The Jacobian Determinant   

If p = á du, 0 ñ and q = á 0, dv ñ are the vectors spanning a rectangle at (a,b) in the uv-plane, then the vectors
J( a,b) p = á fu( a,b)du,gu( a,b) du ñ         and        J( a,b) q = á fv( a,b)dv,gv( a,b) dv ñ
span a parallelogram at T(a,b) in the xy-plane.  We let dA denote the area of the parallelogram in the xy-plane.

To compute dA, we write J( a,b) p and J( a,b) q as 3 dimensional vectors

J( a,b) p = á fu( a,b)du,gu( a,b) du,0 ñ     and    J(a,b) q = á fv( a,b) dv,gv(a,b) dv,0 ñ
and then we compute their crossproduct:
J( a,b) p×J( a,b) q = á 0,0,fu( a,b) gv( a,b) -fv(a,b) gu( a,b) ñ dudv
The area dA is equal to the magnitude of the cross product, which is
dA = || J( a,b) p×J( a,b) q|| = | fu( a,b) gv( a,b) -fv(a,b) gu( a,b) | dudv
(2)

The quantity inside the absolute values in (2) is called the Jacobian determinant. If we let x = f( u,v) and y = g( u,v) , then the Jacobian determinant can also be written in the form
 ( x,y)
( u,v)
=  x
u
  y
v
 -  x
v
  y
u
It follows that the area dA of the parallelogram in the xy-plane is given by 
dA = ê
ê 		 ( x,y)
( u,v)
ê
ê 		dudv
As a result, dA is often called the area differential and is the approximate area of the parallelogram determined by J( a,b) p and J( a,b) q.

EXAMPLE 4    Find the Jacobian determinant and the area differential of T( u,v) = áu2-v2,2uv ñ.  What is the approximate area of the image of the rectangle [1,1.4] × [1,1.2] in the uv-plane?       

Solution: The Jacobian determinant is 
 ( x,y)
( u,v)
=
 x
u
  y
v
 -  x
v
  y
u
=
( 2u) ( 2u) -( -2v) ( 2v)
=
4u2+4v2
Thus, the area differential is given by
dA = ê
ê 		 ( x,y)
( u,v)
ê
ê 		dudv = ( 4u2+4v2) dudv
On the rectangle [1,1.4] × [1,1.2], the variable u changes by du = 0.4 and v changes by dv = 0.2. We thus evaluate the Jacobian at (u,v) = ( 1,1) and thus obtain the area
dA = ( 4·12+4·12) ·0.4·0.2 = 0.32
which is the area of the parallelogram implied by J( 1,1) p and J( 1,1) q.

 

 

 

Let's look at another interpretation of the area differential.  If the coordinate curves under a transformation T(u,v) are sufficiently close together, then they form a grid of lines that are "practically straight" over short distances.  As a result, sufficiently small rectangles in the uv-plane are mapped to small regions in the xy-plane that are practically the same as parallelograms.

The area of one of these "apparent parallelograms" is approximately the same as the area differential dA at that point.

 

EXAMPLE 5    Find the Jacobian determinant and the area differential of the coordinate transformation
T( u,v) = á ucos(v), u2sin(v) ñ
What is the approximate area of the image under T( u,v) of the rectangle with lower left vertex at ( u,v) = ( 3,p) and with width du = 0.1 and height dv = 0.01?       

Solution: Since x = ucos( v) and y = u2sin(2v) , the Jacobian determinant is
 ( x,y)
( u,v)
=
 x
u
  y
v
 -  x
v
  y
u
=
cos( v) ·2u2cos( 2v) -( -usin( v) ) ·2usin( 2v)
=
2u2[ cos( v) cos( 2v) +sin(v) sin( 2v) ]
=
2u2cos( 2v-v)
=
2u2cos( v)
Thus, the area differential is
dA = | 2u2cos( v) | dudv
and the approximate area of the image of the rectangle at ( 1,p) with width du = 0.1  in and height dv = 0.01  in is
dA
=
| 2·9·cos( p) | ·0.1·0.01
=
| -18| ·0.1·0.01
=
0.018  in2
       

       

Check your Reading: Is the image parallelogram in example 4 also a rectangle?