Chapter 8 Change of Variables, Parametrizations, Surface Integrals

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Chapter 8 Change of Variables, Parametrizations, Surface Integrals Chapter 8 Change of Variables, Parametrizations, Surface Integrals 8.1 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 b Z 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 '(b) Z 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: 1 Theorem 1. 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∗ 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 the next section. 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. 8.2 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)g ~ :(u; v) 7! (u; g(u; v)) or of the form (P 2) h~ :(u; v) 7! (h(u; v); v); with g; h continuously differentiable (i.e. C1) and @g=@v; @h=@u nowhere vanishing on D. The condition @g=@v 6= 0 ensures thatg ~ is bijective and similarly for h~. If ' is a linear map then it can be written as the composite of primitive linear maps. Indeed a linear map of the form (P1) looks like 1 a (u; v) 7! (u; v) 0 b and a linear map of the form (P2) like a0 0 (u; v) 7! (u; v) b0 1 and we have 1 a a0 0 a0 + ab0 a a0 0 1 a a0 aa0 · = ; · = : 0 b b0 1 bb0 b b0 1 0 b b0 b + ab0 Now given any 2 × 2 matrix M one can always solve for a; b; a0; b0 for at least one order of the factors. Hence M will be a product of primitive linear maps. 2 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), i.e. φ =g ~ ◦ h~ and for f a continuous function. 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: so that we have bijective maps h~ g~ ∗ D −!D1 −!D whose composite is φ. 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) (it is a region of type I in previous terminology). Then we have x +l Zm ∗ area(Rmr) = (g(x; vr + k) − g(x; vr)) dx: xm By the mean value theorem of 1-variable integral calculus and continuity of g, we can write ∗ 0 0 area(Rmr) = l[g(xm; vr + k) − g(xm; vr)]; 0 for some point xm 2 (xm; xm +l): By the mean value theorem of 1-variable differential calculus, we get @g area(R∗ ) = lk (x0 ; v0 ); mr @v m r 0 ∗ for some vr 2 (vr; vr + k). Now for any continuous function f on D we have ZZ ZZ X 0 0 0 ∗ f(x; y) dx dy = lim sP dx dy = lim f(xm; g(xm; vr))area(Rmr) P P D∗ D∗ m;r 0 0 0 ∗ where sP is the step function with constant value f(xm; g(xm; vr)) on Rmr. Indeed, given > 0, by the small span theorem applied to the continuous function (x; v) 7! f(x; g(x; v)) we can find a partition P so that jf(x0; g(x0; v0)) − f(x; g(x; v))j < area(D∗) 3 0 0 for (x; v); (x ; v ) contained in the same subrectangle Rmr of P . But then ZZ ZZ X 0 0 0 (f(x; y) − sP ) dx dy ≤ jf(x; y) − f(x ; g(x ; v ))j dx dy < . m m r ∗ m;r ∗ D Rmr Hence 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 D∗ m;r 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 @g RR @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 the chain rule that @g @h ~ 1 @u (h(u; v); v) @u (u; v) 0 D'(u; v) = Dg~(h(u; v); v) · Dh(u; v) = @g · @h 0 @v (h(u; v); v) @v (u; v) 1 and @g @h det D'(u; v) = (h(u; v); v) · (u; v) @v @u which is by hypothesis > 0. Thus we get RR f(x; y) dx dy = RR 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 finite 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. 4 8.3 Examples (1) Let Φ = f(x; y) 2 R2jx2 +y2 ≤ a2g; with a > 0: We know that A = area(Φ) = πa2: But let us now do this by using polar coordinates. Put 2 D = f(r; θ) 2 R j0 ≤ r ≤ a; 0 ≤ θ < 2πg; and define ' : D! Φ by '(r; θ) = (r cos θ; r sin θ): Then it is easy to check that ' is 1-1, onto, and moreover, @' @' = (cos θ; sin θ) and = (−r sin θ; r cos θ): @r @θ Hence cos θ −r sin θ det(D') = det = r; sin θ r cos θ which is non-negative, and is non-zero iff r > 0: The integral will not change if we integrate over fr ≥ 0; 0 ≤ θ ≤ 2πg or over fr > 0; 0 ≤ θ ≤ 2πg as the set fr = 0; 0 ≤ θ ≤ 2πg has content zero. By the transformation formula, we get a 2π ZZ ZZ Z Z A = dx dy = j det D'(r; θ)j dr dθ = r dr dθ = Φ D 0 0 a 2π a Z Z Z r2 a = r dr dθ = 2π r dr = 2π = πa2: 2 0 0 0 0 (Note that D is a rectangular strip, which is transformed by ' into a circular region.) RR (2) Compute the integral I = R xdxdy where R is the region f(r; φ) j 1 ≤ r ≤ 2; 0 ≤ φ ≤ π=4g. 2 p p R 2 R π=4 π=4 r3 i 2 7 2 We have I = 1 0 r cos(φ)rdrdφ = sin(φ)]0 · 3 = 2 (8=3 − 1=3) = 6 : 1 (3) Let Φ be the region inside the parallelogram bounded by y = 2x; y = 2x−2; y = x and y = x + 1: Evaluate I = RR xy dx dy.
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