Solutions to the Exercises * on Multiple Integrals

Transcription

1 Solutions to the Exercises * on Multiple Integrals Laurenz Wiskott Institut für Neuroinformatik Ruhr-Universität Bochum, Germany, EU 4 February 27 Contents Introduction 2 2 Calculating multiple integrals 2 2. Exercises Exercise: Integral of a function between two curves Changing the order of integration 3 3. Exercises Exercise: Integration in opposite order Exercise: Integration in opposite order Changing the coordinate system of integration 4 4. Exercises Exercise: Jacobian determinant Exercise: Integration with a nonlinear coordinate transformation Laurenz Wiskott (homepage This work (except for all figures from other sources, if present) is licensed under the Creative Commons Attribution-ShareAlike 4. International License. To view a copy of this license, visit Figures from other sources have their own copyright, which is generally indicated. Do not distribute parts of these lecture notes showing figures with non-free copyrights (here usually figures I have the rights to publish but you don t, like my own published figures). Several of my exercises (not necessarily on this topic) were inspired by papers and textbooks by other authors. Unfortunately, I did not document that well, because initially I did not intend to make the exercises publicly available, and now I cannot trace it back anymore. So I cannot give as much credit as I would like to. The concrete versions of the exercises are certainly my own work, though. In cases where I reuse an exercise in different variants, references may be wrong for technical reasons. * These exercises complement my corresponding lecture notes available at Teaching/Material/, where you can also find other teaching material such as programming exercises. The table of contents of the lecture notes is reproduced here to give an orientation when the exercises can be reasonably solved. For best learning effect I recommend to first seriously try to solve the exercises yourself before looking into the solutions.

2 4..3 Exercise: Variable transformation Exercise: Area integral Gauss theorem - integration by parts in multiple dimensions 9 5. Exercises Exercise: Gauss theorem Introduction 2 Calculating multiple integrals 2. Exercises 2.. Exercise: Integral of a function between two curves Integrate the function f(x, y) xy over the region between the curves y x and y x 2 for x 2. Solution: One has to be a bit careful here. Sketching the two functions x and x 2 shows that x > x 2 for < x < and x > x 2 for x >. Thus, we have to split the integral I into two parts, because otherwise one part would yield a negative contribution. I x 2 x2 xy dy dx + x 2 2 x[y 2 /2] x x dx + 2 x x(x 2 /2 x 4 /2) dx + (x 3 x 5 ) dx + 2 xy dy dx () x[y 2 /2] x2 x dx (2) x(x 4 /2 x 2 /2) dx (3) (x 5 x 3 ) dx (4) ([x 4 /4] [x 6 /6] )/2 + ([x 6 /6] 2 [x 4 /4] 2 )/2 (5) ((/4 /4) (/6 /6))/2 + ((64/6 /6) (6/4 /4))/2 (6) /8 /2 + 63/2 5/8 (7) 62/2 4/8 (8) 62/2 2/2 (9) 4/2 () () 2

3 3 Changing the order of integration 3. Exercises 3.. Exercise: Integration in opposite order Sketch the integration region and solve the following integral in the opposite order (first x then y): x y 2 dy dx. () x2 Solution: The integration region is the upper left triangle of the square with x [, 2] and y [, 2] (Sketch not available!). Integrating in the opposite order yields y y 2 dx dy x2 y 2 y dx dy (2) x2 y 2 [ /x] y dy (3) y 2 ( /y ( /)) dy (4) y 2 dy y dy (5) [y 3 /3] 2 [y 2 /2] 2 (6) (2 3 /3 3 /3) (2 2 /2 2 /2) (7) (8/3 /3) (4/2 /2) (8) 7/3 3/2 (9) 5/6. () 3..2 Exercise: Integration in opposite order Sketch the integration region and solve the following integral in the opposite order (first x then y): +x x x y 2 dy dx. () Solution: The integration region is a symmetric triangle (D: gleichschenkliges Dreieck) pointing to the left within the rectangle with x [, 2] and y [ 2, +2] (Sketch not available!). It can equally well be integrated 3

4 over with +2 2 y x y 2 dx dy y 2 x dx dy + +2 y y y 2 x dx dy (2) y 2 x dx dy (3) y y 2 [x 2 /2] 2 y dy (4) y 2 (2 2 y 2 )/2 dy (5) 4y 2 y 4 dy (6) 4[y 3 /3] 2 [y 5 /5] 2 (7) 4(2 3 3 )/3 (2 5 5 )/5 (8) 32/3 32/5 (9) 6/5 96/5 () 64/5. () 4 Changing the coordinate system of integration 4. Exercises 4.. Exercise: Jacobian determinant Consider the following nonlinear coordinate transformation between variables x and y and variables u and v (a and b are parameters): x u + a cos(u) () y v + b sin(u). (2). Visualize the transformation. Solution: Für a und b wären beide Koordinatensysteme identisch und man könnte zur Visualisierung einfach ein regelmäßiges u-v-gitter in die x-y-ebene zeichnen. Wählen wir a ungleich Null, wird das Koordinatensystem mit zunehmendem u periodisch in x-richtung vor- und zurückgeschoben, was zu einer periodischen Stauchung und Streckung führt. Wählen wir b ungleich Null, wird das Koordinatensystem mit zunehmendem u periodisch in y-richtung hoch- und runtergeschoben, was zu einer Wellenform führt. For b and a.5, for instance, we have: Graph: (Wiskott group, 27, unclear) 4

5 2. Calculate the Jacobian determinant. Solution: J x u y u x v y v ( a sin(u)) b cos(u) (3) (4) ( a sin(u)) b cos(u) (5) a sin(u). (6) Now, consider the following extended nonlinear coordinate transformation: x u + a cos(u) + b sin(v) (7) y v + a cos(v) + b sin(u). (8) 3. Calculate the Jacobian determinant. Solution: J x x u v y y u v ( a sin(u)) b b cos(u) cos(v) ( a sin(v)) () ( a sin(u)) ( a sin(v)) b cos(u) b cos(v) () a sin(u) a sin(v) + a 2 sin(u) sin(v) b 2 cos(u) cos(v). (2) (9) 4. Discuss the result in comparison to the first transformation. Solution: In the first part the shift in y-direction dependent on u, i.e. the term b sin(u) in (2), does not have any effect on the Jacobian, because it does not effect the local volume. If the roles of x and y as well as u and v would be swapped in (, 2), one would get a similar coordinate transformation in which a shift in x-direction dependent on v would have no effect either for symmetry reasons. In some sense one can argue that equations (7, 8) are a combination of these two coordinate transformations. Isn t it curious that in the combination the terms b sin(u) and b sin(v) actually have an effect on the Jacobian? 4..2 Exercise: Integration with a nonlinear coordinate transformation Compute the integral I ( ) x 2 + y 2 dx dy, () with D indicating the disc in the xy-plane around (, ) with radius. Hint: Use polar coordinates (r, φ) with x r cos φ, y r sin φ. D 5

6 Solution: I D 2π 2π ( ) x 2 + y 2 dx dy (2) dφ ( ) r 2 dφ x r y r x φ y φ } {{ } r cos φ cos φ+ r sin φ sin φr dr (3) (r r 2 ) dr (4) 2π([r 2 /2] [r 3 /3] ) (5) 2π((/2 /2) (/3 /3)) (6) 2π(3/6 2/6) (7) π/3 (8) (9) Extra question: How does the volume integrated over look? 4..3 Exercise: Variable transformation (a) Calculate the integral of the function f(x, y) over the dashed area of the figure. Give a geometric interpretation for this integral. φ r CC BY-SA 4. Hint: The integral of y 2 is 2 (y y 2 + arcsin(y)). Solution: The easiest way is to realize that the area is the difference between a quarter unit circle of area π/4 minus half a unit square of area /2. 6

7 The most direct way is to simply calculate the integral. y 2 dx dy y [ ] y 2 x dy () y y2 ( y) dy (2) y2 dy dy + [ 2 (y y 2 + arcsin(y)) ] 2 [arcsin() arcsin() ] + }{{}}{{} 2 π/2 y dy (3) [ ] [ ] y + 2 y2 (4) (5) π 4 2 (6) Another way is to split the integral up into the difference of the integral over the quarter circle (in polar coordinates) and the integral over the triangle (in x-y-coordinates). π/2 x x y r φ dx dy y y dr dφ dx dy (7) r φ A }{{} r cos φ cos φ+ r sin φ sin φr π/2 π/2 [ 2 φ [ 2 r2 ] 2 dφ ] π/2 dφ [ x ] y dy (8) y dy (9) [ 2 y2 + y ] π/4 ( /2 + ) π 4 2 () () (b) Calculate the integral of the function f(x, y) /(x 2 +y 2 ) 3/2 over the same area. Use polar coordinates (r, φ) with x r cos φ, y r sin φ. Solution: First determine the boundaries of the area in polar coordinates: Since we look only at the upper right quadrant we have angles from to 9. This means φ [, π/2]. The upper bound of the radius is just the radius. The lower bound is given by the function y x. If x and y are substituted by polar coordinates we get r sin φ r cos φ and therefore r [/(cos φ + sin φ), ]. 7

8 A y π/2 y 2 /(x 2 + y 2 ) 3/2 dx dy (2) x x r φ r 3 y y dr dφ (3) π/2 π/2 π/2 /(cos φ+sin φ) /(cos φ+sin φ) ] [ r /(cos φ+sin φ) r φ } {{ } r cos φ cos φ+ r sin φ sin φr dr dφ (4) r2 dφ (5) + (cos φ + sin φ) dφ (6) [ φ + sin φ cos φ ] π/2 (7) ( π/2 + sin(π/2) cos(π/2)) ( + sin cos ) (8) π/ π/2 (9) 4..4 Exercise: Area integral Consider a shape in the xy-plane with a boundary given by x ( + a cos(nφ)) cos(φ) () y 3( + a cos(nφ)) sin(φ) (2) for any a [, ] and n N + (without ). (a) Sketch the shap for a.5 and n 6. Solution: The correct sketch is shown on the left. If y is scaled by /3 the figure looks nicer (right). Figure (left): (Wiskott group, 27, unclear); Figure (right): (Wiskott group, 27, unclear) (b) Calculate the area of the shape for general a and n. Hint: First apply a very simple geometrical transformation that makes the task much easier. Solution: First we compress everything in the y-direction by a factor of /3 to get a more symmetric boundary with only /3 of the area. x ( + a cos(nφ)) cos(φ) (3) ỹ ( + a cos(nφ)) sin(φ). (4) 8

9 The integral is then best written in polar coordinates. I/3 2π 2π (+a cos(nφ)) r dr dφ (5) [r 2 (+a cos(nφ)) /2] dφ (6) 2π (( + a cos(nφ)) 2 /2 /2) dφ (7) 2 2π ( + 2a cos(nφ) + a 2 cos(nφ) 2) dφ (8) 2π 2π dφ + a cos(nφ) dφ + a2 2 2 } {{ } 2π cos(nφ) 2 dφ } {{ } π π + + a 2 π/2 () ( + a 2 /2)π () I 3( + a 2 /2)π. (2) The last integral in (9) equals π, because 2π π dφ π (sin(φ)2 + cos(φ) 2 ) dφ and for symmetry reasons. (c) Discuss the plausibility of the result. Solution: The result is plausible in several aspects: (i) The area equals 3π for a, because in that case the shape is an ellipse with radii and 3. (ii) The result does not depend on n for symmetry reasons. After compression in y-direction by a factor of /3 one could rearrange infinitesimally small radial slices to get from any n to any other n (except ) without changing the overall area. (iii) The area increases with a, because what gets subtracted from the inner of the ellipse gets added even more to the outer of the ellipse, because of the radial magnification factor. (9) 5 Gauss theorem - integration by parts in multiple dimensions 5. Exercises 5.. Exercise: Gauss theorem Consider the two-dimensional vector field f(x) x (x, x 2 ) T. () (a) Sketch the vector field. Solution: Not available! (b) Calculate the divergence T f(x). 9

10 Solution: T f(x) f x + f 2 x 2 (2) () x + x 2 (3) x x 2 + (4) 2. (5) (c) Calculate the integral over the divergence over the area of a disc of radius r centered at the origin. Solution: Since the divergence is 2 everywhere the integral is simply twice the area of the disc, i.e. 2πr 2. (d) Calculate the flow of the vector field through the boundary of the disc with radius r centered at the origin. Hint: By Gauss theorem the result should be identical to the integral over the divergence. Solution: At radius r the vectors of the vector field have length r and point strictly outwards, away from the origin, i.e. they are orthogonal to the boundary of the disc. Thus the flow through the boundary is simply r times the circumference, i.e. r2πr, which is the same as the integral over the divergence over the disc, see above. (e) Calculate the integral over the divergence over the area of a disc of radius r centered at an arbitrary point x c. Solution: Since the divergence is 2 everywhere in any case the integral is still simply twice the area of the disc, i.e. 2πr 2. (f) Calculate the flow of the vector field through the boundary of the disc with radius r centered at x c. Hint: Because of the linearity of the integral, the flow through a volume boundary of a sum of two vector fields equals the sum of the flows through the volume boundary of the two vector fields individually. Solution: First consider the flow through the boundary of the disc for the vector field f(x) x c. This vector field looks identical to f(x) itself except that it is centered at x c. Thus its flow through the boundary is exactly the one we have calculated for f(x) and the disc at the origin. Next we realize that the flow of the constant vector field x c through the boundary of any closed volume is zero for symmetry reasons. Thus the flow of f(x) through the disc centered at x c is again r2πr. Hint: There is actually not much to calculate here. Use symmetry arguments and Gedankenexperiments instead. Extra question: How would you calculate the flow through the boundary of a rectangle?