Consider a transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2.

Suppose T is a linear transformation T(u,v) = (au + bv , cu + dv) .

Then the derivative matrix of T is DT(u,v) =

x x— —u v

=y y— —u v

a bc d

We now explore what the transformation does to the rectangle defined by points (vectors) (0 , 0), (u0 , 0), (u0 , v0), (0 , v0).

u

v

(0 , 0)

(0 , v0) (u0 , v0)

(u0 , 0)x

y

(0 , 0)

u

v

(0 , 0)

(0 , v0) (u0 , v0)

(u0 , 0)x

y

T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0)

T(0 , v0) =

(0 , 0)

(bv0 , dv0)(au0 + bv0 , cu0 + dv0)

(au0 , cu0)

(a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0)

T(u0 , v0) = (au0 + bv0 , cu0 + dv0)

T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0)

The area of the rectangle D in the uv plane is u0v0 .

The area of the parallelogram T(D) in the xy plane is

D

T(D)

T(0,0) = (a(0) + b(0) , c(0) + d(0)) = (0,0)

T(0 , v0) = (a(0) + b(v0) , c(0) + d(v0)) = (bv0 , dv0)

T(u0 , v0) = (au0 + bv0 , cu0 + dv0)

T(u0 , 0) = (a(u0) + b(0) , c(u0) + d(0)) = (au0 , cu0)

The area of the rectangle D in the uv plane is u0v0 .

The area of the parallelogram T(D) in the xy plane is

||(au0 , cu0 , 0)(bv0 , dv0 , 0)|| = ||(0 , 0 , adu0v0 – bcu0v0)|| =

u0v0|ad – bc| .

D

du dv =Now we know that u0v0

T(D)

dx dy = u0v0|ad – bc|.and

Consequently,

T(D)

dx dy =

D

|ad – bc| du dv =

D

|det[DT]| du dv .

This is a special case of a more general result. Suppose transformation T(u,v) = (x(u,v) , y(u,v)) from R2 to R2 is one-to-one. Then

T(D)

f(x,y) dx dy =

D

f(x(u,v),y(u,v)) |det[DT]| du dv =

D

(x,y)f(x(u,v),y(u,v)) ——– du dv .

(u,v)

This notation is used to representx x— —u v

y y— —u v

which is called the Jacobian determinant of T.One useful fact about one-to-one linear transformations is that a triangle

is always mapped to a triangle, and a parallelogram is always mapped to a parallelogram.

det

Example Consider the integral of xy over the parallelogram P formed by the lines y = 2x , y = 2x – 2 , y = x , and y = x + 1 .

(a) Sketch the region P in the xy plane, and with T(u,v) = find the region D in the uv plane so that T(D) = P.

xy

(3,4)

(2,2)(1,2)

u

v

T(D) = P

x = (u – v)/2

y = u – v/2

u =

v =

2(y – x)

2y – 4x

(x,y) = (0,0) (u,v) = (0,0)

(x,y) = (1,2) (u,v) = (2,0)

(x,y) = (3,4) (u,v) = (2,–4)

(x,y) = (2,2) (u,v) = (0,–4)

(2,0)

(2,–4)(0,–4)

D

u – v —— , 2

vu – — 2

(b) Evaluate by making the change of variables

P

xy dx dyx = (u – v)/2

y = u – v/2

(x,y)——– =(u,v)

x x— —u v

=y y— —u v

det det1/2 – 1/2

1 – 1/2= 1/4

P

xy dx dy =

D

(u – v)(2u – v) (x,y)—————— ——– du dv = 4 (u,v)

D

(u – v)(2u – v)—————– du dv = 16

(2u2 – 3uv + v2)——————– du dv = 16

0

20

–4

7

Example Consider the following integral:

(a)

D*

xy3

——— dx dy(1 – x)4

yx

D* (3/4 , 1/4)(0 , 1/4)

(0 , 1/2) (1/2 , 1/2)

where D is the region displayed in the graph.

With the transformation T(u,v) = (u/v , v – u), there will be a trapezoidal region D in the uv plane so that T(D) = D*. Find this trapezoidal region D in the uv plane.

ux = — u = v

y = v – u v =

xy——1 – x y——1 – x

v

u

(0 , 1/4)

(3/4 , 1)

D(0 , 1/2)

(1/2 , 1)

(b) Evaluate by making the change of variables

D*x = u / v

y = v – u

(x,y)——– =(u.v)

x x— —u v

=y y— —u v

det det1/v – u/v2

–1 1=

v – u—— v2

D*

xy3

——— dx dy(1 – x)4

xy3

——— dx dy =(1 – x)4

D

uv du dv = Since v > u on D, we see that this is equal to its absolute value.

D*

xy3

——— dx dy =(1 – x)4

D

uv du dv =

239——6144

1/2

0

1/2 + u

uv dv du +

1/21/4 + u

3/4

1/4 + u

1

uv dv du =

vu

(0 , 1/4)

(3/4 , 1)

D(0 , 1/2)

(1/2 , 1)

239——6144

1/2

1/4

v – 1/4

uv du dv +

1/20

1

v – 1/2

v – 1/4

uv du dv =

For each point (x,y) in R2, the polar coordinates (r,) are defined by

x = r cos and y = r sin , where

r = x2 + y2 is the length of the vector (x,y) , and

= the angle that the vector (x,y) makes with the positive x axis .

x

y

(x,y) (r,)

r

We have that 0 r and 0 < 2 .

Recall:

ExampleDescribe the region x2 + y2 36 in terms of rectangular coordinates and in terms of polar coordinates.

r

<

6

0 2

x y

6 6

– 6 x2 6 x2

or

y x

6 6

– 6 y2 6 y2

ExampleDescribe the region inside the triangle with vertices (0,0), (2,0), and (2,–23) in terms of rectangular coordinates and in terms of polar coordinates.

r ( or

–/3 0

r 2 / cos ) x y

0 2

– (3)x 0

y x

–23 0

– y / 3 2or

(2,0)

(2,–23)

5/3 2

2 / cos

ExampleDescribe the region (x + 5)2 + y2 25 in terms of rectangular coordinates and in terms of polar coordinates.

r

/ 2 3 / 2

0 – 10 cos

ExampleDescribe the region interior to both circles x2 + y2 = 1 and x2 + (y + 1)2 = 1 in terms of rectangular coordinates and in terms of polar coordinates.

r = 1

r = – 2 sin

r = 1

=7— 6

r = 1

=11—— 6

r

r

r

7— 6

0 – 2 sin

7— 6

11—— 6

0

11—— 6

2

0 – 2 sin

1

Example Suppose we want to integrate the function f(x,y) over the region D* = T(D) in the xy plane, where D is the same region described in the r plane and T(r,) = (r cos , r sin), that is, T is the polar coordinates transformation. Then, we know that

D

f(x,y) dx dy =

D*

(x,y)f( x(r,) , y(r,) ) ——— dr d

(r,)

(x,y)Find ——— .

(r,)

r

x

y

D

(x,y)——— = (r,)

detcos – r sin

sin r cos = | r cos2 + r sin2 | = r

D*

D*

f(x,y) dx dy =

D

f( r cos , r sin ) r dr d .

We see then how we can change from rectangular coordinates to polar coordinates:

Example Consider the integral of the function f(x,y) = e over D* defined to be the region in the first quadrant of the xy plane between the two circles x2 + y2 = 4 and x2 + y2 = 25.

(a)

x2 + y2

Sketch and describe the region of integration D* in the xy plane and the corresponding region D in the r plane.

r

x

y

x2 + y2 = 4

x2 + y2 = 25

(2 , 0) (5 , 0)

(2 , /2) (5 , /2)

r

0 / 2

2 5

D*D

(b) Evaluate by making the change of variables

D*

e dx dyx = r cosy = r sin

x2 + y2

D*

e dx dy =x2 + y2

D

e r dr d =r2e r dr d =r2

2

5

0

/2

r2e— d = 2

0

/2

r = 2

5

e25 – e4

——— d = 2

0

/2

(e25 – e4)———— 4

Example Find e dx by squaring and using polar coordinates.

–

–x2

–

–x2

–

–y2

e dx e dy = – x2e dx e dy =

– –

– y2

–(x2 + y2)e dx dy =

– –

–r2e r dr d =

2

0

0

–r2 e– —— d = 2

0

2

r = 0

0

2

1— d = 2

We see then that

–

–x2

e dx =