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The Jacobian and Change of Variables

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The Jacobian and Change of Variables The Jacobian and Change of Variables Icon Placement: Section 3.3, p. 204 This set of exercises examines how a particular determinant called the Jacobian may be used to allow us to change variables in double and triple integrals. First, let us review how a change of variables is done in a single integral: this change is usually called a substitution. If we are attempting to integrate f(x) over the interval a ≤ x ≤ b,we can let x = g(u) and substitute u for x, giving us Z b Z d f(x) dx = f(g(u))g0(u) du a c 0 dx where a = g(c)andb = g(d). Notice the presence of the factor g (u)= du . A similar factor will be introduced when we study substitutions in double and triple integrals, and this factor will involve the Jacobian. We will now study changing variables in double and triple integrals. These intergrals are typically studied in third or fourth semester calculus classes. For more information on these integrals, consult your calculus text; e.g., Reference 2, Chapter 13, \Multiple Integrals." We assume that the reader has a passing familiarity with these concepts. To change variables in double integrals, we will need to change points (u; v)topoints(x; y). 2 2 → Ê That is, we will have a transformation T : Ê with T (u; v)=(x; y). Notice that x and y are functions of u and v;thatis,x = x(u; v)andy = y(u; v). This transformation T may or may not be linear. Example 1: A well-known change of variables is the change from rectangular to polar coordinates, which is accomplished by x = r cos θ, y = r sin θ.Herewemayidentifyu with r and v with θ. Thus the transformation may be written as T (r; θ)=(r cos θ, r sin θ). Example 2: The transformation T (u; v)=(3u − 2v;u + v) is a linear transformation, and describes the change of variables x =3u − 2v, y = u + v. Example 3: The transformation T (u; v)=(u2 − v2; 2uv) is accomplished with the changes x = u2 − v2, y =2uv. It is not a linear transformation. It is helpful to consider how transformations change regions in the uv plane to regions in the xy plane. Example 4:LetT be the transformation in Example 3, and consider the region S in the rθ plane given by S = {(u; v)|0 ≤ u ≤ 1; 0 ≤ v ≤ 1}. See Figure 1. Our goal is to find its image R = T (S)inthexy plane. We begin by finding the images of the sides of S. The side 1 2 2 S1 is given by v =0and0≤ u ≤ 1. When v = 0, the equations x = u − v and y =2uv become x = u2;y =0; with 0 ≤ x ≤ 1: Thus the image of S1 under T is the line segment from (0; 0) to (1; 0) in the xy plane. The 2 2 side S2 is given by u =1and0≤ v ≤ 1. When u = 1, the equations x = u −v and y =2uv become x =1− v2 and y =2v,with0≤ v ≤ 1. Eliminating v we have y2 x =1− ; with 0 ≤ x ≤ 1: 4 Thus the image of S2 a portion of a parabola. In like manner, we may find that the image of S3 is y2 x = − 1; and − 1 ≤ x ≤ 0; 4 and the image of S4 is the line segment from (−1; 0) to (0; 0) in the xy plane. The complete image of S (which we call T (S)) is also shown in Figure 1. v y 2 2 S3 THS3L THS2L 1 1 S4 S2 u x -2 -1 S1 1 2 -2 -1 THS4L THS1L 1 2 -1 -1 -2 -2 Figure 1: The region S and its image T (S) We return to Example 2 above and think about changing the variables of integration from x and y to u and v, respectively. To begin, consider the integral ZZ 1 dA; R where R is a region in the xy plane and dA = dx dy is the differential of area in the xy plane. Suppose that R is the image of the region S in the uv plane. Since the integral is just the area of the region R, Theorem 10 on p. 202 of Linear Algebra and its Applications and the observation in the colored box on p. 203 tell us that ZZ ZZ 1 dx dy = 1 dA R R = {area of T (S)} 2 | |·{ } = det B ZZarea of S = |det B| 1 du dv ZZ S = |det B| du dv S RR RR In this case, then, R 1 dx dy = S 5 du dv. If we were to integrate a function f(x; y)over R, it now seems plausible that we could change variables to find ZZ ZZ f(x; y) dx dy = f(x(u; v);y(u; v))|detB| du dv R S This is a `Change of Variables' theorem when the change of variables is a linear transforma- tion. But what if change of variables transformation is not linear? Note that for the general linear transformation x = au + bv, y = cu + dv, the determinant of the standard matrix is @x @x ab @u @v = @y @y cd @u @v It may be shown (see Reference 1, Sections 6.2 and 6.3) that the absolute value of this determinant is the factor to use in changing variables. This determinant is given a special name. Definition:TheJacobian of the transformation T given by x = x(u; v), y = y(u; v)is @x @x @(x; y) @u @v @x @y @x @y = @y @y = − @(u; v) @u @v @u @v @v @u This definition leads directly to the following theorem. Change of Variables Theorem for Double Integrals: Suppose that T is a one-to-one transformation given by x = x(u; v)andy = y(u; v), and that the first partial derivatives of x and y with respect to u and v are continuous functions. If the Jacobian of T is nonzero and if T maps the region S in the uv plane onto the region R = T (S)inthexy plane, if the function f is continuous on R and if the boundaries of R and S are sufficiently regular, then ZZ Z Z @(x; y) f(x; y) dx dy = f(x(u; v);y(u; v)) du dv R S @(u; v) Example 5: Integrate the function x + y over the region R in the first quadrant of xy plane bounded by the x-axis, the y-axis, and the arc x2 + y2 = 4 (See Figure 2). 3 y 2 1 R x 1 2 Figure 2: The region R Solution: To integrate, we change to polar coordinates: the region S so that R = T (S)is { | ≤ ≤ ≤ ≤ π } seen to be S = (r; θ) 0 r 2; 0 θ 2 . Thus the desired integral is ZZ ZZ @(x; y) x + ydxdy = (r cos θ + r sin θ) dr dθ R S @(u; v) Z π Z 2 2 = (r cos θ + r sin θ) rdrdθ 0 0 Z π Z 2 2 = 2(cosθ +sinθ) r2 dr dθ 0 0 Z π 8 2 = 2(cosθ +sinθ) dθ 3 0 16 = 3 There is a similar change of variables theorem for triple integrals. Let T be a transformation that maps a region S in uvw spacetoaregionR = T (S)inxyz space, by the equations x = x(u; v; w), y = y(u; v; w), and z = z(u; v; w). Then we may define the Jacobian as follows: Definition:TheJacobian of the transformation T given by x = x(u; v; w), y = y(u; v; w), and z = z(u; v; w)is @x @x @x @u @v @w @(x; y; z) @y @y @y = @u @v @w @(u; v; w) @z @z @z @u @v @w Then, under conditions similar to those the Change of Variables Theorem for Double Inte- grals, we have the following formula: 4 ZZZ ZZZ @(x; y; z) f(x; y; z) dx dy dz = f(x(u; v; w);y(u; v; w);z(u; v; w)) du dv dw R S @(u; v; w) Questions: 1. Show that the transformations in Examples 1 and 3 are not linear by considering whether T (u1 + u2;v1 + v2)=T (u1;v1)+T (u2;v2) for all choices of u1, u2, v1,andv2. 2. Show that the transformation in Example 2 is linear by the method suggested for Exercises 17-20 in Section 1.8 of the text. 2 3. In each part, draw a picture in Ê of the image set T (S). { | ≤ ≤ ≤ ≤ π } a) S = (r; θ) 0 r 1; 0 θ 2 , T (r; θ)=(r cos θ, r sin θ). { | ≤ ≤ − π ≤ ≤ π } b) S = (r; θ) 1 r 2; 4 θ 4 , T (r; θ)=(r cos θ, r sin θ). c) S = {(u; v)|0 ≤ u ≤ 1; 0 ≤ v ≤ 2}, T (u; v)=(3u − 2v;u + v). d) S = {(u; v)|−1 ≤ u ≤ 1; 1 ≤ v ≤ 2}, T (u; v)=(3u − 2v;u + v). 4. Use the transformation in Example 2 to evaluate the integral ZZ 2x − ydxdy R where R is the region bounded by the lines x +2y =0,x +2y = 10, −x +3y =0,and −x +3y =5. 5. Use polar coordinates to evaluate the integral ZZ 2 2 ex +y dx dy R where R is the region enclosed by the unit circle x2 + y2 =1. 6. Use the transformation in Example 3 to evaluate the integral ZZ ydxdy R where R is the region bounded by the x-axis and the parabolas y2 =4− 4x and y2 =4+4x.
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