MA 242 Test 4 Solutions 1 8 points Given f x y xy 2x find the gradient vector field of f Solution f fx i fy j y 2 i x j x2 y zdz where the curve C is given by the parametric equa Z 2 15 points Evaluate the line integral C tions x t3 y t z t2 over the interval 0 t 1 Solution Z Z 1 Z t3 2 t t2 2tdt x2 y zdz C 0 1 2t9 dt 0 t10 1 1 5 0 5 3 20 points Given F x y x3 y 4 i x4 y 3 j where C is the curve defined by r t t i 1 t3 j on the interval 0 t 1 a Find a function f such that F f Z b Use part a to evaluate the line integral F d r along the given curve C C Use the Fundamental Theorem for Line Integrals Solution 1 a fx x y x3 y 4 implies f x y x4 y 4 g y and fy x y x4 y 3 g 0 y 4 But fy x y x4 y 3 so g 0 y 0 which means that g y K We can take K 0 so 1 f x y x4 y 4 4 b The initial point of C is r 0 0 1 and the terminal point is r 1 1 2 so Z F d r f 1 2 f 0 1 4 0 4 C 4 15 points Given F x y z yz i xz j xy k 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 i x yz b Since F component functions a F j k y z x x i y y j z z k 0 xz xy 0 and F is defined on all of R3 and the partial derivatives of the are continuous so F is conservative c F 0 0 0 0 5 7 Z points Given a conservative vector field F P i Q j Use Green s Theorem to prove that P dx Q dy 0 for any positively oriented piecewise smooth simple closed curve C C Solution P Q Since F P i Q j is conservative y x Then by Green s Theorem ZZ ZZ Z ZZ Q Q Q P dA dA 0dA 0 P dx Q dy y x D x D C D x 6 15 points Set up but do not compute the double integral needed to find the area of the part of the surface z cos x2 y 2 that lies inside the cylinder x2 y 2 1 Be sure to use polar coordinates Recall that the surface area is found by computing the following double integral ZZ s 1 A S D dz 2 dz 2 dA dx dy Solution z f x y cos x2 y 2 fx 2x sin x2 y 2 fy 2y sin x2 y 2 ZZ p So A S 1 2x sin x2 y 2 2 2y sin x2 y 2 2 dA D ZZ q 1 4x2 sin2 x2 y 2 4y 2 sin2 x2 y 2 dA D ZZ q 1 4 x2 y 2 sin2 x2 y 2 dA D D is the circle x2 y 2 1 in the xy plane So in polar coordinates D r 0 r 1 0 2 Then the integral we must calculate in order to find the surface area is Z 2 Z A S 0 0 1 q 1 4r2 sin2 r2 rdrd