As mentioned above, the fundamental concept of an integral is a sum of things of the form f
(x )
dV where
dV is an “infinitesimal” chunk of volume located at the point x. Up to now, this infinitesimal chunk of
volume has had the form of a box with sides dx_{1},
⋅⋅⋅
,dx_{n} so dV = dx_{1}dx_{2}
⋅⋅⋅
dx_{n} but its form is not
important. It could just as well be an infinitesimal parallelepiped for example. In what follows, this is what
it will be.
First recall the definition of a parallelepiped.
Definition 15.5.1Let u_{1},
⋅⋅⋅
,u_{p}be vectors in ℝ^{k}. The parallelepiped determined by these vectors will bedenoted by P
The dot product is used to determine this volume of a parallelepiped spanned by the given vectors and
you should note that it is only the dot product that matters. Let
x = f1(u1,u2,u3),y = f2(u1,u2,u3),z = f3(u1,u2,u3) (15.3)
(15.3)
where u ∈ U an open set in ℝ^{3}and corresponding to such a u ∈ U there exists a unique point
(x,y,z)
∈ V
as above. Suppose at the point u_{0}∈ U, there is an infinitesimal box having sides du_{1},du_{2},du_{3}. Then this
little box would correspond to something in V . What? Consider the mapping from U to V defined
by
( ) ( )
| x | | f1(u1,u2,u3)|
x = ( y ) = ( f2(u1,u2,u3)) = f (u) (15.4)
z f3(u1,u2,u3)
(15.4)
which takes a point u in U and sends it to the point in V which is identified as
(x,y,z)
^{T}≡ x. What
happens to a point of the infinitesimal box? Such a point is of the form
(u01 + s1du1,u02 + s2du2,u03 + s3du3),
where s_{i}≥ 0 and ∑_{i}s_{i}≤ 1. Also, from the definition of the derivative,
f (u10 + s1du1,u20 + s2du2,u30 + s3du3) − f (u01,u02,u03) =
where the last term may be taken equal to 0 because the vector
(s1du1,s2du2,s3du3)
^{T} is infinitesimal,
meaning nothing precise, but conveying the idea that it is surpassingly small. Therefore, a point of this
infinitesimal box is sent to the vector
in which there is no sum on the repeated index. Now in general, if there are n vectors in ℝ^{n},
{v1,⋅⋅⋅,vn}
,
det(vi ⋅vj)1∕2 = |det(v1,⋅⋅⋅,vn)| (15.6)
(15.6)
where this last matrix is the n×n matrix which has the i^{th} column equal to v_{i}. The reason for this is that
the matrix whose ij^{th} entry is v_{i}⋅ v_{j} is just the product of the two matrices,
( )
vT1
|| .. || (v1,⋅⋅⋅,vn)
( .T )
vn
where the first on the left is the matrix having the i^{th} row equal to v_{i}^{T} while the matrix on the right is just
the matrix having the i^{th} column equal to v_{i}. Therefore, since the determinant of a matrix equals the
determinant of its transpose,
(( ) )
vT1
det(v ⋅v) = det|||| .. || (v ,⋅⋅⋅,v )|| = det(v ,⋅⋅⋅,v )2
i j (( . ) 1 n ) 1 n
vTn
and so taking square roots yields (15.6). Therefore, from the properties of determinants, (15.5)
equals
This is the infinitesimal chunk of volume corresponding to the point f
(u0)
in V .
Definition 15.5.2Letx = f
(u )
be as described above. Then the symbol
∂(x1,⋅⋅⋅xn)
------------,
∂ (u1,⋅⋅⋅,un )
called the Jacobian determinant, is defined by
( ∂x(u0) ∂x(u0)) ∂(x1,⋅⋅⋅xn)
det -∂u1--,⋅⋅⋅,-∂un--- ≡ ∂-(u1,⋅⋅⋅,un-).
Also, the symbol
||∂(x1,⋅⋅⋅xn)||
|∂(u1,⋅⋅⋅,un)|
du_{1}
⋅⋅⋅
du_{n}is called the volume elementor increment of volume, or incrementof area.
This has given motivation for the following fundamental procedure often called the change ofvariables formula which holds under fairly general conditions.
Procedure 15.5.3Suppose U is an open subset of ℝ^{n}for n > 0 and suppose f : U → f