QMUL, School of Physics and Astronomy Date: 18/01/2019

Transcription

1 QMUL, School of Physics and stronomy Date: 8//9 PHY Mathematical Techniques Solutions for Exercise Class Script : Coordinate Systems and Double Integrals. Calculate the integral: where the region is defined as: xy dx { (x, y) R : x >, y >, x + y < } Include a sketch of the region and use an appropriate change of variables to calculate this integral. [Comment: this is a type of exercise you could find in a sub-question of section B of the final MT exam.] Demonstrators should remind students that () first of all they need to consider the region over which they need to integrate, understand it and sketch it, () then think about possible symmetries/simplification that can be done based on the shape of this region, () finally consider the integrand (function to integrate) with its own symmetries. Remind the students not to forget the integrand. They need to stu the integral region, but they need to remember what the problem asks. In this case the integrand is quite simple, but we need to think about the integral region. The first two conditions are telling us that we are in the first quadrant of the plane. The condition x + y < tells us that we need to consider the area limited by the ellipse: x + y The natural change of variables can use elliptic coordinates that are just a minor modification of the polar ones. In general if we have an ellipse with equation: the elliptic coordinates are: x a + y b { x ar cos θ y br sin θ Demonstrators should write this above on the whiteboard for the students. Students might not have seen this before so take some time to explain/clarify.

2 Demonstrators should also remind the students that, while doing the change of variables, they need to translate also the infinitesimal area: from d dx to d J dudv, where J is the absolute value of the determinant of the Jacobian matrix (or the Jacobian). Hence once decided the change of variables above, students need to calculate the Jacobian: x x det J r θ In this case we have: r θ x r θ x θ r a cos θ(b r cos θ) ( a r sin θ)b sin θ ab r cos θ + ab r sin θ ab r x r cos θ y r sin θ and hence the Jacobian will be r. The next thing we need to think about are the integration limits: we need to stay in the first quadrant hence: < r < < θ < π Demonstrators should encourage the students to think about the initial region and also about how this transforms (as ). Now we can rewrite the initial integral with this set of variables: π xy dx r cos θ sin θ r drdθ

3 now as we can separate the variables we can separate the integral into the product of the two integrals in r and θ: ( ) ( ) π r dr cos θ sin θ dθ [ ] r [ ] sin π θ 6. Calculate the integral: where the region Ω is defined as: Ω y dx Ω { (x, y) R : x >, y >, x y x, xy } Sketch the region Ω and use an appropriate change of variables to calculate this integral. [Comment: this is a type of exercise you could find in the section B of the final MT exam and it is on the hard side. In this case you are given this integral to have the opportunity to try and understand the proposed method of solving it via a change of variables.] gain we need to think about the integral region. The condition x y x tells us that we need to consider the area between the two lines y x and y x. The other condition xy tells us that we have to consider the area between the two curves xy and xy which are hyperbolae. Demonstrators should introduce this specific change of variable and ask the students to try to use it. The useful change of variables we can think of is the following: u xy v y x This simplifies greatly the area as it is now a rectangular area: x y x y x v xy u So now we need to substitute in our integral the new set of variables and we need to invert the system as well: u x v y uv

4 Thus the integral becomes: ( ) (x, y) uv (u, v) dudv We need to calculate the determinant of the Jacobian matrix which is defined as: (x, y) J(u, v) (u, v) x x u v We can use the property of the Jacobian: Thus we get: J(x, y) (u, v) (x, y) which corresponds to: J(u, v) Back to the integral, we get: u x v x u v (x, y) (u, v) J(x, y) (u,v) (x,y) u v J(u, v) y x y ( x xy ) y x x v y x x J(x, y) v x y (uv) v dudv dv [ u u dudv ] u du [v] where in the last step we have separated the two variables and integrated independently. ( )( ). Integrate the function f(x, y) x over the domain Ω defined as: Ω { (x, y) R : y x } Sketch of the area Ω. Solve this integral using both the methods by vertical lines and by horizontal lines, verifying that both methods bring the same result. [Comment: this is a type of exercise you could find in the section B of the final MT exam.] Let s first understand the area Ω. If x <, the absolute value means that we need to have a such that x + x x for x <. If x >, we have x x + x. So we have a triangular shape area as shown in the figure.

5 In this case we would choose to integrate by horizontal lines so that we can do it in one single integral. y x dx x dx [ x ] y y y ( ( y) y ) ( 8 y + 6y y y ) ( 6y + y y ) ] [y y + y y ( ) + ( ) ( ) Now we need to use also the method by vertical lines, then we have to write two integrals: x x x dx x dx + x dx x [y] x dx + x dx + x [y] x dx x ( x)dx [ ] x + (x x )dx [ ] + x x ( + + )

6 s expected we got the same result from the two methods.. Integrate the function f(x, y) (x + y) over the domain Ω defined as: { Ω (x, y) R : < y <, y < x < } y Perform the integration using: (a) Cartesian coordinates. Integration range is an 5 slice of a circle (from ). In this case, it is different if we fix x to first integrate over y (by vertical lines) or vice-versa (by horizontal lines). In the case of fixing x, for y we have to consider two parts: the first part corresponds simply to a triangle with vertices (, ), ( /, ), (, ). The second part corresponds to the circle part. So the integral I can be obtain from the sum of two integrals, I + I. What we can choose is to fix y and integrate first on x so that we can consider one unique integral. So, as given in the problem, for y we have the limits [, ] and for x we have [y, y ]. Demonstrators should encourage the students to think about these issues above before starting to solve the integral. By horizontal lines, the integral is: / / / / / y y dx(x + y) [xy + y x xy [ x y ] + xy x y xy [ y y + y ] y y [ y + y ] y [ y y ] ( y ) / / [ ] 6 + ] (b) polar coordinates. 6

7 Sketch the area Ω. [Comment: this is a type of exercise you could find in the section B of the final MT exam.] If we want to use the polar coordinate, we need to think about the new integration range first. r should be within [, ], θ should be within [, π/] Then transform the function: f(x, y) (x + y) r cos θ + r sin θ r(cos θ + sin θ) So reminding that the area element in polar coordinate is rdrdθ, then the integral is: π/ dr dθr (cos θ + sin θ) ( ) ( π/ ) r dr (cos θ + sin θ)dθ [ r ] r [sin θ cos θ] θπ/ θ r 7