MTH 234 Solutions to Exam 2 April 3, 25 Multiple Choice. Circle the best answer. No work needed. No partial credit available.. (5 points) Parametrize of the part of the plane 3x+2y +z = that lies above the ellipse x 2 +4y 2 = 4. A. r(s,t) = 2scost,ssint, 6scost 2ssint with t [,2π] and s [,]. B. r(s,t) = scost,ssint, 3scost 2ssint with t [,2π] and s [,2]. C. r(s,t) = s,t, 3s 2t, with s [ 2,2], and t [,]. D. r(s,t) = s,t, 3s 2t, with s [ 2,2], and t [,]. E. None of the above. 2. (5 points) Which of the following vector field plots could be F = yi xyj? D. Extra Work Space. Page 2 of 3
MTH 234 Solutions to Exam 2 April 3, 25 Fill in the Blanks. No work needed. No partial credit available. 3. Let F = (5x 3 z)i+(xyz)j+( 6yz 2 )k. (a) (5 points) divf = 5x 2 z +xz 2yz. (b) (5 points) curlf = i ( xy +6z 2) + j ( 5x 3) + k(yz). 4. (5 points) If a smooth surface S is given parametrically as r(u,v) with (u,v) D, then its surface area is given by the formula: A(S) = S ds = D r u r v da. 5. ( points) The volume in the first octant between the spheres shown to the right is given by: z ˆ θ2 ˆ φ2 ˆ ρ2 ρ ρ 2 sinφ dρ dφ dθ (,,3) (,,) where ρ = y ρ 2 = 3 φ 2 = π/2 x θ 2 = π/2 Extra Work Space. Page 3 of 3
MTH 234 Solutions to Exam 2 April 3, 25 Standard Response Questions. Show all work to receive credit. Please BOX your final answer. 6. (2 points) Let f(x,y,z) = xsiny +e xy lnz. (a) Find f at P (3,,2). f x = siny +ye xy = f x (P ) = f y = xcosy +xe xy = f y (P ) = 6 f z = z = f z(p ) = ( 2 ) Thus ( f) P = i+6j+ ( ) k 2 (b) Find the derivative of f at P in the direction of A =,2, 2. Let u = A A = (,2, 2 ). Then 3 D u f = f u = 3 = 3 3 ( i+6j+ ( ) ) k (,2, 2 ) 2 Page 4 of 3
MTH 234 Solutions to Exam 2 April 3, 25 7. (2 points) Let f(x,y) = 4x 2 + y 3 6xy. Find and classify each critical point of f as a local maximum, a local minimum, or a saddle point. i. Find the critical points. Notice that f x = 8x 6y and f y = 3y 2 6x. So f has a critical points at P(,) and Q(9/8,3/2). ii. Now let D(x,y) = f xx f yy f 2 xy = 48y 36. Notice that f xx = 8 > and that D(,) = 36 < and D(9/8,3/2) = 36 > It follows that f has a local minimum at Q and a saddle point at P. Page 5 of 3
MTH 234 Solutions to Exam 2 April 3, 25 8. (2 points) Sketch the region of integration for the integral below and write an equivalent integral with the order of integration reversed. Do not evaluate the integral. ˆ ˆ 8 3 x 2 (x+3y 2 )dydx y = 8 = ˆ 8 ˆ y+ (x+3y 2 )dxdy x = y + 9. (2 points) Evaluate the integral below. ˆ 3 ˆ 9 x 2 3 cos ( 4x 2 +4y 2) dydx We switch to polar coordinates = = π = π 8 ˆ π ˆ 3 ˆ 3 ˆ 36 ( cos4r 2 ) rdrdθ ( cos4r 2 ) rdr cosudu = π 36 8 sinu = π 8 sin36 Page 6 of 3
MTH 234 Solutions to Exam 2 April 3, 25. (2 points) Set up but do not evaluate the iterated integral for computing the volume of a region D if D is the right circular cylinder whose base is the disk r = 2cosθ (in the xy-plane) and whose top lies in the plane z = 9 2x. z D dv = ˆ π/2 ˆ 2cosθ ˆ 9 2x π/2 rdzdrdθ z = 9 2x Of course, 9 2x = 9 2rcosθ. r = 2cosθ x y The base of the cylinder is shown in the sketch below. We plot a few points to justify the limits of integration for θ in problems and. From Math 33... θ r = 2cosθ (x,y) Plot Symbol π/2 (, ) π/3 /2 (/2, 3/2) π/4 2 (, ) π/6 3 (3/2, 3/2) 2 (2, ) π/6 3 (3/2, 3/2) π/4 2 (,) π/3 /2 (/2, 3/2) π/2 (, ) y r = 2cosθ π/6 x For example, the length of the purple line segment is 3. Page 7 of 3
MTH 234 Solutions to Exam 2 April 3, 25. ( points) Find the area of the top of the cylinder in Problem. This is straightforward. Let R be the circle r = 2cosθ and its interior. Notice that the area of R is π. Now ( ) z 2 ( ) z 2 SA = + + da R x y = +( 2) 2 da R = 5 da = 5π R }{{} area of R For those that remain unconvinced, we evaluate the double integral (using polar coordinates). SA = 5 da as we observed above. R ˆ 2cosθ = ˆ π/2 5 rdrdθ π/2 ˆ 5 π/2 = r 2 2cosθ dθ 2 π/2 = 2 ˆ π/2 5 cos 2 θdθ π/2 = 5 = 5 ˆ π/2 π/2 (+cos2θ)dθ ( θ + sin2θ 2 ) π/2 π/2 = 5π Page 8 of 3
MTH 234 Solutions to Exam 2 April 3, 25 2. ( points) Find the work done by the force F = y,4x along the straight line segment from (3,5) to (2,). Let C be the indicated line segment. Now let r(t) = (3 t)i+(5 5t)j, t Then dr = (( )i+( 5)j) dt and F dr = ( 7t)dt it follows that ˆ C F dr = ˆ ( 7t)dt = 5(22t 7t 2 ) = 75 Page 9 of 3
MTH 234 Solutions to Exam 2 April 3, 25 3. (a) (6 points) Find a function f so that f = y 3 i+(3xy 2 5z 2 )j+( yz)k. Let f (x,y,z) = xy 3 5yz 2 then f = y 3 i+(3xy 2 5z 2 )j+( yz)k (b) (5 points) Evaluate the integral below. ˆ y 3 dx+(3xy 2 5z 2 )dy +( yz)dz C Here C is any path from (,2,) to (2,, ). Observe that df = y 3 dx+(3xy 2 5z 2 )dy +( yz)dz It follows that ˆ (2,, ) (,2,) y 3 dx+(3xy 2 5z 2 )dy +( yz)dz = ˆ (2,, ) (,2,) df = f (x,y,z) (2,, ) (,2,) = 3 8 Page of 3
MTH 234 Solutions to Exam 2 April 3, 25 4. (2 points) Let E = {(x,y,z) x z, y 5, y z 5}. Rewrite the triple integral below as an iterated integral and evaluate. E dv E dv = = ˆ 5 ˆ 5 ˆ z y ˆ 5 ˆ 5 y ˆ 5 dxdzdy zdzdy = (25 y 2 )dy 2 = ) 5 (25y y3 2 3 = 88 3 Equivalently, E dv = = ˆ 5 ˆ z ˆ z ˆ 5 ˆ 5 (z ) dydxdz ˆ z = (z )zdz ( ) z 3 5 = 3 z2 2 = 88 3 dxdz Page of 3
MTH 234 Solutions to Exam 2 April 3, 25 5. (2 points) Find the work done by the force F = 6xyi+ ( 3x 2 +2x ) j when moving a particle around the circle x 2 +y 2 = 6 starting and ending at (4,) traveling in the counterclockwise direction. Let C be the circle of radius 4 centered at the origin. Method : Direct calculation. Now C is given by the parametric equation r(t) = (4cost)i+(4sint)j, t 2π dr = (( 4sint)i+(4cost)j)dt Hence F dr = (6xy)( 4sint)dt+ ( 3x 2 +2x ) (4cost)dt =. = 6 [ 24sin 2 tcost+2cos 3 t+2cos 2 t ] dt It follows that the work done is ffi ˆ 2π [ F dr = 6 24sin 2 tcost+2cos 3 t+2cos 2 t ] dt C =. = 6 = 32π ˆ 2π (+cos2t)dt Method 2: Green s Theorem Let R be the interior of the circle C. Then the area of R is 6π and by Green s Theorem ffi ( ( 3x 2 +2x ) ) F dr = (6xy) da C R x y = (2)dA as we saw above. R = 2 Area of R = 32π Clearly, this is the easier of the two calculations. Page 2 of 3