MATH 228: Calculus III (FALL 216) Sample Problems for FINAL EXAM SOLUTIONS
MATH 228 Page 2 Problem 1. (2pts) Evaluate the line integral C xy dx + (x + y) dy along the parabola y x2 from ( 1, 1) to (2, 4). Let r(t) t, t 2 with 1 t 2. Then ˆ C xy dx + (x + y) dy ˆ 2 1 ˆ 2 1 (t)(t 2 )(1) dt + (t + t 2 )(2t) dt (3t 3 + 2t 2 ) dt ( 3 4 t4 + 2 3 t3 45 4 + 2 9 3 16.25 ) 2 1
MATH 228 Page 3 Problem 2. (2pts) Determine whether or not the vector field F (y + z)i + (x + z)j + (x + y)k is conservative, and if so, find a potential function for it. First, let us compute the curl of F y + z, x + z, x + y : i j k F x y z y + z x + z x + y 1 1, (1 1), 1 1 Since the domain of F is simply connected, we know a potential function exists, i.e. F is conservative. To find the potential function f(x, y, z), note that f x implies f(x, y, z) x(y + z) + g(y, z) where g(y, z) needs to satisfy from which we deduce g(y, z) yz works. Hence, is a potential function, as can easily be checked. g y z and g z y f(x, y, z) xy + xz + yz
MATH 228 Page 4 Problem 3. (2pts) Find the counter-clockwise circulation of F (x y)i + (y x)j along the curve C around the square bounded by x, x 1, y, y 1. We can apply Green s theorem with P x y and Q y x. Note that Q x 1 P y. Hence, ˆ ˆ F dr (x y) dx + (y x) dy C C da D where D is the unit square bounded by C.
MATH 228 Page 5 Problem 4. (2pts) Find the surface area of the lower portion cut from the sphere x 2 + y 2 + z 2 2 by the cone z x 2 + y 2. The radius of the sphere is R 2. The portion inside the cone is determined by φ π. Its surface area 4 is ˆ 2π ˆ π/4 R 2 sin φ dφ dθ 4π ( cos φ) π/4 ( 4π 1 1 ). 2 The surface area of the sphere is 4πR 2 8π. Hence, the surface area of the lower portion is ( 8π 4π 1 1 ) ( 4π 1 + 1 ). 2 2
MATH 228 Page 6 Problem 5. (2pts) Find the outward flux of F (y x)i + (z y)j + (y x)k along the boundary of the cube E bounded by the planes x ±1, y ±1, z ±1. We can apply Gauss s theorem. Note that F 1 1 + 2 so that the outward flux is since the volume of E is 8. F σ ( 2) dv 16 E E
MATH 228 Page 7 Problem 6. (2pts) Find the outward flux of F x 2 i + y 2 j + z 2 k across the boundary of the region D cut from the solid cylinder x 2 + y 2 4 by the planes z and z 1. Again, we can apply Gauss s theorem. Note that so that the outward flux is D F σ F 2x + 2y + 2z 2(x + y + z) dv D ˆ 2π ˆ 2 ˆ 1 ˆ 2π ˆ 2 ˆ 2π ˆ 2 ˆ 2π ˆ 2π 2(r cos θ + r sin θ + z)r dz dr dθ (2r 2 (cos θ + sin θ)z + 2rz2 2 ( 2r 2 (cos θ + sin θ) + r ) dr dθ ( 2r 3 r2 (cos θ + sin θ) + 3 2 ( ) 16 (cos θ + sin θ) + 2 3 ( ) 2π 8 (sin θ cos θ) + 2θ 3 4π. ) 2 dθ dθ ) 1 dr dθ
MATH 228 Page 8 Problem 7. (2pts) A space probe in the shape of the ellipsoid 4x 2 + y 2 + 4z 2 16 enters Earth s atmosphere and its surface begins to heat. After 1 hour, the temperature at point (x, y, z) on the probe s surface is T (x, y, z) 8x 2 + 4yz 16z + 6. Find the hottest point on the probe s surface. We need to solve T λ g where g(x, y, z) 4x 2 + y 2 + 4z 2. Since T 16x, 4z, 4y 16 and g 8x, 2y, 8z this translates into the system of equations 16x 8λx 4z 2λy 4y 16 8λz From the first equation, we see λ 2 (unless x, we ll get back to this). Since λ 2, the second equation implies y z, while the third equation tells us the common value of y and z is y z 4 3. Plugging back into the constraint g(x, 4/3, 4/3) 16 tells us 4x 2 16 16 9 416 9 64 9 so that x ± 4 3. This gives 2 critical points where T (x, y, z) 8 16 9 + 416 9 16 16 448 + 6 9 9 + 6. This turns out to be the maximum temperature at the hottest point (see below); however, to deduce this rigorously, we also need to check for potential critical points satisfying the condition x.
In this case, the system of equations we need to solve is: y 2 + 4z 2 16 4z 2λy 4y 16 8λz Since 2z λy, plugging into the first equation we get y 2 + λ 2 y 2 y 2 (1 + λ 2 ) 16 while plugging into the third equation we get 4y 16 4λ 2 y or It follows that 4y(1 λ 2 ) 16. y 2 (1 + λ 2 ) 16 4y(1 λ 2 ) and in particular, that y, so that y(1 + λ 2 ) 4(1 λ 2 ) or y 4(1 λ2 ) 1 + λ 2. Since z λy 2, we have z 2λ(1 λ2 ) 1 + λ 2. Plugging into the constraint, we get 16(1 λ 2 ) 2 (1 + λ 2 ) 2 + 16λ2 (1 λ 2 ) 2 (1 + λ 2 ) 2 16 which reduces to (1 λ 2 ) 2 1 1 + λ 2 or λ 4 2λ 2 + 1 1 + λ 2, or λ 2 (λ 2 3), i.e. λ, ± 3. The case λ implies z and y 4, given T 6. The cases λ ± 3 implies y 2 and z 3, giving T 6 + (4y 16)z 6 ± 24 3. Since 24 3 < 448 448, we conclude that 6 + is the maximum temperature. 9 9 Remark 1: From this analysis, we can also conclude that 6 24 3 is the minimum temperature on the ellipsoid. Remark 2: This problem is too long to be suitable for a final exam question, but it is instructive to work through to see how to use Lagrange multipliers.
MATH 228 Page 1 Problem 8. (3pts) Short Answers. 1. For what values of α is r α r conservative? (Here, r x 2 + y 2 + z 2.) Since r α r ( 1 α+1 rα+1), the vector field is conservative for α 1. 2. TRUE/FALSE: Every conservative vector field is irrotational. TRUE, because F conservative means F f for some scalar f, so that F f by the equivalence of mixed partials. 3. TRUE/FALSE: Every irrotational vector field defined on all of 3-space is conservative TRUE, because F and domain is simply-connected. 4. Give an example of an irrotational vector field that is not conservative. Be sure to specify the domain. θ yi+xj is irrotational, but not conservative because the circulation about x 2 + y 2 1, z is r x 2 +y 2 2π. Domain is the complement of the z-axis. ˆ 5. Which of the following conditions guarantees that F dr depends only on the endpoints of the path C. C (A) F (B) F (C) F f for some scalar-valued function f (D) F G for some vector field G 6. Which of the following conditions guarantees that boundary of S. ˆ S F dr depends only on the values of F on the (A) F (B) F (C) F f for some scalar-valued function f (D) F G for some vector field G
MATH 228 Page 11 Problem 9. (3pts) Stokes Theorem (a) Let r xi x2 + y 2 + yj x2 + y 2 and θ yi x2 + y 2 + xj x2 + y 2. Let C be the circle of radius 3 centered at the origin. Compute the flux of the vector fields r and θ across C with respect to the outward normal? Since θ is tangent to C at every point, hence perpendicular to the outward normal, the flux of θ across C vanishes. Since r coincides with the outward unit normal at every point on C, the flux of r is the length of C, or 6π. (b) IDENTIFY the true statement(s) among the following. (A) F dσ ( F) dv (B) E E (C) ( F) dσ F dv (D) E E (E) F dσ ( F) dr (F ) S S ( F) dσ F dr S S F dr ( F) dσ S S ( F) dσ F dv E E (A) is Gauss Theorem while (D) is Stokes Theorem.