550 CHAPTER 27. THE INTEGRAL IN OTHER COORDINATES

where this last matrix is the p× p matrix which has the ith column equal to vi. Therefore,from the properties of determinants, 27.5 equals∣∣∣∣det

(∂x(u0)

∂u1du1, · · · ,

∂x(u0)

∂updup

)∣∣∣∣= ∣∣∣∣det(

∂x(u0)

∂u1, · · · , ∂x(u0)

∂up

)∣∣∣∣ du1 · · · dup

This is the infinitesimal chunk of volume corresponding to the point f (u0) in V . The

advantage of(

det(

∂x(u0)∂ui

dui · ∂x(u0)∂u j

du j

))1/2is that it goes on making sense even if the

vectors ∂x(u0)∂u j

are in Rq for q > p thus allowing the consideration of integrals on p dimen-sional surfaces in Rq. However, this will not be pursued much further in this book.

Definition 27.5.2 Let x= f (u) be as described above. Then the symbol

∂ (x1, · · ·xp)

∂ (u1, · · · ,up),

called the Jacobian determinant, is defined by

det(

∂x(u0)

∂u1, · · · , ∂x(u0)

∂up

)≡

∂ (x1, · · ·xp)

∂ (u1, · · · ,up).

Also, the symbol∣∣∣∣ ∂(x1,···xp)

∂(u1,··· ,up)

∣∣∣∣ du1 · · · dup is called the volume element or increment of vol-

ume, or increment of area.

This has given motivation for the following fundamental procedure often called thechange of variables formula which holds under fairly general conditions.

Procedure 27.5.3 Suppose U is an open subset of Rp for p > 0 and suppose f :U → f (U) is a C1 function which is one to one, x= f (u). 2 Then if h : f (U)→ R, isintegrable, ∫

Uh(f (u))

∣∣∣∣ ∂ (x1, · · · ,xp)

∂ (u1, · · · ,up)

∣∣∣∣ dV =∫f(U)

h(x) dV.

Example 27.5.4 Find the area of the region in R2 which is determined by the lines y =2x,y = (1/2)x,x+ y = 1,x+ y = 3.

You might sketch this region. You will find it is an ugly quadrilateral. Let u = x+y andv = y

x . The reason for this is that the given region corresponds to (u,v) ∈ [1,3]×[ 1

2 ,2], a

nice rectangle. Now we need to solve for x,y to obtain the Jacobian. A little computationshows that

x =u

v+1, y =

uvv+1

2This will cause non overlapping infinitesimal boxes in U to be mapped to non overlapping infinitesimalparallelepipeds in V .

Also, in the context of the Riemann integral we should say more about the set U in any case the functionh. These conditions are mainly technical however, and since a mathematically respectable treatment will notbe attempted for this theorem in this part of the book, I think it best to give a memorable version of it which isessentially correct in all examples of interest. The simple statement above is just fine if you are using the Lebesgueintegral. This integral and a slightly less general theorem is proved in a special chapter on the integral.