Fall 2014 Math 2E Second Midterm Solutions Problem 1 a Use Green s theorem to evaluate cid 82 cid 68 2x2ye x2 from 0 0 to 1 1 to 1 0 to 0 0 and F is the vector eld given by ln 2x 1 xe x2 C F dr where C is the triangle F x y z cid 69 y Let D be the domain inside the triangle C D x y 0 x 1 0 yx We have e x2 2x2e x2 Q x 2x2e x2 P y cid 90 cid 90 cid 90 1 F dr C is negatively oriented Thus Green s theorem gives us cid 90 b Use Green s theorem to nd the area of the domain enclosed by the curve cid 10 cos3 t sin2 t cid 11 e x2 dydx xe x2 dx e x2 dA e 1 1 cid 90 x cid 90 1 1 2 D C 0 0 0 0 t and the segment 1 0 1 0 A x dy cid 90 0 cid 90 C cid 90 0 2 cos4 t sin t dt cid 104 cos5 t cos3 t 2 sin t cos t dt cid 105 cid 28 2 5 0 4 5 xy yz 1 xz 2 1 1 xz cid 90 1 1 x 0 dx cid 29 1 xz 2 2z where x y z Problem 2 a Let F x y z are positive i Find divF ii Show that F is conservative divF 2yz2 1 xz 3 0 2x2y 1 xz 3 2 2 2y x2 z2 1 xy 3 curlF cid 126 0 x 1 xz 2 y 1 xz 2xyz x 1 xz 2 y 1 xz 2xyz 1 xz 3 z 1 xz 2 1 xz 3 z 1 xz 2 1 iii Find a scalar function f such that F f f veri es f x f y f z yz 1 xz 2 1 1 xz xy 1 xz 2 2z f x y z y 1 xz h y z f y 1 1 xz h y We integrate the rst equation with respect to x and get We take the partial derivative of this expression with respect to y and get The second equation gives us then h y f has the form f x y z y 1 xz respect to z and get 0 and thus h y z h z h z We take the partial derivative with f z xy 1 xz 2 h z The third equation gives us h z f x y z y z2 1 xz 2z We choose h z z2 We then have iv Evaluate cid 82 C F dr where C is the line segment from 0 1 3 to 1 2 2 2 The fundamental theorem for line integrals gives us cid 90 C F dr f 2 2 f 0 1 3 5 1 2 b i Let f be a scalar function whose partial derivatives exist on an open domain D of R3 Let F be a vector eld on R3 whose components admit partial derivatives Prove that on D div f F f divF F f 2 Let F P cid 126 i Q cid 126 j R cid 126 k Thanks to the product rule we get div f F f P f Q x cid 18 P f x y Q y cid 19 z f R cid 18 f x R z cid 19 P Q f y f z R f divF f F ii Let f be a differentiable function on 0 Let cid 126 r x cid 126 i y cid 126 j z cid 126 k and r cid 126 r Use the previous identity to nd div f r cid 126 r Do not forget to use the chain rule when computing the gradient of f We apply the previous formula with F cid 126 r and f f r We have divF x x y y z z 3 and by the chain rule f cid 126 i cid 126 j f y f x f r cid 126 k f z r y cid 32 r x f cid 48 r cid 126 i f r cid 126 j f r r z cid 126 k x cid 112 x2 y2 z2 cid 126 i Therefore y cid 112 x2 y2 z2 cid 126 j z cid 112 x2 y2 z2 cid 126 k f cid 48 r r cid 126 r cid 33 div f r cid 126 r f r div cid 126 r f cid 126 r 3 f r Problem 3 Let S be the part of the cone y y 2 z 0 and z x 1 3 f cid 48 r r cid 126 r cid 126 r 3 f r r f cid 48 r x2 z2 between the planes y 0 a Give two parametrizations of the surface S one seeing S as the graph of a function of x and z and one using spherical or cylindrical coordinates If we see S as the graph of a function of x and z where x z D and D is the projection of S onto the xz plane D is the part of the disk x2 z2 4 between the lines z 0 and z x x x y z z x2 z2 1 3 3 Using cylindrical coordinates we get the parametrization x r cos y r z r sin 1 3 The line z 0 in polar coordinates corresponds to 0 The line z x corresponds to r sin r cos and thus tan that 1 3 1 3 is 6 Thus D in polar coordinates is expressed as D r cos r sin 0 r 2 0 6 b Give an equation of the tangent plane to S at point The two tangent vectors and their cross product are given by 1 x x2 z2 0 rx rz 2 5 1 1 5 x x2 y2 1 z x2 y2 0 z x2 z2 1 2 1 1 1 1 5 5 rx rz 2 5 1 1 5 2 5 1 5 rx rz 2 5 1 5 Therefore at point 1 a normal vector to the tangent plane is given by An equation to the tangent plane is thus 2 x 2 5 5 y 1 1 5 z 1 5 0 c Find the area of S We have thanks to the previous question cid 115 rx rz x2 x2 z2 1 x2 z2 2 z2 4 Therefore by swithing to polar coordinates in the double integral over D we get cid 104 r2 cid 105 2 0 6 2 3 r d dr enclosed by the cylinder y2 z2 4 and the planes x 1 and x 3 S z xy2 dS where S is the boundary of the region S consists of 3 surfaces S1 the lateral surface of the cylinder S2 the disk on the 0 0 S D cid 90 cid 90 dS 2 dA 2 cid 90 2 …