Chapter 8 Change of Variables, Parametrizations, Surface Integrals x0. The transformation formula In evaluating any integral, if the integral depends on an auxiliary function of the variables involved, it is often a good idea to change variables and try to simplify the integral. The formula which allows one to pass from the original integral to the new one is called the transformation formula (or change of variables formula). It should be noted that certain conditions need to be met before one can achieve this, and we begin by reviewing the one variable situation. Let D be an open interval, say (a; b); in R , and let ' : D! R be a 1-1 , C1 mapping (function) such that '0 6= 0 on D: Put D¤ = '(D): By the hypothesis on '; it's either increasing or decreasing everywhere on D: In the former case D¤ = ('(a);'(b)); and in the latter case, D¤ = ('(b);'(a)): Now suppose we have to evaluate the integral Zb I = f('(u))'0(u) du; a for a nice function f: Now put x = '(u); so that dx = '0(u) du: This change of variable allows us to express the integral as Z'(b) Z I = f(x) dx = sgn('0) f(x) dx; '(a) D¤ where sgn('0) denotes the sign of '0 on D: We then get the transformation formula Z Z f('(u))j'0(u)j du = f(x) dx D D¤ This generalizes to higher dimensions as follows: Theorem Let D be a bounded open set in Rn;' : D! Rn a C1, 1-1 mapping whose Jacobian determinant det(D') is everywhere non-vanishing on D; D¤ = '(D); and f an integrable function on D¤: Then we have the transformation formula Z Z Z Z ¢ ¢ ¢ f('(u))j det D'(u)j du1::: dun = ¢ ¢ ¢ f(x) dx1::: dxn: D D¤ 1 Of course, when n = 1; det D'(u) is simply '0(u); and we recover the old formula. This theorem is quite hard to prove, and we will discuss the 2-dimensional case in detail in x1. In any case, this is one of the most useful things one can learn in Calculus, and one should feel free to (properly) use it as much as possible. It is helpful to note that if ' is linear, i.e., given by a linear transformation with associated matrix M, then D'(u) is just M. In other words, the Jacobian determinant is constant in this case. Note also that ' is C1, even C1, in this case, and moreover ' is 1 ¡ 1 i® M is invertible, i.e., has non-zero determinant. Hence ' is 1 ¡ 1 i® det D'(u) 6= 0. But this is special to the linear situation. Many of the cases where change of variables is used to perform integration are when ' is not linear. x1. The formula in the plane Let D be an open set in R2: We will call a mapping ' : D! R2 as above primitive if it is either of the form (P 1) (u; v) ! (u; g(u; v)) or of the form (P 2) (u; v) ! (h(u; v); v); with @g=@v; @h=@u nowhere vanishing on D: (If @g=@v or @h=@u vanishes at a ¯nite set of points, the argument below can be easily extended.) We will now prove the transformation formula when ' is a composition of two primitive transformations, one of type (P1) and the other of type (P2). For simplicity, let us assume that the functions @g=@v(u; v) and @h=@u(u; v) are always positive. (If either of them is negative, it is elementary to modify the argu- ¤ ment.) Put D1 = f(h(u; v); v)j(u; v) 2 Dg and D = f(x; g(x; v))j(x; v) 2 D1g: By hypothesis, D¤ = '(D): Enclose D1 in a closed rectangle R; and look at the intersection P of D1 with a partition of R; which is bounded by the lines x = xm; m = 1; 2; :::; and v = vr; r = ¤ 1; 2; :::; with the subrectangles Rmr being of sides 4x = l and 4v = k: Let R ; ¤ respectively Rmr, denote the image of R, respectively Rmr; under (u; v) ! (u; g(u; v)): ¤ Then each Rmr is bounded by the parallel lines x = xm and x = xm + l and by the arcs of the two curves y = g(x; vr) and y = g(x; vr + k): Then we have xZm+l ¤ area(Rmr) = (g(x; vr + k) ¡ g(x; vr)) dx: xm By the mean value theorem of 1-variable integral calculus, we can write ¤ 0 0 area(Rmr) = l[g(xm; vr + k) ¡ g(xm; vr)]; 2 0 for some point xm in (xm; xm +l): By the mean value theorem of 1-variable di®erential calculus, we get @g area(R¤ ) = lk (x0 ; v0 ); mr @v m r 0 0 for some xm 2 (xm; xm + l) and vr 2 (vr; vr + k): So, for any function f which is integrable on D¤; we obtain ZZ X 0 0 0 @g 0 0 f(x; y) dx dy = lim klf(xm; g(xm; vr)) (xm; vr): P @v m;r D¤ RR @g The expression on the right tends to the integral f(x; g(x; v)) @v (x; v) dx dv: Thus D1 we get the identity RR RR @g f(x; y) dx dy = f(x; g(x; v)) @v (x; v) dx dv ¤ D D1 By applying the same argument, with the roles of x; y (respectively g; h) switched, we obtain RR RR @g @g @h f(x; g(x; v)) @v (x; v) dx dy = f(h(u; v); g(h(u; v); v)) @v (h(u; v); v) @u (u; v) du dv D1 D Since ' = g ± h we get by chain rule that @h @g det D'(u; v) = (u; v) (h(u; v); v); @u @v which is by hypothesis > 0: Thus we get RR RR f(x; y) dx dy = f('(u; v))j det D'(u; v)j du dv D¤ D as asserted in the Theorem. How to do the general case of '? The fact is, we can subdivide D into a ¯nite union of subregions, on each of which ' can be realized as a composition of primitive trans- formations. We refer the interested reader to chapter 3, volume 2, of "Introduction to Calculus and Analysis" by R. Courant and F. John. x2. Examples 3 (1) Let © = f(x; y) 2 R2jx2 +y2 · a2g; with a > 0: We know that A = area(©) = ¼a2: But let us now do this by using polar coordinates. Put D = f(r; θ) 2 R2j0 < r < a; 0 < θ < 2¼g; and de¯ne ' : D! © by '(r; θ) = (r cos θ; r sin θ): Then D is a connected, bounded open set, and ' is C1, with image D¤ which is the complement of a negligible set in ©; hence any integration over D¤ will be the same as doing it over ©. Moreover, @' @' = (cos θ; sin θ) and = (¡r sin θ; r cos θ): @r @θ Hence µ ¶ cos θ ¡r sin θ det(D') = det = r; sin θ r cos θ which is positive on D. So ' is 1 ¡ 1. By the transformation formula, we get ZZ ZZ Za Z2¼ A = dx dy = j det D'(r; θ)j dr dθ = r dr dθ = © D 0 0 Za Z2¼ Za ¸ r2 a = r dr dθ = 2¼ r dr = 2¼ = ¼a2: 2 0 0 0 0 We can justify the iterated integration above by noting that the coordinate func- tion (r; θ) ! r on the open rectangular region D is continuous, thus allowing us to use Fubini. RR (2) Compute the integral I = R xdxdy where R is the region f(r; Á) j 1 · r · 2; 0 · Á · ¼=4g. R R i2 p p 2 ¼=4 ¼=4 r3 2 7 2 We have I = 1 0 r cos(Á)rdrdÁ = sin(Á)]0 ¢ 3 = 2 (8=3 ¡ 1=3) = 6 : 1 (3) Let © be the region insideRR the parallelogram bounded by y = 2x; y = 2x¡2; y = x and y = x + 1: Evaluate I = xy dx dy. © The parallelogram is spanned by the vectors (1; 2) and (2; 2), so it seems reasonable to make the linear change of variable µ ¶ µ ¶ µ ¶ x 1 2 u = y 2 2 v 4 since then we'll have to integrate over the box [0; 1] £ [0; 1] in the u ¡ v-region. The Jacobian matrix of this transformation is of course constant and has determinant ¡2. So we get Z 1 Z 1 Z 1 Z 1 I = (u + 2v)(2u + 2v)j ¡ 2jdudv = 2 (2u2 + 6uv + 4v2)dudv 0 0 0 0 Z 1 Z 1 3 2 2 ¤1 2 = 2 2u =3 + 3u v + 4v u 0 dv = 2 (2=3 + 3v + 4v )dv 0 0 2 3 ¤1 = 2(2v=3 + 3v =2 + 4v =3 0) = 2(2=3 + 3=2 + 4=3) = 7: (4) Find the volume of the cylindrical region in R3 de¯ned by W = f(x; y; z)jx2 + y2 · a2; 0 · z · hg; where a; h are positive real numbers. We need to compute ZZZ I = vol(W ) = dx dy dz: W It is convenient here to introduce the cylindrical coordinates given by the trans- formation ' : D! R3 given by '(r; θ; z) = (r cos θ; r sin θ; z); where D = f0 < r < a; 0 < θ < 2¼; 0 < z < hg: It is easy to see ' is 1-1, C1 and onto the interior D¤ of W: Again, since the boundary of W is negligible, we may just integrate over D¤.