In addition to the problems below, here are appropriate study problems from the Miscellaneous Exercises for Chapter 6, page 410: Problems

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1 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 1 of 9 22M:28 Spring 05 J. Simon Study Guide for Final Exam Chapter 6 Portion How to use this guide: I am not going to list a lot of problems of each kind. Some are "one of a kind" thinking type problems. For calculation type problems (the majority), I will try to list one problem of each kind that might be on the exam. It is your job to make up (and then solve, of course) a bunch of similar problems. Try studying with others in the class, challenging each other with problems. Caution: If you only "study" by reading through the solutions that I give below to some of the problems, and don't make up/solve problems on your own, you will kid yourself about being prepared for the exam. In addition to the problems below, here are appropriate study problems from the Miscellaneous Exercises for Chapter 6, page 410: Problems 1bc, 2ab, 11, 12 (note clockwise vs. counter-clockwise), 14 (with given functions, e.g. f(x) = exp(x)*cos(x), g(y) = arctan(y), and given C via a sketch. The key is to calculate Nx-My and invoke Green's Theorem.) **21** (use the "divergence form" of Green's Theorem) %%%%%%%%%%%%%%%%%%%%%%% Here are samples of the kinds of problems you should be able to do. This "Guide" may include suggestions/hints/reminders for solving the problems; but, of course, there won't be such hints on the Exam. A typical problem might involve any or all of the following three steps: a) parameterize some curve or region in R^2; b) set up a definite integral (in one variable) and/or an iterated integral (in 2 dimensions); c) evaluate the integral(s) from step (b). d) Invoke Green's Theorem (in either the "curl form" or the "divergence form") as needed. Remark on evaluating integrals: Some of you have expressed anxiety (or amnesia) regarding evaluating integrals. The blanket answer to the question, "Which integrals should I know?" is "Everything you were expected to know in Calc II." In practice, I expect you to be able to evaluate "simple" integrals for the exam. What is "simple"? A good sample would be: whatever integrals we actually worked out by hand in class this semester

2 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 2 of 9 Problem 1. Set up, but do not evaluate, a definite integral whose value is the line integral of the function f ( x, y, z ) = 3 x + y 2 along the path C, where C is the upper half of the circle x 2 + y 2 = 25 Solution: You first need to parameterize the curve, then set up the integral. x := t y := 25 t 2 X := [ t, 25 t 2 ] vel := 1, speed := 1 + t 25 t 2 t 2 25 t 2 Integrand := ( 3 t + 25 t 2 ) 1 + Integral := 5-5 ( 3 t + 25 t 2 ) 1 + t 2 25 t 2 t 2 dt 25 t 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

3 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 3 of 9 Problem 2 Calculate the line integral of the vector field f(x,y,z) = (x+y) i+(z) j - (xy) k over the curve X(t ) = ( t, cos(t), sin(t) ), t=0..4*pi. (Express your answer as a definite integral, but do not evaluate it.) This might also be phrased as Calculate Int ( (x+y) dx + (z) dy - (xy) dz ) over that curve. ( x + y) dx + ( z) dy + ( xy) dz Solution: C x := t y := cos( t ) z := sin( t ) dx := 1 dy := sin( t ) dz := cos( t ) (note - I'm abusing notation, using Maple to do the arithmetic - there should be a "dt" after each of the above dx, dy, dz) M := t + cos( t ) N := sin( t ) P := t cos( t ) LineIntegral := 0 4 π t + cos( t ) sin( t ) 2 t cos( t ) 2 dt The integral above is the end of the solution of Problem 2. % % % % Variation: Here is a version of the preceding problem with a function you should be able to integrate explicitly on the exam: Suppose the given vector field is. F(x,y,z) = (x+y)i +(z) j + (y) k, the final integral would be 0 4 π t + cos( t ) sin( t ) 2 + cos( t ) 2 dt which you *are* expected to be able to evaluate. The third term in the original integral, (t * cos(t) ^2 ), IS something you can work out, but I won't ask you to work out an integral that long on the exam.. d t cos( t ) 2 t = t cos( t ) sin( t ) t 1 cos( t ) t2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

4 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 4 of 9 Problem 3 Suppose for some vector field F(x,y,z) and the curve x(t) = (t, t^2, t^3) (t = 0..3), we calculate that Int(F dot ds) along C = -Pi/sqrt(2). (i) What would be the value if we used the parameterization x (t) = (3-t, (3-t)^2, (3-t)^3), (t = 0..3)? (ii) What would be the value if we reparameterize C as x (t) = (2t, 4t^2, 8t^3) (t = 0..3/2) Remarks on Solution: This is the question: "What happens to the line integral Int (F dot ds) over some curve C if we change the parameterization of the curve? See Theorems 1.4, 1.5 in Section 6.1. This problem deals with line integrals of vector fields. Note that line integrals of scalar functions (e.g. the scalar function is density of a filament at each point and the line integral is the total mass of the filament) are not supposed to care what sense of orientation we give the filament, whereas vector line integrals *do*. On the other hand, neither kind of line integral is supposed to care about the speed of parameterization, just so long as the same curve is being traversed. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

5 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 5 of 9 Problem 4. Let D be the region in the plane bounded by the curve = 4 x2 9 y2 1 Let C denote the boundary of D, oriented in the usual way. Let F be the vector field F(x,y) = (x+y) i + (xy) j. STATE Green's Theorem for this situation. Express your answer in terms of one or more iterated integrals and/or one or more definite integrals. Remarks on Solution: You are supposed to know that Green's theorem says that a certain line integral is equal to a certain double integral, then set up a definite integral in one variable for the line integral and an iterated integral in 2 variables for the double integral. M := x + y N := x y The integrand for the double integral is Nx-My = y-1. The double integral is the integral of (Nx-My) over the elliptical region D. To finish setting up the iterated integral, you need to parameterize the set D (e.g. x goes from -2 to 2, y goes from -(3/2)*sqrt(4-x^2) to +(3/2)*sqrt(4-x^2), to get 2-2 3/ 2 4 x 2 3/ 2 4 x 2 y 1 dy dx The other integral is the line integral of F along the boundary curve. We can parameterize that curve in two pieces, the top half of the ellipse, C1 and the bottom half, C2. For C1, we can use x=t, y = (3/2)*sqrt(4-x^2), t goes from +2 to -2. For C2, we can use x =t, y = -(3/2)*sqrt(4-x^2), t goes from -2 to +2. The line integral of M dx + N dy along C1 is obtained as follows. You also need to set up the integral along C2, and then add the two integrals. This total is what Green's Theorem says is equal to the double integral above. x := t y := dx := 1 dy := t2 + 4 t t M := t + 2 t N := 2 t t Mdx_plus_Ndy := t + 2 t t2 LineIntegralC1 := t + d 2 t t2 t

6 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 6 of 9 Problem 5 Let F be the vector field F(x,y) = (x^2) i + (x-y) j. Let D be the unit square in the xy-plane, that is the set of points {(x,y) x = 0..3, y = 0..3 }. Calculate the (outward) flux of F across the boundary of D in two ways (and, hopefully, get the same number), as follows: (a) Use Green s theorem. (b) Calculate the flux directly from the definition of flux. Solution: (a) According to the divergence form of Green s Theorem, the flux of F across the boundary of D is equal to the integral over D of the divergence of F. Here, div(f) = Mx+Ny = 2x-1. So IntegralOfDivergence := Int(Int(2*x-1, x = (0.. 3)), y = (0.. 3)) = 18. (b) On the other hand, we can compute the flux directly by looking at the boundary of D (here four separate segments), finding an outward pointing unit normal vector N (for each piece), calculating the dot product F N, and integrating that dot product (as a scalar function) on the segment. Bottom edge: x=0..3, y=0. F(x,y) = (x^2) i + (x) j. N = - j constant. So F N = -x. Integrate (-x) for x=0..3 to get (-9/2) as the first contribution. Right side edge: x = 3, y= F(x,y) = (9) i + (3-y) j. N = i constant. So F N = 9. Integrate (9) for y=0..3 to get (27) as the second contribution. Top edge: x = 0..3, y = 3. F(x,y) = (x^2) i + (x-3) j. N = j constant. So F N = x-3. Integrate (9) for y=0..3 to get ( 9/2) as the third contribution. Left edge: x = 0, y=0..3. F(x,y) = (0) i + (-y) j. N = -i, constant. So F N = 0. Integrate (0) for y=0..3 to get (0) as the fourth contribution. Add the four contributions: (-9/2) (-9/2) + (0) = 18. Observe: the flux integral = 18 and the double integral of div(f) = 18 also. %%%%%%%%%%%

7 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 7 of 9 Problem 6 Here is a picture of a vector field F and a closed curve C (made of several smooth pieces). Estimate visually whether the (outward) flux of F across C is positive, negative, or zero. Solution: The flux across the right edge seems to be large, outward (so positive). The flux across the bottom seems to be small, in (so negative), likewise for the top. The flux across the left edge appears to be 0. So it looks as if the outward flux across the right edge dominates, and the total flux is positive. Bonus remark: This picture is precisely the vector field and box from Problem 5.

8 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 8 of 9 Problem 7 Let F be the vector field F(x,y) = (x+y) i + cos(y) j + 3 k. Decide whether or not F is conservative. Solution: Since F is defined (and smooth) on all of R^3, the question, Is F conservative? (i.e. Does F have path-independent work integrals? ) is equivalent to the question: Is the curl of F equal to 0 everywhere? In this case, F = determinant of the matrix which equals [0, 0, -1]. Since the curl of F is not the zero vector, we conclude F is NOT conservative. Remarks: The fact that F is not conservative means that there are some paths between some pairs of points along which the work integrals of F depend on the choice of paths. For example, consider the paths x(t) = (t,t,t) and y(t) = (t, t^2, t^3). Each path goes from (0,0,0) to (1,1,1) as t goes from 0 to 1. Along the first path, the work integral of F is 4+sin(1). Along the second path, the work integral is (23/6) + sin(1). The two work integrals are different (which is another proof that the vector field F is not conservative. However, there may be some paths that product the same value work integrals: That doesn t prove that the vector field is conservative, just that more testing is necessary before a decision can be made. For example, along the path x(t) = (t, t, t^3), we still get 4+sin(1).

9 22M:28 Spring 05 J. Simon Ch. 6 Study Guide for Final Exam page 9 of 9 Problem 8 Let C be the boundary of the triangle with vertices (0,0), (2,0), and (2,3). Verify that Green s Theorem is correct for this triangle and the vector field F(x,y) = (2y) i + (xy) j. Solution: Green s Theorem states that a certain double integral over the 2-dimensional triangle shaped region equals the circulation integral of F along the boundary triangle The double integral is R (N x -M y ) da = x=0..2 y=0..(3/2)x (y-2) dy dx = -3. On the other hand, we have to calculate the line integral (2y) dx + (xy) dy along the three line segments and add the contributions. On the base of the triangle, (2y) = 0, and also dy = 0; so both parts of the integrand are zero. On the vertical segment, dx = 0 and we are left with y=0..3 (2y) dy = 9. On the hypotenuse, we can use the parameterization (being careful about orientation!!) x = 2-t, y=(3/2)x = 3-(3/2) t, t=0..2. This gives dx = -1, dy = -3/2 (both times dt). So we have (2y) dx + (xy) dy = t=0..2 2(3-(3t/2)(-1) + (2-t)(3-3t/2)(-3/2) dt = -12. Add 9 + (-12) to get 3, the same as we got from the double integral.