MATH 201 Final, Fall 2009

American University of Beirut MATH 201 Calculus and Analytic Geometry III Fall 2009-2010 Final Exam - solution Exercise 1 a. Find the directional derivative of f(x, y) = x2e−2y at P(1, 0) in the direction of the vector v = 3i + 4j b. The equation 1 − x − y2 − sin(xy) defines y as a differentiable function of x. Find dy/dx at the point P(0, 1). c. Find the points on the surface xy + yz + zx − x − z2 = 0 , where the tangent plane is parallel to the xy-plane. Exercise 2 Find the absolute minimum and maximum values of f(x, y) = x2+xy+y2−3x+3y on the triangular region R cut from the first quadrant by the line x + y = 4. Exercise 3 Use Lagrange Multipliers to find the maximum and the minimum values of f(x, y) = xy subject to the constraint x2 + y2 = 10. Exercise 4 Reverse the order of integration, then evaluate the integral I = ∫ 2 0 ∫ 4−x2 0 xe2y 4 − y dydx solution: to express the integral in the order dxdy, we sketch the region of integration R in the xy-plane. 6 - y y = 4 − x2 R 2 0 4 x I = ∫ 4 0 ∫ √4−y 0 xe2y 4 − y dx dy = ∫ 4 0 e2y 4 − y [x2 2 ]√4−y 0 dy = ∫ 4 0 e2y 2 dy = [e2y 8 ]4 0 = e8 − 1 4 Exercise 5 Let V be the volume of the region D that is bounded below by the xy-plane, above by the paraboloid z = 9 − x2 − y2, and lying outside the cylinder x2 + y2 = 1. solution: 6 - � x y z 1 D 3 a) V = ∫ 2π 0 ∫ 3 1 ∫ 9−r2 0 r dzdrdθ = 2π ∫ 3 1 r(9 − r2) dr = 2π [ 9 2 r2 − r4 4 ]3 1 = 32π b) to express the integral in the order drdzdθ, we sketch the region of integration R in the rz-plane. 6 - r z z = 9 − r2 1 R 3 V = ∫ 2π 0 ∫ 8 0 ∫ √9−z 1 r dr dz dθ ..... 8 Exercise 6 Let V be the volume of the region that is bounded form below by the sphere x2 + y2 + (z − 1)2 = 1 and from above by the cone z = √ x2 + y2. Express V as an iterated triple integral in spherical coordinates, then evaluate the resulting integral (sketch the region of integration). solution: The equation of the sphere x2 +y2 +(z −1)2 = 1 in spherical coordinates is ρ = 2 cos ϕ 6 - � x y z x2 + y2 + (z − 1)2 = 1 z = √ x2 + y2 1 D V = ∫ 2π 0 ∫ π/2 π/4 ∫ 2 cos ϕ 0 ρ2 sin ϕ dρ dϕ dθ = ∫ 2π 0 ∫ π/2 π/4 8 3 cos3 ϕ sin ϕ dϕ dθ = ∫ 2π 0 2 3 [ − cos4 ϕ ]π/2 π/4 dθ = ∫ 2π 0 1 6 dθ = π 3 Exercise 7 Write six different iterated triple integrals for the volume of the tetrahedron cut from the first octant by the plane 6x + 3y + 2z = 6. (do not evaluate any of the integrals) Exercise 8 a. Find the line integral of f(x, y) = (x+y2)/ √ 1 + x2 along the curve C : y = x2/2 from (1, 1/2) to (0, 0). solution: - r(t) = (1 − t)i + (1−t)2 2 j , 0 ≤ t ≤ 1; - v(t) = dr dt = −i − (1 − t)j, and |v(t)| = √ 1 + (1 − t)2 ; - f(t) = (1−t)+ (1−t)4 4 √ 1+(1−t)2 , hence ∫ C f(s)ds = ∫ 1 0 f(t).|v(t)|dt = ∫ 1 0 [ (1 − t) + (1 − t)4 4 ] dt = 11/20 b. Show that the differential form 2 cos ydx + ( 1 y − 2x sin y)dy + (1/z)dz is exact, then evaluate ∫ (1,π/2,2) (0,2,1) 2 cos ydx + (1 y − 2x sin y ) dy + (1/z)dz solution: f(x, y, z) = 2x cos y + ln(yz) + C is a potential function (check it!), hence ∫ (1,π/2,2) (0,2,1) 2 cos ydx + (1 y − 2x sin y ) dy + (1/z)dz = [2x cos y + ln(yz) + C](1,π/2,2) (0,2,1) = ln(π/2) c. Find the counterclockwise circulation of the field F = 2xy3i + 4x2y2j across the curve C in the first quadrant, bounded by the lines y = 0, x = 1 and the curve y = x3. i) direct calculation: cicculation = � C Mdx + Ndy C1 : r1(t) = ti , 0 ≤ t ≤ 1; Mdx + Ndy = 0, and ∫ C1 dx + Ndy = 0 C2 : r2(t) = i + tj , 0 ≤ t ≤ 1; Mdx + Ndy = 4t2dt, and ∫ C2 Mdx + Ndy = ∫ 1 0 4t2dt = 4/3 C3 : r3(t) = (1 − t)i + (1 − t)3j , 0 ≤ t ≤ 1; Mdx + Ndy = −14(1 − t)10dt, and ∫ C3 Mdx + Ndy = ∫ 1 0 −14(1 − t)10dt = −14/11 circulation(F) = � C Mdx + Ndy = 0 + 4/3 − 14/11 = 2/33 ii) Green’s theorem: circulation(F) = ∫ ∫ R (curlF) · k dA (curlF) · k = ∂N ∂x − ∂M ∂y = 8xy2 − 6xy2 = 2xy2 ∫ ∫ R (curlF) · k dA = ∫ 1 0 ∫ x3 0 2xy2 dy dx = ∫ 1 0 2x [y3 3 ]x3 0 dx = 2/3 ∫ 1 0 x10dx = 2/33