MTH 234 Solutions to Exam 2 April 13, Multiple Choice. Circle the best answer. No work needed. No partial credit available.

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1 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

2 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

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

4 MTH 234 Solutions to Exam 2 April 3, (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

5 MTH 234 Solutions to Exam 2 April 3, (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

6 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 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

7 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

8 MTH 234 Solutions to Exam 2 April 3, ( 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

9 MTH 234 Solutions to Exam 2 April 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

10 MTH 234 Solutions to Exam 2 April 3, (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

11 MTH 234 Solutions to Exam 2 April 3, (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