MATH 2203 Final Exam Solutions December 14, 2005 S. F. Ellermeyer Name

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1 MATH 223 Final Exam Solutions ecember 14, 25 S. F. Ellermeyer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su cient to just write down an answer with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator but you may not use any books or notes. This test contains eight problems. You only have to do six of them. Write o not grade on the two that you do not want me to grade. I will grade only six! 1. Find the surface area of the portion of the paraboloid z x 2 + y 2 that lies between the two planes z 4 and z 9. (See the gure below which should help you to visualize this surface.) Solution: The surface area is q (z x ) 2 + (z y ) da p 4x2 + 4y da where is the region consisting of all points (x; y) such that 4 x 2 + y 2 9. We are best o using polar coordinates. In this case, the region consists of all points (r; ) such that 2 r 3 and 2. The surface area is thus p 4r2 + 1r dr d 37p p

2 2. A wire is in the shape of the upper half of the unit circle (and is positioned along the upper half of the unit circle as pictured) and has density function (x; y) x + 1. Find the total mass, m, and the center of mass, (x; y), of this wire. Solution: We parameterize the wire as and observe that and hence that jr (t)j 1. The total mass of the wire is r (t) cos (t) i + sin (t) j r (t) m. sin (t) i + cos (t) j (x; y) ds (r (t)) jr (t)j dt (cos (t) + 1) dt The x coordinate of the center of mass of the wire is x 1 x (x; y) ds m and the y coordinate of the center of mass is y 1 y (x; y) ds m 1 2. cos 2 (t) + cos (t) dt (sin (t) cos (t) + sin (t)) dt Thus, the center of mass of the wire is located at the point 1 2 ; 2. 2

3 3. Let F be the vector eld F (x; y) (2x 3y) i 3xj and let be the part of the unit circle (oriented in the counterclockwise direction) that lies in the rst quadrant. (a) Write down a parameterization for : r (t) (b) Use the parameterization for that you wrote in part a to evaluate the line integral F dr. Solution: A parameterization for is r (t) cos (t) i + sin (t) j t 2. Since and and we see that we obtain F dr 2 F (r (t)) r (t) dt F (r (t)) (2 cos (t) 3 sin (t)) i 3 cos (t) j r (t) sin (t) i + cos (t) j, F (r (t)) r (t) 2 sin (t) cos (t) + 3 sin 2 (t) 3 cos 2 (t) 2 sin (t) cos (t) 3 cos (2t), F dr F (r (t)) r (t) dt ( 2 sin (t) cos (t) 3 cos (2t)) dt 3

4 4. onsider the same vector eld, that was given in problem 3. F (x; y) (2x 3y) i 3xj, (a) Show that this vector eld is conservative and nd a function, f, such that F rf. (b) Now use the Fundamental Theorem of Line Integrals to evaluate the line integral F dr (same integral as in problem 3). Solution: Since the functions P (x; y) 2x 3y and Q (x; y) 3x are de ned on all of < 2 we conclude that F is conservative. Since the potential function, f, must satisfy we must have However, f must 2x 3y, f (x; y) x 2 3xy + g (y). 3x + g (y). Therefore g (y) which means that g must be a constant function. Since we only need a single function f, it is convenient to choose this constant to be. Thus our potential function is f (x; y) x 2 3xy. By the Fundamental Theorem of Line Integrals, F dr rf dr f (; 1) f (1; ) Let be the triangle with vertices at the points (; ), (1; ), and (; 1) oriented in the counterclockwise direction and let F be the vector eld F (x; y) xyi + j. In this problem, you will evaluate the line integral P dx + Q dy in two di erent ways: not using Green s Theorem and using Green s Theorem. 4

5 (a) Evaluate this line integral directly (without using Green s Theorem). (b) Evaluate this line integral by using Green s Theorem. Solution: The component functions of F are P (x; y) xy and Q (x; y) 1. Referring to the picture of, we have three curves, 1, 2, and 3, along which we will compute separate line integrals. On the curve 1, we have with x 1. Thus On the curve 2, we have with t 1. Thus 1 P dx + Q dy x x dx dx y dy 1 xy dx + dy x 1 t dx dt y t dy dt 2 P dx + Q dy xy dx + dy t 2 t 2 x () dx t dt + dt t + 1 dt On the curve 3, we have x dx y 1 t dy dt 5

6 with t 1. Thus Therefore, 3 P dx + Q dy 1 P dx + Q dy dt Here is the solution using Green s Theorem: Green s Theorem tells P dx + Q ( x) da x x dy dx 6. Let F be the vector eld (a) ompute curl(f). F (x; y; z) (x + y + z) i xyzj 2zk. (b) Is F a conservative vector eld? Explain why or why not. (c) Without computing div(curl (F)), what do you know that div(curl (F)) should be equal to? (Hint: There is a theorem that answers this question.) (d) ompute div(curl (F)). Solution: i j k x + y + z xyz 2z ( + xy) i ( 1) j + ( yz 1) k xyi + j (yz + 1) k. Since curl(f) 6, we know that F is not conservative. However, we know that it must be true that div(curl (F)) (whether F is conservative or not). Let us verify this: div (curl (F)) div (xyi + j (yz + (1) ( yz y + y. 6

7 7. Let F be the vector eld F (x; y; z) xj and let S be the upper half of the unit sphere (with upward orientation) and let be the unit circle with positive orientation relative to S. Note that is the boundary curve of S. The purpose of this problem is to compute the line integral F dr in two di erent ways: not using Stokes Theorem and using Stokes Theorem. (a) ompute the line integral F dr directly (without using Stokes Theorem). (b) Use Stokes Theorem to compute the above line integral. Solution: The unit circle (with positive orientation relative to the upper half of the unit sphere) can be parameterized as r (t) cos (t) i + sin (t) j + k. Thus and from which we obtain Therefore r (t) sin (t) i + cos (t) j + k F (r (t)) i + cos (t) j + k F (r (t)) r (t) cos 2 (t). F dr 2 2. F (r (t)) r (t) dt cos 2 (t) dt Here is the solution using Stokes Theorem: Since curl(f) k, we have F dr curl (F) ds S ( P z x Qz y + R) da 1 da 7

8 where is the unit disk. The last integral above is the area of this disk, which we know to be (1) 2. Therefore the answer is. 8. The picture shown below is that of a cross section (in the xy plane) of a vector eld, F (x; y; z), that does not depend on z. In other words, the vector eld looks exactly the same in all planes that are parallel to the xy plane. Yet in other words, F is of the form F (x; y; z) P (x; y) i + Q (x; y) j + k. (a) oes curl(f) point in the direction of k direction or in the direction of curl(f)? (You must justify your answer.) (b) Is div(f) positive, negative, or zero? (You must justify your answer.) k or is (c) Answer questions a and b above for the vector eld pictured below. 8

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