Fall 2014 Math 2E First Midterm Problem 1 a Change from rectangular to spherical coordinates the point 3 1 2 3 Solutions cid 113 z 3 2 1 2 2 3 4 3 2 cos cid 113 x2 y2 z2 We have and Thus Moreover 6 3 2 16 4 x 3 4 1 2 3 2 cos sin 0 Hence 6 y and sin sin b Sketch the vector eld F x y y cid 126 i x cid 126 j We have F cid 112 x2 y2 and thus the length of the vectors are increasing the farer we are from the origin Let r x cid 126 i y cid 126 j We have F r 0 Therefore at each point P x y F is orthogonal to r We then get the following drawing c Find a differentiable function R R such that the vector eld de ned by is conservative Then evaluate the line integral 4y2 1 positively oriented F d cid 126 r where C is the ellipse x2 F x y 2xy 1 x2 cid 126 i x cid 126 j cid 90 C 1 Let P x y 2xy 1 x2 and Q x y x We want F to be conservative that is Q x P y cid 48 x 2x 1 x2 cid 90 cid 90 cid 90 and Thus we choose x ln 1 x2 C is a closed curve so its nal and initial points coincide Therefore by the fundamental theorem for line integrals since F is conser vative the line integral of F along C is 0 Problem 2 Evaluate the integral tered at the origin and with radius 2 and outside the cone z2 3 x2 y2 xyz dV where E lies inside the sphere cen E The equation of the sphere in spherical coordinates is 2 The equation of the cone in spherical coordinates is 2 cos2 3 2 sin2 cos2 2 sin2 sin2 that is and thus and 6 6 Therefore E in spherical coordinates is given by tan2 1 3 The integral is therefore cid 90 cid 90 cid 90 xyz dV E E 0 2 6 cid 90 2 cid 90 6 cid 18 cid 90 2 cid 90 2 cid 19 cid 18 cid 90 2 6 0 0 5 d 0 0 6 0 2 sin cos sin sin cos 2 sin d d d cid 19 cid 18 cid 90 6 6 cid 19 cos sin d sin3 cos d 0 Problem 3 Find the volume of the solid that lies under the surface 1 x2 y2 2 cid 112 x2 y2 y z and above the region D in the xy plane given by The volume of the solid is given by D x y x x2 y2 1 cid 90 cid 90 1 x2 y2 2 cid 112 x2 y2 y D V dA 2 D is the region above the x axis because we want the volume above the xy plane and thus y 0 if z 0 outside the circle r cos and inside the circle r 1 Therefore we can divide D into D1 the region in the rst quadrant given by D1 r cos r 1 0 2 and D2 the region in the second quadrant given by D2 r 0 r 1 2 Thus the integral becomes sin 1 r2 2 r d dr 2 cid 90 cid 90 1 cid 20 cid 90 cid 90 1 2 2 0 2 cid 21 1 sin 1 r2 2 r d dr 1 sin cid 90 2 1 r2 2 cid 20 d 0 1 2 sin d 0 1 2 cos 2 1 2 cid 90 2 cid 90 2 0 cid 90 1 cid 20 0 sin 1 cos2 cid 21 1 cos sin 1 r2 sin 2 arctan cos cos 2 0 d cos d cid 21 2 V 1 4 1 4 1 4 1 2 8 1 2 16 Problem 4 Evaluate the integral cid 90 cid 90 y 2x y 2x dA e R where R is the trapezoidal region with vertices 0 2 1 0 2 0 and 0 4 R is a parallelogram bounded by the lines R1 y 2x 2 R2 y 0 3 R3 y 2x 4 R4 x 0 4 3 2 1 0 We choose the following change of variables u y 2x v y 2x 1 2 x y v u 4 v u 2 R is the image of the region S given by S1 v 2 S2 v u S3 v 4 S4 v u 4 2 2 4 S is then the region given by S u v v u v 2 v 4 4 2 4 We compute the Jacobian of the transformation 1 2 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 1 4 1 4 dudv cid 21 4 cid 90 4 1 2 1 4 2 cid 12 cid 12 cid 12 cid 12 cid 12 cid 12 1 4 cid 105 v cid 104 ve u v v x y u v cid 90 4 cid 90 v 2 1 4 v v 1 e u 4 cid 20 v2 e e 1 e e 1 3 2 2 2 Thus cid 90 cid 90 R y 2x y 2x dA e cid 90 4 2 1 4 dv v e e 1 dv Problem 5 Evaluate the following line integrals a xz3 sin y ds where C is given by the parametric equations x 3 cos t y t z 3 sin t xz3 sin y ds 3 cos t 3 sin t 3 sin t 3 cos2 t 1 3 sin2 t dt x cid 48 t y cid 48 t 1 3 sin t cid 90 2 cid 90 2 0 18 0 sin4 t cos t dt 18 We have And thus cid 90 C cid 90 b 0 0 to 1 2 C yex xy2 dx ex x2y y dy where C consists of the line segment from Let F yex xy2 cid 126 i ex x2y y cid 126 j We check if F is conservative cid 90 C 0 t 2 3 cos t z cid 48 t cid 112 cid 34 cid 35 2 sin5 t 5 0 18 5 F is then conservative We want to nd f such that F f f veri es Q x ex 2xy ex 2xy P y f x f y yex xy2 ex x2y y f x y yex h y x2y2 2 5 We integrate the rst equation with respect to x and get We take the partial derivative with respect to y of this expression and get The second equation gives us h cid 48 y y and so we choose h y Therefore y2 2 ex x2y h cid 48 y f y f x y yex x2y2 2 y2 2 cid 90 C We apply the fundamental theorem for line integrals and get F dr f 2 1 f 0 0 e2 2 e2 1 2 5 2 …