consider a transformation t ( u,v ) = ( x ( u,v ) , y ( u,v )) from r 2 to r 2
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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 =
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