MTH 234 Solutions to Exam 2 November 23, 2015Name:Section: Recitation Instructor:READ THE FOLLOWING INSTRUCTIONS.• Do not open your exam until told to do so.• No calculators, cell phones or any other electronic devices can be used on this exam .• Clear your desk of everything excepts pens, pencils and erasers.• If you need scratch paper, use the back of the previous page.• Without fully opening the exam, check that you have pages 1 through 9.• Fill in your name, etc. on this first page.• Show all your work. Write your answers clearl y ! Include enough steps for the grader tobe able to follow your work. Don’t s k ip limits or equal signs, etc. Include words to clarifyyour reasoning.• Do first all of the problems you know how to do immediately. Do not spend too muchtime on any particular problem. Return to difficult problems later.• If you have any questions please raise your hand and a proctor wi ll come to you.• There is no talking allowed during the exam.• You will be given exactly 90 minutes for this exam.I have read and understand the above instructions:. SIGNATUREPage 1 of 9MTH 234 Solutions to Exam 2 November 23, 2015Multiple Choice. Circle the best answer. No work needed. No partial credit availab le.1. (5 points) Parameterize the part of the plane 6x + 3y + z = 12 that li e s in the first octant.A. r(s, t) = h6s, 3t, 12 −s − ti with s ∈ [0, 2] and t ∈ [0, 4 − 2s] .B. r(s, t) = hs, t, 1 2 − 6s − 3ti, with s ∈ [0, 2], and t ∈ [0, 4 − 2s].C. r(s, t) = hs, t, 12 − s − ti with s ∈ [0, 2] and t ∈ [0, 4].D. r(s, t) = hs, t, 12 − 6s − 3t i, with s ∈ [0, 2], and t ∈ [0, 4].E. None of the above.2. (5 points) Which of the following vector field plots could be F = xy i − y2j?B.Extra Work Space.Page 2 of 9MTH 234 Solutions to Exam 2 November 23, 20153. (5 points) Let F =x2, xy, 3z. Which of the following is true?A. curl F = y and div F = 3x + 3B. curl F = y and div F = h2x, x, 3iC. curl F = h0, 0, yi and div F = 3x + 3D. curl F = h2x, x, 3i and d iv F = yE. None of the above.Fill in the Blanks. No work needed. No parti al c re dit available.4. (5 points) Write the spherical equation, ρ = cos φ, in rectangular coordinates.z = x2+ y2+ z25. (10 points) Convert the integralˆ1−1ˆ√1−y20ˆx0(x2+ y2) dz dx dyto an equivalent integral in cylindrica l coordinates.ˆθ2θ1ˆr20ˆz20r3dz dr dθwhereθ1=−π/2θ2= π/2r2= 1z2= r cos θExtra Work Space.Page 3 of 9MTH 234 Solutions to Exam 2 November 23, 2015Standard Response Questions. Show all work to receive credit. Please BOX your final answer.6. (10 points) Let f(x, y, z) = x ex+ yz2.(a) Find ∇f at P0(0, 2 , −3).Solution:fx= (x + 1)ex=⇒ fx(P0) =fy= z2=⇒ fy(P0) = 9fz= 2yz =⇒ fz(P0) = −12Thus(∇f)P0= i + 9 j + −12 k(b) Find the derivative of f at P0in the direction of the vector A = 2 i − j + 2 k.Solution:Let u =A|A|=13(2 i − j + 2 k). ThenDuf = ∇f · u =13( i + 9 j + −12 k) · (2 i − j + 2 k) =−3137. (14 points) Let f (x, y) = x3− 12xy + 8y3. Find and classify each critical point of f as a local minimum, alocal max imum, or a saddle point.Solution:i. Find the critical points. Notice t ha t fx= 3x2− 12y and fy= 24y2− 12x. So f has acritical points at P (0, 0) and Q(2, 1).ii. Now let D(x, y) = fxxfyy− f2xy= 144( 2x y − 1). Notice that fxx(Q) = 12 > 0 and thatD(P ) = −144 < 0 and D(Q) = 432 > 0It follows that f has a local minimum at Q and a saddle point at P .Page 4 of 9MTH 234 Solutions to Exam 2 November 23, 20158. (12 points) Sketch the region o f integration for the integral below and evaluate the integral by reversing theorder of integration.ˆ10ˆ1ysin(πx2) dx dy11y = xSolution:=ˆ10ˆx0sin(πx2) dy dx=ˆ10x sin (πx2) dx=−12πcosπx210=1π9. (12 points) Evaluate the integral below.ˆ20ˆ√4−x2−√4−x21p1 + x2+ y2dy dxSolution:We switch to polar coordinates=ˆπ/2−π/2ˆ201√1 + r2r dr dθ= πˆ201√1 + r2r dr=π2ˆ511√udu = π√u51= π√5 − 1Page 5 of 9MTH 234 Solutions to Exam 2 November 23, 201510. (12 points) Find the volume of the part of the sphere ρ ≤ 2 t ha t lies between the cones φ = π/4 and φ = π/3.Solution:V =ˆ2π0ˆπ/3π/4ˆ20ρ2sin φ dρ dφ dθ= 2πˆπ/3π/4ˆ20ρ2sin φ dρ dφ=16π3ˆπ/3π/4sin φ dφ=−16π3cosπ3− cosπ4=8π3√2 − 111. (12 points) Find the area of the surface z = xy that lies inside the cylinder x2+ y2= 4.Solution:Let r(x, y) = x i + y j + xy k. Thenrx= i + y kry= j + x kSo thatrx× ry= −y i − x j + kand|rx× ry|2= 1 + x2+ y2ThusSA =¨x2+y2≤4|rx× ry| dA=¨x2+y2≤4p1 + x2+ y2dA= 2πˆ20p1 + r2r dr=2π353/2− 1Page 6 of 9MTH 234 Solutions to Exam 2 November 23, 201512. (12 points) Find the work done by the force F = h4y, −2xi alo ng the straight l ine segment from (2, 8) to (1, −2).Solution:Let C be the indicated line segment. Now letr(t) = (2 − t) i + (8 − 10t) j, 0 ≤ t ≤ 1Thendr = (−i − 10 j) dt and F · dr = (8 + 20t) dtit follows thatˆCF · dr =ˆ10(8 + 20t) dt=8t + 10t210= 1813. Let F = (2 + 2xyez) i + x2ezj + x2yezk and answer the questions below.(a) (6 points) Find a function f so t ha t ∇f = F.Solution:Letf (x, y, z) = x2yez+ 2xthen∇f = (2 + 2xyez) i + x2ezj + x2yezk(b) (6 points) Let C be any path from (2, 1, 0) to (3, 0, 1). E valuate the integralˆC(2 + 2xyez) dx + x2ezdy + x2yezdz = ISolution:Observe thatdf = (2 + 2xyez) dx + x2ezdy + x2yezdzIt follows thatI =ˆ(3,0,1)(2,1,0)df = f (x, y, z)(3,0,1)(2,1,0)= 6 − 8Page 7 of 9MTH 234 Solutions to Exam 2 November 23, 201514. (14 points) Let E = {(x, y, z) |1 ≤ y ≤ 4, y ≤ z ≤ 4, 0 ≤ x ≤ z}. Rewrite the triple integral below as aniterated integral and evaluate.˚Ezx2+ z2dVSolution:˚Ezx2+ z2dV =ˆ41ˆ4yˆz0zx2+ z2dx dz dyWe try trig substitutio n. So let x = z t a n θ. Then θ(0) = 0, θ(z) = π/4 and dx = z sec2θ dθ.Thus˚Ezx2+ z2dV =ˆ41ˆ4yˆtan−1(z/z)tan−10z2sec2θ dθz2sec2θdz dy=ˆ41ˆ4yˆπ/40dθ dz dy=ˆ41ˆ4yπ4dz dy=π4ˆ41(4 − y) dy=π4(4y − y2/2)41=9π815. (10 points) Use Green’s Theorem to evaluate the line integralˆCF · dr. Here C is the circle x2+ y2= 4positively oriented andF(x, y) = h3y − c os y, x sin yiSolution:Let C be the circle of radius 2 centered at the origin and let R be its interior. Then the area ofR is 4π …