� � � 25. Change of coordinates: IDefinition 25.1. A function f : U −→ V between two open subsets of Rn is called a diffeomorphism if: (1) f is a bijection, (2) f is differentiable, and (3) f−1 is differentiable. Almost be definition of the inverse function, f ◦ f−1 : V −→ V and f−1 ◦ f : U −→ U are both the identity function, so that (ff−1)(�y)= �y and (f−1 f)(�x)= �x.◦◦ It follows that Df(�x)Df−1(�y)= In and Df−1(�y)Df(�x)= In, by the chain rule. Taking determinants, we see that det(Df)det(Df−1) = det In =1. Therefore, det(Df−1) = (det(Df))−1 . It follows that det(Df)=0. Theorem 25.2 (Inverse function theorem). Let U ⊂ Rn be an open subset and let f : U −→ R be a function. Suppose that (1) f is injective, (2) f is C1, and (3) Df(�x)=0 �for all �x ∈ U. Then V = f(U) ⊂ Rn is open and the induced map f : U −→ V is a diffeomorphism. Example 25.3. Let f(r,θ)=(r cos θ,r sin θ). Then cos θ sin θ Df(r,θ)= ,−r sin θr cos θ so that det Df(r,θ)= r. It follows that f defines a diffeomorphism f : U −→ V between U = (0, ∞)×(0, 2π) and V = R2 \{ (x,y) ∈ R2 | y =0,x ≥ 0 }. 1 �� �� � � � Theorem 25.4. Let g : U −→ V be a diffeomorphism between open subsets of R2 , g(u,v)=(x(u,v),y(u,v)). Let D∗ ⊂ U be a region and let D = f(D∗) ⊂ V . Let f : D −→ R be a function. Then f(x,y)dx dy = f(x(u,v),y(u,v))| det Dg(u,v)| du dv. DD∗ It is convenient to use the following notation: ∂(x,y)(u,v) = det Dg(u,v). ∂(u,v)The LHS is called the Jacobian. Note that ∂(x,y) ∂(u,v) −1 (u,v)= (x,y) . ∂(u,v)∂(x,y)2 Example 25.5. There is no simple expression for the integral of e−x. However it is possible to compute the following integral ∞ 2 I = e−xdx. −∞ (In what follows, we will ignore issues relating to the fact that the integrals are improper; in practice all integrals converge). Instead of 2 �� � �� � � � � computing I, we compute I2 , �� ���� ∞ 2 ∞ 2 I2 = e−xdxe−ydy � −∞�� −∞� ∞∞ 22 = e−x−ydx dy −∞ −∞ 22 = e−x−ydx dy �� R2 = re−r2 dr dθ R2 � �� 2π � ∞ 2 = re−rdθ dr 00 ∞ 2 2π = re−rdθ dr 00 ∞ 2 =2π re−rdr �0 �e−r2 ∞ =2π − 2 0 = π. So I = √π. Example 25.6. Find the area of the region D bounded by the four curves xy =1, xy =3,y = x 3 , and y =2x 3 . Define two new variables, 3xu = and v = xy. y Then D is a rectangle in uv-coordinates, D∗ = [1/2, 1] × [1, 3] Now for the Jacobian we have ∂(u,v)(x,y)= ��� 3xy 2 −xy23 ��� =4x3 =4u. ∂(x,y)yx y It follows that ∂(x,y)1 (u,v)= . ∂(u,v)4u 3 �� This is nowhere zero. In fact note that we can solve for x and y expliccitl in terms of u and v. uv = x 4 and y = x. v So x =(uv)1/4 and y = u−1/4 v 3/4 . Therefore area(D)= dx dy �� D 1 =du dv 4u �D∗ 3 �� 1 � 11 =du dv 4 11/2 u � 31 1 = 4 [ln u]1/2 dv 1� 31 = ln2dv 4 1 1 = ln2. 2 Theorem 25.7. Let g : U −→ V be a diffeomorphism between open subsets of R3 , g(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w)). Let W ∗ ⊂ U be a region and let W = f(W ∗) ⊂ V . Let f : W −→ R be a function. Then��� ��� f(x,y,z)dx dy dz = f(x(u,v,w),y(u,v,w),z(u,v,w))| det Dg(u,v,w)| du dv dw. WW ∗ As before, it is convenient to introduce more notation:∂(x,y,z)(u,v,w) = det Dg(u,v,w). ∂(u,v,w)4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.