Math 244 Spring 2017 (Practice) Final 5/11/2017 Time Limit: 2 hours Name: No calculators or notes are allowed. One side of each sheet is blank and may be used as scratch paper. heck your answers whenever possible. Show your work clearly. Grade Table (for instructor use only) Question Points Score 1 20 2 20 3 20 4 20 5 20 6 20 7 20 8 20 9 20 10 20 Total: 200
Math 244 (Practice) Final - Page 2 of 14 5/11/2017 onverting between coordinate systems ρ = x 2 + y 2 + z 2 r = x 2 + y 2 = ρ sin φ x = r cos θ y = r sin θ z = ρ cos φ Integrals in cylindrical and spherical coordinates f(ρ, φ, θ) dv = f(ρ, φ, θ) ρ 2 sin φ dρ dφ dθ. f(r, θ, z) dv = f(r, θ, z) r dr dθ dz. omputing work or circulation is curve parametrized by r(t) where a t b. irculation or work along (in the direction of increasing t) b = F T ds = M dx + N dy = F d r = b t=a t=a F d r dt dt. Substitution in double integrals f(x, y) dx dy = f(x(u, v), y(u, v)) J(u, v) du dv, where S J(u, v) = and S is the region in the uv-plane that corresponds to. url Suppose F = M, N, P. The curl of F is F î ĵ k =. x y z M N P omputing flux (in the plane) is a closed loop, traversed counterclockwise. F = M, N. Outward flux across = F n ds = M dy N dx. x u y u x v y v,
Math 244 (Practice) Final - Page 3 of 14 5/11/2017 Green s Theorem is a closed loop in the xy-plane, traversed counterclockwise, and enclosing a region. F = M, N, 0. Outward flux across = F n ds = F da. ounterclockwise circulation around = Stokes Theorem F T ds = ( F ) k da. is a closed loop in 3-dimensional space, and it forms the boundary of a surface 1 S. F is a vector field. F T ds = ( F ) n dσ, where the normal vectors to S are chosen using the right-hand rule, and the direction in which we compute the circulation around. Divergence Theorem S is a closed surface 1, enclosing a 3-dimensional region and F is a vector field. Outward flux across S = F n dσ = F dv. S S 1 Legalese: We assume S is piecewise-smooth and oriented.
Math 244 (Practice) Final - Page 4 of 14 5/11/2017 1. (a) (10 points) Give a parametrization for the part of the plane z = x + 3 that is inside the cylinder x 2 + y 2 = 1. Solution: Using cylindrical coordinates: r(θ, r) = r cos θ, r sin θ, r cos θ + 3, 0 θ 2π, 0 r 1. (b) (10 points) Give a parametrization for the part of the surface of the sphere x 2 + y 2 + z 2 = 1 that is inside the cone z = 3(x 2 + y 2 ). Solution: Some basic trig reveals that the angle made by the surface of the cone with the z-axis is π/6. Using spherical coordinates: r(θ, φ) = sin φ cos θ, sin φ sin θ, cos φ, 0 θ 2π, 0 φ π/6.
Math 244 (Practice) Final - Page 5 of 14 5/11/2017 2. (20 points) ompute the volume of the region in the first octant that is inside the sphere x 2 + y 2 + z 2 = 9 and below the cone z = x 2 + y 2. Solution: V = π/2 π/2 = 9π 2 θ=0 φ=π/4 3 ρ=0 ρ 2 sin φ dρ dφ dθ [ cos φ]π/2 φ=π/4 = 9π 2 2.
Math 244 (Practice) Final - Page 6 of 14 5/11/2017 3. onsider the integral 2 (y+4)/2 0 y/2 y 3 (2x y)e (2x y)2 dx dy. (a) (3 points) Sketch the region of integration in the xy-plane. Solution: 3 y 2 1 1 1 2 3 4 5 1 x (b) (7 points) Letting x = u + v/2 and y = v, sketch the region in the uv-plane that corresponds to the above region in the xy-plane via this coordinate transformation. Solution: Solve for u and v in terms of x and y to get u = x y/2 and v = y. The corresponding region in the uv-plane is then: 3 v 2 1 u 1 1 2 3 4 5 1
Math 244 (Practice) Final - Page 7 of 14 5/11/2017 (c) (5 points) ewrite the integral entirely in terms of u and v. Solution: J(u, v) = 1, so the integral becomes 2 2 u=0 v=0 2uv 3 e (2u)2 dv du. (d) (5 points) Evaluate the integral. Solution: 2 u=0 2 v=0 2 2uv 3 e (2u)2 dv du. = 4 = u=0 [e 4u2] 2 0 2ue 4u2 du = e 16 1
Math 244 (Practice) Final - Page 8 of 14 5/11/2017 4. (20 points) Find the outward flux of F = x 2 + y 2, 2xy across the boundary of the square whose vertices are (0, 0), (1, 0), (1, 1), and (0, 1). Solution: F = 4x, and by Green s Theorem, if is the region enclosed by square, then the outward flux is F da = 1 y=0 1 x=0 4x dx dy = 2.
Math 244 (Practice) Final - Page 9 of 14 5/11/2017 5. (20 points) Let f(x, y, z) = x2 z. Find S f dσ where S is the part of the paraboloid z = x2 +y 2 between z = 2 and z = 6. Solution: We parametrize the surface: r(r, θ) = r cos θ, r sin θ, r 2, 0 θ 2π, 2 r 6. Then r r r θ = 2r 2 cos θ, 2r 2 sin θ, r, whose magnitude is r 1 + 4r 2. So the surface integral becomes 2π 6 θ=0 r= 2 r 2 cos 2 θ r 2π 6 1 + 4r r 2 dr dθ = cos 2 θr 1 + 4r 2 θ=0 r= 2 dr dθ 2 2π [ ] (1 + 4r = cos 2 2 ) 3/2 6 θ dθ 12 2 = π θ=0 125 27 12 = 49π 6.
Math 244 (Practice) Final - Page 10 of 14 5/11/2017 6. (20 points) Let F = y 2 cos x, 2y sin x sin y. Let be the straight line path that goes from (0, 0) to (π, π). Find F T ds. Solution: This is in fact a conservative field, and f(x, y) = y 2 sin x + cos y is a potential function for it. So the work integral is just f(π, π) f(0, 0) = 1 1 = 2.
Math 244 (Practice) Final - Page 11 of 14 5/11/2017 7. Let F = 3y, 3x z 2, (2yz + 3z 2 ). Let be the path where 1 t 2. r(t) = t 2, 2 t, 3, (a) (10 points) Find the work done by F on this path by evaluating the line integral directly. Solution: F T ds = = = = 6 2 t=1 2 t=1 F d r dt dt 6 3t, 3t 2 9, (12 6t + 27) 2t, 1, 0 dt 9t 2 + 12t + 9 dt (b) (10 points) Find the work done by F on this path using a potential function for F. Solution: f(x, y, z) = 3xy z 2 y z 3 is a potential function. r(1) = 1, 1, 3 and r2 = 4, 0, 3. So the work done is f(4, 0, 3) f(1, 1, 3) = 6.
Math 244 (Practice) Final - Page 12 of 14 5/11/2017 8. (20 points) Let S be the part of the cone z = 2 x 2 + y 2 where 0 z 2. ompute the inward flux of F = x, y, 2z across S (normal vectors point towards the z-axis). Solution: Parametrize the surface as: r(r, θ) = r cos θ, r sin θ, 2r, 0 θ 2π, 0 r 1. Sketching a picture reveals that the vector r r r θ points towards the z-axis. So we need to evaluate 2π 1 θ=0 r=0 F ( r r r θ ) dr dθ = = 2π 1 θ=0 r=0 2π 1 θ=0 = 4π. r=0 r cos θ, r sin θ, 4r 2r cos θ, 2r sin θ, r dr dθ 6r 2 dr dθ
Math 244 (Practice) Final - Page 13 of 14 5/11/2017 9. (20 points) Let S be the part of the surface of the sphere x 2 + y 2 + z 2 = 16 that lies below the plane z = 2. Use Stokes Theorem to find ( F ) n dσ, S where F = yz, 3xz, z 2 and the normal vectors to S point away from the origin. Solution: Let S 1 be the part of the plane z = 2 inside the sphere. S 1 and S have a common boundary, which is the circle of radius 2 3 centered around the z-axis and in the plane z = 2. By Stokes Theorem, ( F ) n dσ = F T ds, S if the integral on the right is computed via a parametrization that traverses in a clockwise direction when viewed from above. Also by Stokes Theorem, this equals S 1 ( F ) ( k) dσ, where we choose k instead of k for the normal vector using the right hand rule and the direction in which we are going around. F = 3x, y, 2z, and so the integral becomes: S 1 2z dσ, which because z = 2 on S 1 is just S 1 4 dσ, in other words, 4 times the area of S 1, which is 48π.
Math 244 (Practice) Final - Page 14 of 14 5/11/2017 10. (20 points) Let F = 3y 2 z 3, 9x 2 yz 2, 4xy. Use the divergence theorem to compute the outward flux of F across the surface of the cube 1 x, y, z 1. Solution: The divergence of F = F = 9x 2 z 2. By the divergence theorem, the outward flux is 1 1 1 9x 2 z 2 dz dy dz = 8. x= 1 y= 1 z= 1