(t)√t2(2tdt) =Z102t9dt = [t105]10=153. [20 points] Given−→F (x, y) = x3y4ˆi + x4y3ˆj where C is the curve defined by−→r (t) =√tˆi + (1 + t3)ˆjon the interval 0 ≤ t ≤ 1.(a) Find a function f such that−→F = ∇f.(b) Use part (a) to evaluate the line integral,ZC−→F • d−→r along the given curve C.(Use the Fundamental Theorem for Line Integrals).Solution:(a) fx(x, y) = x3y4implies f(x, y) =14x4y4+ g(y) and fy(x, y) = x4y3+ g0(y).But fy(x, y) = x4y3so g0(y) = 0 which means that g(y) = K. We can take K = 0, sof(x, y) =14x4y4.(b) The initial point of C is−→r (0) = (0, 1) and the terminal point is−→r (1) = (1, 2), soZC−→F • d−→r = f(1, 2) − f (0, 1) = 4 − 0 = 44. [15 points] Given−→F (x, y, z) = yzˆi + xzˆj + xyˆk(a) Compute ∇ ×−→F (the curl of−→F ).(b) Based on part (a), is the vector field conservative? Explain.(c) Compute ∇ •−→F (the divergence of−→F ).Solution:(a) ∇ ×−→F =ˆiˆjˆk∂/∂x ∂/∂y ∂/∂zyz xz xy= (x − x)ˆi − (y − y)ˆj + (z −z)ˆk =−→0(b) Since ∇ ×−→F =−→0 and−→F is defined on all of R3and the partial derivatives of thecomponent functions are continuous so−→F is conservative.(c) ∇ •−→F = 0 + 0 + 0 = 05. [7 points] Given a conservative vector field−→F = Pˆi + Qˆj. Use Green’s Theorem to prove thatZCP dx + Q dy = 0 for any positively oriented, piecewise-smooth, simple closed curve C.Solution:Since−→F = Pˆi + Qˆj is conservative,∂P∂y=∂Q∂xThen by Green’s Theorem,ZCP dx + Q dy =ZZD[∂Q∂x−∂P∂y]dA =ZZD[∂Q∂x−∂Q∂x]dA =ZZD0dA = 06. [15 points] Set up (but do not compute) the double integral needed to find the area of the part of thesurface z = cos(x2+ y2) that lies inside the cylinder x2+ y2= 1. Be sure to use polar coordinates.Recall that the surface area is found by computing the following double integral,A(S) =ZZDs1 + (dzdx)2+ (dzdy)2dASolution:z = f (x, y) = cos(x2+ y2), fx= −2x sin(x2+ y2), fy= −2y sin(x2+ y2)So A(S) =ZZDp1 + (−2x sin(x2+ y2))2+ (−2y sin(x2+ y2))2dA=ZZDq1 + 4x2sin2(x2+ y2) + 4y2sin2(x2+ y2)dA=ZZDq1 + 4(x2+ y2) sin2(x2+ y2)dAD is the circle x2+ y2= 1 in the xy-plane.So in polar coordinates D = {(r, θ)|0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π}Then the integral we must calculate in order to find the surface area isA(S) =Z2π0Z10q1 + 4r2sin2(r2)

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