(x + y) ds. 2 (1) dt = p Find the work done by the force eld. yzk

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1 MATH Final Exam (Version 1) Solutions May 4, 11 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. 1. Evaluate (x + y) ds where is the straight line segment joining the point (; 1) to the point (1; ). Solution: The line segment can be parameterized by r (t) ti + (1 t) j, t 1. From this we obtain r (t) i j and jr (t)j p. Therefore (x + y) ds p 1 (1) dt p.. Find the work done by the force eld F (x; y; z) xyi + yj in moving an object from the point (; ; ) to the point (1; 1; 1) along the curve Solution: The curve of integration,, is x t y t z t. r (t) ti + t j + tk t 1. yzk Thus r (t) i + tj + k and the work done is 1 F dr F (x (t) ; y (t) ; z (t)) r (t) dt t i + t j t + t t dt t k (i + tj + k) dt t dt 1

2 . Find a potential function for the vector eld F (x; y) x y i + 1 x y j. Include all steps in your reasoning and check to see that the answer that you get is correct. Solution: We want to nd f such that rf F. Since i j, we must have and this tells us that f has the x y By di erentiation we obtain but we must also have and hence f (x; y) x y + 1 x y + h (y) 1 Therefore h (y) y and thus h (y) conclude that any function of the form x y + h (y) y x x 1 x y y y. y 1 + (where can be any constant). We f (x; y) x y 1 y + x 1 + y is a potential function for F. Let us check that this correct. Letting f above, we x y and which shows that our answer is 1 y x be as de ned 4. Let be the square (oriented counterclockwise) with vertices at the points (; ), (; ), (; ), and (; ). Find the circulation of the velocity eld F (x; y) sin (y) i + x cos (y) j

3 around. You can do this either directly (without using Green s Theorem) of do it by using Green s Theorem. Whichever way you choose, make sure to include all details. Solution: First let us do this directly. enoting the bottom, right, top, and left sides of the square by 1 ; ; ; and 4 respectively, we have the parameterizations 1 : x x; y ; dx dx; dy x, : x ; y y; dx ; dy dy y, The circulation over 1 is The circulation over is The circulation over is The circulation over 4 is : x x; y ; dx dx; dy x, 4 : x ; y y; dx ; dy dy y : 1 M dx + N dy M dx + N dy M dx + N dy 4 M dx + N dy ( sin ()) dx. cos (y) dy. sin dx. ( cos (y)) dy. The overall circulation is thus M dx + N dy Now let us do this using Green s Theorem. By Green s Theorem, the circulation around is @y

4 where is the interior of the square. Since @M cos (y) cos (y). cos (y) dy dx sin (y)j y y dx dx 5. Find a parameterization for the spherical band that lies on the sphere x + y + z between the planes z p and z p. Solution: We use spherical coordinates to parameterize the sphere (which is centered at the origin and has radius p ): x p sin () cos () y p sin () sin () z p cos (). In order to obtain the band in question, we set z p to obtain or p cos () p cos () 1 which has solution. Likewise we nd that the plane z to. We thus use the parameterization given above with p corresponds. 6. Find the surface area of the spherical band described in problem 5. Solution: The surface area is S 1 d 4 jr r j da

5 where r is the parameterization given in problem 5. We saw in class that jr r j sin (). (In general, for a sphere of radius a we saw that jr r j a sin ().) surface area of the band is jr r j da sin () d d 6. cos ()j d cos d + cos d Thus the It is interesting to note that the entire sphere has surface area 4 p 1 and thus the band has exactly half the surface area of the entire sphere. 5

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

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