### Change of Variables: Jacobians

```Change of Variables:
Jacobians
By Dr. Julia Arnold
Courtesy of a CDPD grant
If you haven’t read the text yet, you are probably
wondering “What is a Jacobian”?
Wikipedia: Jacobian
In vector calculus, the Jacobian is shorthand for either the
Jacobian matrix or its determinant, the Jacobian determinant.
These concepts are all named after the mathematician
Carl Gustav Jacobi. The term "Jacobian" is normally pronounced
[ja’kobiən].
Born: 10 Dec
1804 in Potsdam,
Prussia (now
Germany)
Died: 18 Feb
1851 in Berlin,
Germany
http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Jacobi.html
Our two objectives in this lesson are:
1. Understand the concept of a Jacobian
2. Use a Jacobian to change variables in a double integral
Objective 1
Understand the concept of a
Jacobian
The Jacobian, which turns out to be a 2 x 2 determinant,
is a natural outgrowth of the changing of variables in an
integral problem.
In a single integral you can change variables by letting x = g(u),
dx = g’(u)du and obtain
b

a
d
f ( x ) dx 
 g ( u ) g '( u ) du
w ith a  g ( c ), and b  g ( d )
c
It’s the double integral where we see the Jacobian.

f ( x , y ) dA 
R

f ( g ( u , v ), h ( u , v ))
S
x
x
u
v
y
y
u
v
dudv
This is the Jacobian
where
 ( x, y )
 (u , v )

x
x
u
v
y
y
u
v

x y
u v

y x
u v
Let’s work out the Jacobian where we do the conversion of
rectangular to polar coordinates.
x  r cos 
y  r sin 
x
y
r
x

 cos 
 sin 
r
y
  r sin 

cos 
 r sin 
sin 
r cos 
 r cos 
 r cos   r sin   r
2
2
H ence ,

R
f ( x , y ) dA 

S
f ( r cos  , r sin  ) r dr d 
Objective 2
Use a Jacobian to change variables
in a double integral
We will start off with a simple problem just to illustrate
the basic method of solving.
First we need to show that we can take any defined
region in the xy plane and create a transformation onto a
simpler region in the uv-plane.
There are many ways to do this, and what you choose may
have a lot to do with the pattern of the boundaries of the
region.
We call this a transformation of R in the xy-plane to S
in the uv-plane, or sometimes it may be called a
mapping.
Let’s suppose R is bounded by the lines
X – 2y = 0
X – 2y = -4

X+y=4

X+y=1

This creates the
parallelogram as shown
in the figure.
y

x













Let’s suppose R is bounded by the lines
X – 2y = 0
X – 2y = -4
X+y=4
X+y=1
Let u = x + y and v = x – 2y
Thus the new bounds would be:
X – 2y = 0 v = 0
X – 2y = -4 v = -4
X+y=4
u=4
X+y=1
u=1
y


This creates the
rectangle as shown in
the figure.


x












uv-plane

Let’s suppose R is bounded by the lines
X – 2y = 0, X – 2y = -4, X + y = 4, X + y = 1
Let u = x + y and v = x – 2y
Now to find the partials we need to solve for x in terms of u and v
and solve for y in terms of u and v.
u  x y
v  x  2y
u  v  y  2y
u  v  3y
2u  2 x  2 y
v
2u  v  3 x
2u  v
3
uv
3
 y
x  2y
 x
u  x y
Solving
2u  2 x  2 y
v  x  2y
simultaneously
v
u  v  y  2y
x  2y
2u  v  3 x
2u  v
u  v  3y
 x
3
uv
 y
3
x
2u  v
uv
, y 
3
x
u
y
u


2
3
;
x
3
v
1
y
;
3
 ( x, y )
 (u , v )

v

1
1
So the Jacobian for this
change of variables is
3
-1/3
3

2
1
3
3
1
1
3
3

2  1  1  1  2 1
3
1







 
3 3  33
9
9
9
3
Since we’ve done all this work let’s make a problem using
that particular R.
Solve
 2u  v   u  v    1 
 3 xydA   3  3   3   3  dudv
R
S
Where R is the area
bounded by the lines
X – 2y = 0
X – 2y = -4
X+y=4
X+y=1
S is the area in the uv
plane bounded by
v = 0, v = -4, u = 4 and u = 1
 2u  v   u  v    1 
3
   3   3   3  dudv
4 1
0 4
 3 xydA 
R
1
9
0 4
   2 u  v   u  v  dudv 
1
9
4 1
4
  2u
2
 uv  v
2
 du 
1
 
2u
3
3
0 4
   2u

2
 uv  v
2
dudv
4 1
2
u v
2
v u
2
4
1
 
 2 43
  2 13

2
2
4
v
1
v
2
2


 v 4  

 v 1 
 3
  3

2
2

 

15

2 
42

v

3
v


2


0
15

2
42

v

3
v
 
2
4
1
9
  164 
164
9
0
15 2

3
dv

42
v

v

v
  164

4
4

Some transformations may create problems which are not
difficult, just tedious because of the algebra that has to be
done. At that point you may wish to use a TI89 or a math
program like math cad, mathematica, or maple to simplify
the problem.
For comments on this presentation you may email the
author
Dr. Julia Arnold at [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */
```