Ma 227 Final Exam Solutions 5/8/03

Transcription

1 Ma 7 Final Eam Solutions 5/8/3 Name: Lecture Section: I pledge m honor that I have abided b the Stevens Honor Sstem. ID: Directions: Answer all questions. The point value of each problem is indicated. If ou need more work space, continue the problem ou are doing on the other side of the page it is on. Youmanot use a calculator on this eam. Score on Problem # # #3 #4 #5 #6 #7 #8 Total

2 Problem a) ( points) Calculate the iterated integral dzdd Be sure to show all steps. Solution: dzdd dd d d b) (5 points) Give an epression in clindrical coordinates for the volume of the solid T bounded above b the plane z and below b the paraboloid z.sketcht. Donot evaluate this integral. Solution: In polar coordinates the plane has the equation z rsin and the paraboloid has the equation z r. The two surfaces intersect when, that is the circle or. However, it is onl the part of this circle that is in first and second quadrants that 4 is the projection of the solid onto the, plane, since the plane z goes through the ais. The equation of this circle is r sin sin r sin Volume r rdzdrd Problem a) ( points)

3 Give two triple integral epressions for the volume under the surface z and above the triangle in the, plane with vertices,,,,4,. Sketch the triangle in the, plane. Do not evaluate the epression. Solution:,,,,4,,, The line joining, and, has equation and the line joining, to 4, has equation 4. Thus b)( points) 4 Volume dzdd 4 4 dzdd dzdd Calculate the surface integral F nds, where S F,,z 3 i 3 j z 3 k and S is the closed surface of the solid bounded b the clinder and the planes z and z. Solution: We can use the Divergence Theorem, since S is a closed surface. F 3 z so 3

4 S F nds F dv V 3 z dv 3 r z rdzdrd 3 r 3 z r z3 drd 3 3 r r d Problem 3 a) (5 points) Find the eigenvalues and eigenvectors of A 3 Solution: r 3 r r r3 r r so the eigenvalues are r,,3. The sstem of equations that determines the eigenvectors is r 3 r 3 r 3 For r, we have 3 is arbitrar, and 3 Thus 3 and 3. Thus we have the eigenvector sstem implies, 3,, is arbitrar. Thus.For r, the. For r 3 the sstem implies 4

5 3, is arbitrar, and. Thus b) ( points) Solve the initial value problem t At where A is the matri above. Solution: The general solution to the homogeneous sstem is t c e t c e t c 3 e 3t Then c c c 3 c c c 3 c c 3 c Thus c,c 3,c. The solution is Problem 4 t e t a) (5 points) Verif Green s theorem is true for the line integral d d C where C consists of the parabola from, to, and the line segment from, to,. SketchC. Solution: 5

6 Let C be the parabola and C the line. Then C : rt ti t j t. Thus d d C Using Green s Theorem we have d d C R t t 4 t t t dt t6 6 t R da 4dd t andc : rt ti j t dt da d 5 d b) ( points) Put the matri in row reduced echelon form

7 Solution: R R R R R R R 3 R R R 3 3 R 3 3 R 3 R R3 R 3 Problem 5 a) ( points) Let Calculate Solution: F i j k z F,,z i j z k F curlf z i j k b) (5 points) Verif that Stokes Theorem is true for the vector field in part a) where S is the part of the paraboloid z that lies above the, plane, and S has up orientation. Solution: Since F from part a) S F nds The bottom of the paraboloid in the, plane is the circle. We let C : t cost,t sin t,z so rt costi sintj k, t r t sinti costj F t cos ti sin tj k Then F dr C 3 sin tcos t costsin t dt cos3 t sin 3 t 7

8 Problem 6 a)(5 points) If F,,z 3 z i 4j 5 4 3z k find a function f such that f F. f 4 so Then so Now we have f g,z f z g z 5 4 3z g,z 5 4 z z 3 h f 5 4 z z 3 h f 3 z h 3 z Hence h andh K. Finall, f,,z 5 4 z z 3 K b)( points) Evaluate C F dr where F is the vector field in part a) and C is the curve given b the vector equation rt t i t 5 j 3t 6 k t Solution: Since there eists a function f,,z such that f F, the line integral is independent of path. The curve C begins at,, and ends at,3,4. Therefore C F dr f,3,4 f,, f,3,4 4 K and f,, 8 K. Thus C F dr f,3,4 f,, 4 Problem 7 a) ( points) Let A be a constant matri and r an eigenvalue of A with corresponding eigenvector u. Show that t t r u is a solution of the sstem t t At 8

9 Solution: We have that Since t t r u,then t t t rt r u Au ru t r ru t r Au At r u At b) (5 points) Solve the sstem t t 3 5 t t Solution: det r r 3 5 r 8 6r r r 4r so the eigenvalues are,4. r 3 5 r 3 3 so 3 and the eigenvector is 3. The other eigenvector is the solution is t c t 3 c t 4 4. From part a) Problem 8 a)(5 points) Evaluate the sinda R where R is the region in the first quadrant that is outside the circle r and inside the cardioid r cos. SketchR and shade it. Solution: 9

10 sin da cos sinrdrd r cos sin d R cos sin sin d cos3 3 cos b)( points) Give two iterated integrals for the area of the region R in the first quadrant that lies above the hperbola and the line and below the line. Sketch R and shade it. Do not evaluate these integrals. Solution: The hperbola and the line intersect when, that is at in the first quadrant. Thus at,. The line intersects the hperbola at,thatisat,. The line intersects the line at,.

11 Area da R dd dd dd