The Change-of-Variables Formula for Double Integrals

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1 The Change-of-Variables Formula for Double Integrals Minh-Tam Trinh In this document, I suggest ideas about how to solve some of the exercises in ection 15.9 of tewart, Multivariable Calculus, 8th Ed.

2 The Change-of-Variables Formula for Double Integrals 2 1. Exercises 7-10 We are given a set in the uv-plane and a transformation from the uv-plane to the xy-plane. We are asked to find the image of under the transformation. Exercise 7. = {(u,v) : 0 u 3, 0 v 2} and the transformation is (x, y) = (2u + 3v, u v). Graphing shows that it is a rectangle. To determine what happens to under the transformation, we just need to find what happens to each of its sides. For example, the rightmost side of the rectangle is the line segment between (3, 0) and (3, 2) in the uv-plane, i.e., the set of points where u = 3 and 0 v 2. If u = 3, then (1.1) (x, y) = (6 + 3v, 3 v) = (6, 3) + (3, 1)v. Now, if v runs from 0 to 2, then (x, y) runs from (6, 3) to (12, 1). The image is a straight line segment between these points, because (1.1) is the vector equation of a line. (In other words, x = 6 + 3v and y = 3 v are both linear functions of v.) To finish the problem, one has to do this procedure for the other three sides. You should find that the image of is a parallelogram. Exercise 9. is the triangle with vertices (0, 0), (1, 1), (0, 1) and the transformation is (x, y) = (u 2,v). Again, is a polygon, so we need to find what happens to each of its sides. The most interesting case is the side from (0, 0) to (1, 1). This is the set of points in the uv-plane of the form (t, t), where 0 t 1. Plugging (u,v) = (t, t) into (x, y) = (u 2,v) gives (1.2) (x, y) = (t 2, t). Looking at the second coordinate, we see that y = t, i.e., we can write y wherever we see t. Then the first coordinate gives x = t 2 = y 2, so the image is the graph of x = y 2 as y runs from 0 to 1. Again, to finish, one has to repeat this analysis for the other two sides of the triangle. You should find that the other sides don t change, and only the side discussed above gets smushed by the transformation.

3 The Change-of-Variables Formula for Double Integrals 3 2. Exercises We are given a region in the xy-plane. We are asked to find a cartesian rectangle in the uv-plane and a transformation T that maps onto. emark. In each of these problems, is some kind of smushed rectangle, bounded by four sides. We need T to transform lines of the form u = a and u = b onto one pair of sides, and transform lines of the form v = c and v = d onto the other pair of sides. The constants a, b, c, d will come from the equations given to us in the problem. Exercise 11. is bounded by the four lines (2.1) y = 2x 1, y = 2x + 1, y = 1 x, y = 3 x. Graphing shows that this is a parallelogram. The lines y = 2x 1 and y = 2x + 1 contain one pair of opposite sides, while y = 1 x and y = 3 x contain the other pair. It s reasonable to guess that T sends opposite sides of the rectangle to opposite sides of the parallelogram. o let us assume T transforms u = a and u = b into the first pair of lines, while it transforms v = c and v = d into the second pair of lines. 1 We observe that y = 2x 1 and y = 2x + 1 only differ in their constant term. As u runs from 1 to 1, the line y = 2x + u runs from y = 2x 1 to y = 2x + 1, which is another way of saying that: The relation y = 2x + u transforms the lines u = 1 and u = 1 onto the lines y = 2x 1 and y = 2x + 1, respectively. In a similar way, The relation y = v x transforms the lines v = 1 and v = 3 onto the lines y = 1 x and y = 3 x, respectively. Therefore, to find T, we need to solve the system of equations { y = 2x + u (2.2) y = v x for x and y. The rectangle will be the set bounded by u = 1 and u = 1 and v = 1 and v = 3. Exercise 13. is the region in the first quadrant bounded by the circles x 2 + y 2 = 1 and x 2 + y 2 = 2. Graphing shows that this is a polar rectangle, namely, a quarter-ring shape. Its inner curved side corresponds to x 2 + y 2 = 1 and its outer curved side corresponds to x 2 + y 2 = 2. Its other two sides are on the x- and y-axes, i.e., on the lines y = 0 and x = 0, respectively. Again, we assume that T transforms opposite sides of onto opposite sides of. More precisely, suppose T takes u = a and u = b onto the two curved sides. We notice that x 2 + y 2 = 1 and x 2 + y 2 = 2 only differ in their constant term. As u runs from 1 to 2, the circle x 2 + y 2 = u 2 expands from x 2 + y 2 = 1 to x 2 + y 2 = 2, i.e.: 1 If you switch the roles of u and v here, then T will change by a flip in orientation, but it will still be a valid answer. The only important thing is that u = a and u = b have to go to opposite sides, as do v = c and v = d.

4 The Change-of-Variables Formula for Double Integrals 4 The relation x 2 + y 2 = u 2 transforms the lines u = 1 and u = 2 onto the circles x 2 + y 2 = 1 and x 2 + y 2 = 2, respectively. (Notice that we could have used x 2 + y 2 = u instead of x 2 + y 2 = u 2, in which case u would run from 1 to 2, instead of from 1 to 2. But I will stick with x 2 + y 2 = u 2, because now, it s really natural to think of u as a radius.) It is more tricky to see how to deal with the other two lines, y = 0 and x = 0. The easiest way to finish the problem is to realize, from the fact that is a polar rectangle, that we ultimately want T to be a polar substitution: (2.3) x = u cosv and y = u sinv. Thinking of v as an angle, we see that this substitution takes the line v = 0 onto the line y = 0 (i.e., x-axis, which is angled at 0 radians), and the line v = π/2 onto the line x = 0 (i.e., the y-axis, which is angled at π/2 radians).

5 The Change-of-Variables Formula for Double Integrals 5 3. Exercises We are given a double integral over a region in the xy-plane and a transformation T from the uv-plane to the xy-plane. We are asked to evaluate the integral using T. Exercise 15. is the triangular region with vertices (0, 0), (2, 1), (1, 2), the integral is (3.1) (x 3y) da, and T is (x, y) = (2u + v, u + 2v). First, we need to find the region that gets mapped onto by the transformation T. One way to do this is to find the transformation U from the xy-plane to the uv-plane that reverses what T does. To find U, we back-solve { x = 2u + v (3.2) y = u + 2v for u and v. Doing so, you should get: { u = (2x y)/3 (3.3) v = (2y x)/3 Now, will be the image of under U, so we can find it in exactly the same way we would solve Exercises 7-10 (except now we re going from the xy-plane to the uv-plane). By plugging the vertices of into (3.3), you should find that is the triangle in the uv-plane with vertices (0, 0), (1, 0), (0, 1). The change-of-variables formula now says (3.4) The Jacobian is (x 3y) da = (x, y) ((2u + v) 3(u + 2v)) (u,v) da. (3.5) x u y v x v y u = 2(2) 1(1) = 3, so we end up evaluating (3.6) ( u 5v) 3 da. ince is both a Type I and a Type I I region in tewart s terminology, you can finish the problem using the ideas in ection Exercise 18. is the interior of the ellipse x 2 xy + y 2 = 2, the integral is (3.7) (x 2 xy + y 2 ) da, and T is (x, y) = ( 2u 2/3v, 2u + 2/3v).

6 The Change-of-Variables Formula for Double Integrals 6 Again, we need to find the region that gets mapped onto by T. In the previous problem, we did this by finding the inverse transformation to T. Here, we re given the equation of the ellipse that bounds, so a much simpler way to find is to plug x = 2u 2/3v and y = 2u + 2/3v into this equation to see what it does to the ellipse. You should find that (3.8) ( 2u 2/3v) 2 ( 2u 2/3v)( 2u + 2/3v) + ( 2u + 2/3v) 2 = 2u 2 + 2v 2, so the ellipse x 2 xy + y 2 = 2 corresponds to the circle 2u 2 + 2v 2 = 2 in the uv-plane, i.e., the unit circle x 2 + y 2 = 1. The change-of-variables formula now says (3.9) (x 2 xy + y 2 ) da = (2u 2 + 2v 2 (x, y) ) (u,v) da, where is the interior of the unit circle. The Jacobian is (3.10) x u y v x v y u = 2( 2/3) ( 2/3)( 2) = 2/ 3 + 2/ 3 = 4/ 3, so we end up evaluating (3.11) (2u 2 + 2v 2 ) 4 3 da. ince is a circular region, perhaps the easiest way to finish the problem is to apply a polar substitution to.