Change of Variables, Parametrizations, Surface Integrals

Transcription

1 Chapter 8 Change of Variables, Parametrizations, Surface Integrals 8. he 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. he 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 ϕ : D R be a -, C mapping (function) such that ϕ on D. Put ϕ(d). By the hypothesis on ϕ, it s either increasing or decreasing everywhere on D. In the former case (ϕ(a), ϕ(b)), and in the latter case, (ϕ(b), ϕ(a)). Now suppose we have to evaluate the integral I b a f(ϕ(u))ϕ (u) du, for a nice function f. Now put x ϕ(u), so that dx ϕ (u) du. his change of variable allows us to express the integral as I ϕ(b) f(x) dx sgn(ϕ ) f(x) dx, ϕ(a) where sgn(ϕ ) denotes the sign of ϕ on D. We then get the transformation formula f(ϕ(u)) ϕ (u) du f(x) dx D his generalizes to higher dimensions as follows:

2 heorem. Let D be a bounded open set in R n, ϕ : D R n a C, - mapping whose Jacobian determinant det(dϕ) is everywhere non-vanishing on D, ϕ(d), and f an integrable function on. hen we have the transformation formula f(ϕ(u)) det Dϕ(u) du... du n f(x) dx... dx n. D Of course, when n, det Dϕ(u) is simply ϕ (u), and we recover the old formula. his 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 he formula in the plane Let D be an open set in R 2. We will call a mapping ϕ : D R 2 as above primitive if it is either of the form (P ) g : (u, v) (u, g(u, v)) or of the form (P 2) h : (u, v) (h(u, v), v), with g, h continuously differentiable (i.e. C ) and g/, h/ nowhere vanishing on D. he condition g/ ensures that g is bijective and similarly for h. If ϕ is a linear map then it can be written as the composite of primitive linear maps. Indeed a linear map of the form (P) looks like ( ) a (u, v) (u, v) b and a linear map of the form (P2) like (u, v) (u, v) ( a ) b and we have ( ) a b ( ) a b ( a + ab a bb b ), ( ) a b ( ) ( ) a a aa b b b + ab. Now given any 2 2 matrix M one can always solve for a, b, a, b for at least one order of the factors. Hence M will be a product of primitive linear maps. 2

3 We will now prove the transformation formula when ϕ is a composition of two primitive transformations, one of type (P) 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/(u, v) and h/(u, v) are always positive. (If either of them is negative, it is elementary to modify the argument.) Put D {(h(u, v), v) (u, v) D} and {(x, g(x, v)) (x, v) D }. so that we have bijective maps D h g D D whose composite is φ. Enclose D in a closed rectangle R, and look at the intersection P of D with a partition of R, which is bounded by the lines x x m, m, 2,..., and v v r, r, 2,..., with the subrectangles R mr being of sides x l and v k. Let R, respectively R mr, denote the image of R, respectively R mr, under (u, v) (u, g(u, v)). hen each R mr is bounded by the parallel lines x x m and x x m + l and by the arcs of the two curves y g(x, v r ) and y g(x, v r + k) (it is a region of type I in previous terminology). hen we have area(r mr) x m+l (g(x, v r + k) g(x, v r )) dx. x m By the mean value theorem of -variable integral calculus and continuity of g, we can write area(r mr) l[g(x m, v r + k) g(x m, v r )], for some point x m (x m, x m +l). By the mean value theorem of -variable differential calculus, we get area(rmr) lk g (x m, v r), for some v r (v r, v r + k). Now for any continuous function f on we have f(x, y) dx dy lim s P dx dy lim f(x P P m, g(x m, v r))area(r mr) where s P is the step function with constant value f(x m, g(x m, v r)) on Rmr. Indeed, given ɛ >, by the small span theorem applied to the continuous function (x, v) f(x, g(x, v)) we can find a partition P so that m,r f(x, g(x, v )) f(x, g(x, v)) < ɛ area( ) 3

4 for (x, v), (x, v ) contained in the same subrectangle R mr of P. But then (f(x, y) s P ) dx dy f(x, y) f(x m, g(x m, v r)) dx dy < ɛ. m,r Hence we obtain f(x, y) dx dy lim P R mr m,r he expression on the right tends to the integral we get the identity klf(x m, g(x m, v r)) g (x m, v r). D f(x, g(x, v)) g (x, v) dx dv. hus f(x, y) dx dy D f(x, g(x, v)) g (x, v) dx dv By applying the same argument, with the roles of x, y (respectively g, h) switched, we obtain D f(x, g(x, v)) g (x, v) dx dy D f(h(u, v), g(h(u, v), v)) g h (h(u, v), v) (u, v) du dv Since ϕ g h we get by the chain rule that Dϕ(u, v) D g(h(u, v), v) D h(u, v) ( g g ) (h(u, v), v) (h(u, v), v) and det Dϕ(u, v) g h (h(u, v), v) (u, v) which is by hypothesis >. hus we get ( h h ) (u, v) (u, v) f(x, y) dx dy D f(ϕ(u, v)) det Dϕ(u, v) du dv as asserted in the heorem. How to do the general case of ϕ? he fact is, we can subdivide D into a finite union of subregions, on each of which ϕ can be realized as a composition of primitive transformations. We refer the interested reader to chapter 3, volume 2, of Introduction to Calculus and Analysis by R. Courant and F. John. 4

5 8.3 Examples () Let {(x, y) R 2 x 2 +y 2 a 2 }, with a >. We know that A area() πa 2. But let us now do this by using polar coordinates. Put and define ϕ : D by D {(r, θ) R 2 r a, θ < 2π}, ϕ(r, θ) (r cos θ, r sin θ). hen it is easy to check that ϕ is -, onto, and moreover, Hence r (cos θ, sin θ) and θ ( r sin θ, r cos θ). ( ) cos θ r sin θ det(dϕ) det r, sin θ r cos θ which is non-negative, and is non-zero iff r >. he integral will not change if we integrate over {r, θ 2π} or over {r >, θ 2π} as the set {r, θ 2π} has content zero. By the transformation formula, we get A dx dy a 2π det Dϕ(r, θ) dr dθ r dr dθ a 2π r dr D a dθ 2π ] a r dr 2π r2 πa 2. 2 (Note that D is a rectangular strip, which is transformed by ϕ into a circular region.) (2) Compute the integral I xdxdy where R is the region {(r, φ) r R 2, φ π/4}. We have I 2 ] 2 π/4 r cos(φ)rdrdφ sin(φ)] π/4 r3 2 (8/3 /3) (3) Let be the region inside the parallelogram bounded by y 2x, y 2x 2, y x and y x +. Evaluate I xy dx dy. he parallelogram is spanned by the vectors (, 2) and (2, 2), so it seems reasonable to make the change of variable ( ) ( ) ( ) x 2 u y 2 2 v 5

6 since then we ll have to integrate over the box [, ] [, ] in the u v-region. he Jacobi matrix of this transformation is of course constant and has determinant 2. So we get I 2 (u + 2v)(2u + 2v) 2 dudv 2 (2u 2 + 6uv + 4v 2 )dudv 2u 3 /3 + 3u 2 v + 4v 2 u ] dv 2 (2/3 + 3v + 4v 2 )dv 2(2v/3 + 3v 2 /2 + 4v 3 /3 ] ) 2(2/3 + 3/2 + 4/3) 7. (4) Find the volume of the cylindrical region in R 3 defined by W {(x, y, z) x 2 + y 2 a 2, z h}, where a, h are positive real numbers. We need to compute I vol(w ) dx dy dz. It is convenient here to introduce the cylindrical coordinates given by the transformation W given by ϕ : D R 3 ϕ(r, θ, z) (r cos θ, r sin θ, z), where D { r a 2, θ 2π, z h}. It is easy to see ϕ is -, C and onto W. Moreover, cos θ r sin θ det Dϕ det sin θ r cos θ r Again, the set {(, θ, z)} where det Dϕ has content, and the integral will be unchanged if this set is included. By the transformation formula, I h a 2π r dr dθ dz πa 2 h, as expected. (5) Let W be the unit ball B () in R 3 with center at the origin and radius. Evaluate 6

7 I e (x2 +y 2 +z 2 ) 3/2 dx dy dz. W Here it is best to use the spherical coordinates given by the transformation ψ : D R 3, (ρ, θ, φ) (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ), where D { ρ, θ 2π, φ π}. hen ψ is C, - and onto W. Moreover sin φ cos θ ρ sin φ sin θ ρ cos φ cos θ det(dψ) det sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ cos φ ρ sin φ ( ) ( ) ρ sin φ sin θ ρ cos φ cos θ sin φ cos θ ρ sin φ sin θ cos φ det ρ sin φ det ρ sin φ cos θ ρ cos φ sin θ sin φ sin θ ρ sin φ cos θ ρ 2 (cos φ) 2 sin φ ρ 2 (sin φ) 3 ρ 2 sin φ. Note that sin φ is on [, π]. Hence det(dψ) ρ 2 sin φ, and we get (by the transformation formula): I π 2π e ρ3 ρ 2 sin φ dρ dφ dθ π e ρ3 ρ 2 dρ 2π sin φ dφ dθ 2π e ρ3 ρ 2 dρ ( cos φ) π 4π 3 e u du, where u ρ 3 4π 3 eu ] 4π 3 (e ). 8.4 Parametrizations Let n, k be positive integers with k n. A subset of R n is called a parametrized k-fold iff there exist a bounded, connected region in R k together with a C injective mapping ϕ : R n, u (x (u), x 2 (u),..., x n (u)), such that ϕ( ). It is called a parametrized surface when k 2, and a parametrized curve when k. 7

8 Example: We restrict the spherical coordinates of the above example to the sphere S (a) of radius a >, i.e. we set ρ a. hen the parametrizing domain is {(θ, φ) R 2 θ < 2π, < φ < π} [, 2π) (, π) and the parametrization is given by (θ, φ) (a sin φ cos θ, a sin φ sin θ, a cos φ). Note that we omitted the endpoints θ 2π and φ, π so that the parametrization is injective. his means we are really parametrizing the unit sphere minus its north and south pole (corresponding to φ and φ π and arbitrary θ). If we want to integrate over the sphere, removing those two points doesn t make a difference because they form a set of content zero. 8.5 Surface integrals in R Integrals of scalar valued functions Let be a parametrized surface in R 3, given by a C injective mapping ϕ : R 3, R 2, ϕ(u, v) (x(u, v), y(u, v), z(u, v)). When we say that ϕ is C we mean that it is so on an open set containing. Let ξ ϕ(u, v) (x, y, z) be a point on. Consider the curve C on passing through ξ on which v is constant. hen the tangent vector to C at ξ is simply given by (u, v). Similarly, we may consider the curve C 2 on passing through ξ on which u is constant. hen the tangent vector to C 2 at ξ is given by (u, v). So the surface has a tangent plane J (ξ) at ξ iff and are linearly independent there. From now on we will assume this to be the case (see also the discussion of tangent spaces in Ch. 3 of the class notes). When this happens at every point, we call smooth. In fact, for integration purposes, it suffices to know that and are independent except at a set {ϕ(u, v)} of content zero. By the definition of the cross product, there is a natural choice for a normal vector to at ξ ϕ(u, v) given by: (u, v) (u, v). Definition : Let f be a bounded scalar field on the parametrized surface. he surface integral of f over, denoted f ds, is given by the formula f ds f(ϕ(u, v)) (u, v) (u, v) du dv. 8

9 We say that f is integrable on if f(u, v) (u, v) (u, v) is integrable on. An important special case is when f. In this case, we get area() (u, v) (u, v) du dv Note that this formula is similiar to that for a line integral in that we have to put in a scaling factor which measures how the parametrization changes the (infinitesimal) length of the curve, resp. area of the surface. Note also that unlike the case of curves this formula only covers the case of surfaces in R 3 rather than in a general R n (more about the general case in the last section). Note further that the definition of a surface integral is independent of the parametrization. Indeed, suppose α : is a bijective C -map whose total derivative Dα is everywhere bijective, i.e. det(dα)(u, v ) for all (u, v ). Write α(u, v ) (u(u, v ), v(u, v )) and ϕ (u, v ) : ϕ(α(u, v )). hen the chain rule implies (u, v ) ( (u, v ) ( ( (α(u, v )) (α(u, v )) (α(u, v )) (u, v ) + (α(u, v )) ) (u, v ) ) (α(u, v )) (u, v ) + (α(u, v )) (u, v ) ) ( (u, v ) (u, v ) (u, v ) ) (u, v ) ( (α(u, v )) ) (α(u, v )) det(dα)(u, v ) and the transformation formula of heorem implies f ds f(u, v) (u, v) (u, v) du dv f(α(u, v )) (u, v ) (u, v ) du dv. (8.) Example: Find the area of the standard sphere S S (a) in R 3 given by x 2 + y 2 + z 2 a 2 with a >. Recall the parametrization of S from above given by {(θ, φ) R 2 θ < 2π, < φ < π} [, 2π) (, π) 9

10 and and ϕ(θ, φ) (x(θ, φ), y(θ, φ), z(θ, φ)) (a sin φ cos θ, a sin φ sin θ, a cos φ). So we have φ θ ( a sin φ sin θ, a sin φ cos θ, ) (a cos φ cos θ, a cos φ sin θ, a sin φ). Hence θ i j k φ det a sin φ sin θ a sin φ cos θ a cos φ cos θ a cos φ sin θ a sin φ ( ) ( ) a sin φ cos θ a sin φ sin θ det i det j+ a cos φ sin θ a sin φ a cos φ cos θ a sin φ ( ) a sin φ sin θ a sin φ cos θ + det k a cos φ cos θ a cos φ sin θ a 2 cos θ sin 2 φi a 2 sin θ sin 2 φj a 2 (sin 2 θ sin φ cos φ + cos 2 θ sin φ cos φ)k a 2 cos θ sin 2 φi a 2 sin θ sin 2 φj a 2 sin φ cos φk. Note that for a point in the upper hemisphere z > we have < φ < π with our 2 parametrisation, hence sin φ > and cos φ > which means that the k-component of is negative. So what we ve computed here is an inward pointing normal θ φ vector to the sphere. Moreover So we find θ φ a2 (cos 2 θ sin 4 φ + sin 2 θ sin 4 φ + sin 2 φ cos 2 φ) /2 2π area(s) a 2 a 2 (sin 4 φ + sin 2 φ cos 2 φ) /2 a 2 sin φ. π dθ π sin φ dφ 2πa 2 sin φ dφ 4πa 2. Example: Find the area of the cylinder ϕ( ) where [, 2π] [, ] and ϕ(u, v) (cos(u), sin(u), v). We have and ( sin(u), cos(u), ), (,, ),. (cos(u), sin(u), )

11 Hence area() 2π dudv 2π. Assume now that we are trying to approximate the area of a parametrized surface by a sum over areas of little triangles with vertices lying on, in the same way that we approximated the length of a parametrized curve by a sum over lengths of little line segments with vertices on the curve. his corresponds to a subdivision (partition) of the parameter space into triangles. Continuing the example of the cylinder we subdivide [, 2π] [, ] into the 2km congruent triangles with vertex sets and S ij : S ij : {( 2jπ m, i ) ( (2j + 2)π, k m, i k ) ( (2j + )π, m, i + )} k {( (2j + )π m, i + ) ( (2j + 3)π, k m, i + ) ( (2j + 2)π, k m, i )} k for j m and i k (and taking the first coordinate modulo 2π). Of course, the areas of these 2km triangles sum up to the area of which is (also) 2π. Now let s look at the triangle spanned by ϕ(s ij ) which has vertices {( cos 2jπ 2jπ, sin m m, i k ) (, cos (2j + 2)π m (2j + 2)π, sin m, i ), k ( cos (2j + )π (2j + )π, sin m m, i + k he baseline of has length b : 2 sin( π ) and the height of is m h : ( k + cos π ) 2, 2 m so the area of is area( ) 2 b h sin( π m ) k 2 + ( cos π m) 2. Summing over all 2km triangles, our approximation to the area of the cylinder then becomes A k,m : 2 k m sin( π m ) k 2 + ( cos π m) 2. he first thing to notice is that this set of numbers is unbounded as can be seen from fixing m and letting k tend to infinity. his is in contrast to the curve case where the length of a polygon inscribed into a curve is always bounded by the length of the curve (essentially because of the triangle inequality). he second thing to notice is that if we pick a sequence of subdivisions of whose diameter tends to zero and so that the corresponding sequence of areas A k,m does have a limit (for example )}

12 k m 2 ), this limit need not equal the area of. Indeed, setting x m k we have lim A m m 2,m lim 2 sin(πx) x x 2π lim x 2π x 4 + ( cos(πx)) 2 x 2 x 4 + ( cos(πx)) 2 + π4 4. x 4 2π + lim x ( cos(πx)) 2 x 4 Moreover, this limit will usually depend on the particular sequence of subdivisions we choose. For example, for k m the limit is 2π. So the idea of approximating areas by finer and finer subdivisions into triangles fails in all respects. Here is a useful result: Proposition. Let be a surface in R 3 parametrized by a C, - function ϕ : R 3 of the form ϕ(u, v) (u, v, h(u, v)). In other words, is the graph of z h(x, y). hen for any integrable scalar field f on, we have f ds f(u, v, h(u, v)) ( h )2 + ( h )2 + du dv. Proof. h (,, ) and i j k det h h (,, h ). L h i h j + k. ( h )2 + ( h )2 +. Now the assertion follows by the definition of f ds. Example. Let be the surface in R 3 bounded by the triangle with vertices (,,), (,,) and (,,). Evaluate the surface integral x ds. Note that is a triangular piece of the plane x + y + z. Hence is parametrized by ϕ : R 3, ϕ(u, v) (u, v, h(u, v)), 2

13 where h(u, v) u v, and { v u, u }. h h,, and ( h )2 + ( h ) By the Proposition above, we have: x ds 3 u u du dv 3 u( u) du Integrals of vector valued functions here is also a notion of an integral of a vector field over a surface. As in the case of line integrals this is in fact obtained by integrating a suitable projection of the vector field (which is then a scalar field) over the surface. Whereas for curves the natural direction to project on is the tangent direction, for a surface in R 3 one uses the normal direction to the surface. Note that a unit normal vector to at ξ ϕ(u, v) is given by n (u, v) (u, v) (u, v) (u, v) and that n n(u, v) varies with (u, v). his defines a unit vector field on called the unit normal field. Definition: Let F be a vector field on. hen the surface integral of F over, denoted F n ds, is defined by F n ds F (ϕ(u, v)) n(u, v) du dv F (ϕ(u, v)) ( ) du dv. Again, we say that F is integrable over if F (ϕ(u, v)) n(u, v) is integrable over. It is clear that this definition is independent of the parametrization if we choose a coordinate change map α : with det(dα)(u, v ) > for all (u, v ). hen we have (u, v ) (u, v ) (u, v ) (u, v ) (α(u, v )) (α(u, v )) (α(u, v )) (α(u, v )) using (8.) and det(dα)(u, v ) det(dα)(u, v ). So the surface integral we have defined associates a number to a vector field and an oriented surface. A change of orientation changes the sign of the integral, just as is the case for line integrals. 3

14 he notation F n ds does not make reference to any coordinates whereas the definition of the surface integral explicitly refers to a parametrization. As in the case of line integrals there is a third notation for this integral which makes explicit reference to coordinates (x, y, z) of the ambient space. If F (P, Q, R) we write F n ds P dy dz + Qdz dx + Rdx dy. (8.2) Here the notation v w indicates a product of vectors which is bilinear (i.e. (λ v + λ 2 v 2 ) w λ (v w) + λ 2 (v w) and similarly in the other variable) and antisymmetric (i.e. v w w v, in particular v v ). In this sense is similar to on R 3 except that v w does not lie in the same space where v and w lie. On the positive side v w can be defined for vectors v, w in any vector space. After these lengthy remarks let s see how this formalism works out in practice. If the surface is parametrized by a function ϕ(u, v) (x(u, v), y(u, v), z(u, v)) then dy y y du + dv and by the properties of outlined above we have dy dz ( y z y z )du dv which is the x-coordinate of. So the equation (8.2) does indeed hold if we interpret an integral fdu dv (over some region in R2 ) as an ordinary double integral fdudv whereas fdv du equals fdudv. All of this suggests that one should give meaning to dx, dy, dz as elements of some vector space instead of being purely formal notation as in our definition of multiple integrals. his can be done and is in fact necessary to a complete development of integration in higher dimensions and on spaces more general than R n. 8.6 Integrals over k-folds in R n If is a connected open region of R k with boundary of content zero, k n and ϕ is a parametrization ϕ : R n, u (x (u), x 2 (u),..., x n (u)), such that : ϕ( ) has a well defined tangent space at every point ϕ(u), i.e. so that for any u the vectors (Dϕ)(u)(e i ) are linearly independent for i,..., k, one might wonder what the correct way to integrate over is. In particular, how do we compute the k-dimensional volume of? For that we need a digression on linear algebra. If v,..., v n are linearly independent vectors in R n then the volume of the parallelepiped P spanned by v,..., v n is vol(p ) det(v ij ) where v ij, j,..., n are the coordinates of v i. Here the notion of volume is related to the standard inner product <, > in R n. In fact, all geometric notions that have to 4

15 do with measuring lengths of vectors, angles between vectors, areas and volumes can be expressed in terms of the inner product (and would correspondingly change if we would put a different positive definite bilinear form on R n as the inner product). For the volume the relation is simply the following. If V is the matrix (v ij ) then V V t is the symmetric matrix with entries n ι v iιv jι < v i, v j >< v j, v i >. But since V and the transpose matrix V t have the same determinant we get by multiplicativity of the determinant vol(p ) det(v ) det(< v i, v j >) and this last expression visibly only depends on the inner product. he length of a vector v has a similar expression v < v, v >. If we now have k linearly independent vectors v,..., v k R n then the volume of the k-dimensional parallelepiped P P (v,..., v k ) spanned by v,..., v k is given by the same formula vol(p ) det(< v i, v j >) where the determinant is that of a k k-matrix. One could say that we endow the k-dimensional space spanned by the v i with the inner product of the ambient space R n. Coming back to our k-fold R n the definition of an integral of a scalar valued function f over is then f dv f(u) det(< Dϕ(u)(e i ), Dϕ(u)(e j ) >) du du k ( f(u) det < (u), ) (u) > du du k. (8.3) i j When discussing the cross product in R 3 we have remarked that the length of v v 2 is the area of the parallelogram spanned by v and v 2. Hence for n 3 we have v v 2 det ( ) < v, v > < v, v 2 > v < v 2, v > < v 2, v 2 > 2 v 2 2 < v, v 2 > 2, a formula that can be checked directly. So the case k 2, n 3 of formula (8.3) recovers the definition of surface integrals in R 3 given in section 8.5. he case k of formula (8.3) recovers our definition of line integrals and the case k n of formula (8.3) recovers the transformation formula of heorem. 5