∗ Solutions ∗ Transformations of R2.
Purpose: In class, we’ve seen several different coordinate systems on R2 and R3 beyond the usual
rectangular ones: polar, cylindrical, and spherical. The lectures on Monday and Wednesday will cover
the crucial technique of simplifying hard integrals using a change of coordinates (Section 15.9). The
point of this worksheet is to familiarize you with some basic concepts and examples for this process.
Starting point: Here we consider a variety of transformations T : R2 → R2. Previously, we have used
such functions to describe vector fields on the plane, but we can also use them to describe ways of
distorting the plane:
T
1. Consider the transformation T (x, y)= (x−2y,x+2y).
(a) Compute the image under T of each vertex in the below grid and make a careful plot of
them, which should be fairly large as you will add to it later.
To speed this up, divide the task up among all members of the group.
y
x
(0,0)
(−1,2) (2,2)
(−1,−1) (2,−1)
SOLUTION:
See the image following part (f).
(b) For each pair A and B of vertices of the grid joined by a line, add the line segment joining
T (A) to T (B) to your plot. This gives a rough picture of what T is doing.
SOLUTION:
See the image following part (f).
(c) What is the image of the x-axis under T ? The y-axis?
SOLUTION:
The image of the x-axis is the line y = x. The image of the y-axis is the line y =−x. To see
this, parametrize the x-axis as r(t )= (t ,0),−∞< t <∞. Then T (r(t ))= (t , t ),−∞< t <∞,
which traces out the line y = x. Do the same for the y-axis.
(d) Consider the line L given by x + y = 1. What is the image of L under T ? Is it a circle, an
ellipse, a hyperbole, or something else?
SOLUTION:
Parametrize L by r(t ) = (t ,1− t ),−∞< t <∞. T (L) is parametrized by T (r(t )) = (t −2(1−
t ), t +2(1− t ))= (3t −2,−t +2). These are the parametric equations of a line.
(e) Consider the circleC given by x2+ y2 = 1. What is the image ofC under T ?
SOLUTION:
ParametrizeC by r(t )= (cos t , sin t ),0≤ t ≤ 2π. ThenT (r(t ))= (cos t−2sin t ,cos t+2sin t ),0≤
t ≤ 2π. Note that if we let x = cos t − 2sin t , y = cos t + 2sin t , then y − x = 4sin t and
y + x = 2cos t . So the curve T (C ) satisfies the equation ( y−x4 )2 + ( y+x2 )2 = 1. This is the
equation of an ellipse.
(f) Add T (L), T (C ) and T
( )
to your picture. Check your answer with the instructor.
SOLUTION:
Note: The transformation T is a particularly simple sort called a linear transformation.
2. Consider the transformation T (x, y)= (y,x(1+ y2)). Draw the image of the picture below under
T .
xy
(1,1)
SOLUTION:
Label the 5 line segments as at left below. The image of the left hand picture is the right hand
picture.
We can figure this out as follows. First parametrize the line segments:
rA(t )= (0, t ),0≤ t ≤ 1 rB (t )= (t ,0),0≤ t ≤ 1
rC (t )= (1, t ),0≤ t ≤ 1 rD (t )= (t ,1),0≤ t ≤ 1
rE (t )= (t , t ),0≤ t ≤ 1
Then compute the image under T of each of these:
T (rA(t ))= (t ,0),0≤ t ≤ 1 T (rB (t ))= (0, t ),0≤ t ≤ 1
T (rC (t ))= (t ,1+ t2),0≤ t ≤ 1 T (rD (t ))= (1,2t ),0≤ t ≤ 1
T (rE (t ))= (t , t (1+ t2)),0≤ t ≤ 1
Graphing each of these gives the image above at left.
3. In this problem, you’ll construct a transformation T : R2 →R2 which rotates counter-clockwise
about the origin by π/4, as shown below.
T(a) Give a formula for T in terms of polar coordinates. That is, how does rotation affect r and
θ?
SOLUTION:
T (r,θ)= (r,θ+π/4)
(b) Write down T in terms of the usual rectangular (x, y) coordinates. Hint: first convert into
polar, apply part (a) and then convert back into rectangular coordinates.
SOLUTION:
First convert (x, y) into polar:
(r,θ)= (
√
x2+ y2,arctan(y/x))
Then apply T in polar coordinates:
T (r,θ)= (r,θ+π/4)
Then convert the result to rectangular coordinates:
T (x, y)= (r cos(θ+π/4),r sin(θ+π/4)), where r =√x2+ y2,θ = arctan(y/x).
Recall the double angle formulas cos(a+b)= cos(a)cos(b)− sin(a)sin(b) and sin(a+b)=
sin(a)cos(b)+ sin(b)cos(a). Using these we see that
cos(θ+π/4)= cosθcos(π/4)− sinθ sin(π/4)=p2/2(cosθ− sinθ)
and
sin(θ+π/4)= sinθcos(π/4)+ sin(π/4)cosθ =p2/2(sinθ+cosθ).
Hence we have
r cos(θ+π/4)=p2/2(r cosθ− r sinθ)=p2/2(x− y)
and
r sin(θ+π/4)=p2/2(r sinθ+ r cosθ)=p2/2(x+ y).
So we have
T (x, y)= (p2/2(x− y),p2/2(x+ y)).
4. Consider the region R in R2 shown below at right. In this problem, you will do a change of
coordinates to evaluate: Ï
R
x−2y dA
S
u
v
T
R
x
y
(1,1)
(2,1)
(3,4)
(1,3)
(a) Find a linear transformation T : R2 →R2 which takes the unit square S to R.
Write you answer both as a matrix
(
a b
c d
)
and as T (u,v)= (au+bv, cu+dv), and check your
answer with the instructor.
SOLUTION:
T (u,v)= (2u+ v,u+3v). In matrix form,(
2 1
1 3
)
(b) Compute
Î
R x−2y dA by relating it to an integral over S and evaluating that. Check your
answer with the instructor.
SOLUTION:
The Jacobian of T is
det
(
2 1
1 3
)
= 6−1= 5
So Ï
R
x−2y dA=
Ï
S
[(2u+ v)−2(u+3v)]5dA
=
∫ 1
0
∫ 1
0
−25v dudv =
[−25v2
2
]1
0
=−25/2
5. Another simple type of transformation T : R2 →R2 is a translation, which has the general form
T (u,v)= (u+a,v +b) for a fixed a and b.
(a) If T is a translation, what is its Jacobian matrix? How does it distort area?
SOLUTION:
If T (u,v)= (u+a,v +b) where a and b are constants, then the Jacobian is
det
(
1 0
0 1
)
= 1.
So T does not distort areas.
(b) Consider the region S = {u2 + v2 ≤ 1} in R2 with coordinates (u,v), and the region R ={
(x−2)2+ (y −1)2 ≤ 1} in R2 with coordinates (x, y).
Make separate sketches of S and R.
SOLUTION:
(c) Find a translation T where T (S)=R.
SOLUTION:
T (u,v)= (u+2,v +1)
(d) Use T to reduce Ï
R
x dA
to an integral over S, and then evaluate that new integral using polar coordinates.
SOLUTION:
The Jacobian of T is just 1, as noted in part (a). So we haveÏ
R
x dA=
Ï
S
(u+2)dA
Converting the second integral above to polar we have
Ï
S
(u+2)dA=
∫ 2π
0
∫ 1
0
(r cosθ+2)r dr dθ =
∫ 2π
0
[
r 3 cosθ
3
]1
0
dθ+2π[r 2]10
= 1/3
∫ 2π
0
cosθdθ+2π= 1/3[sinθ]2π0 +2π= 2π
6. Consider the region R shown below. Here the curved left side is given by x = y − y2. In this
problem, you will find a transformation T : R2 →R2 which takes the unit square S = [0,1]×[0,1]
to R.
(2,1)
R
x
y
(2,0)
(0,1)
(a) As a warm up, find a transformation that takes S to the rectangle [0,2]× [0,1] which con-
tains R.
SOLUTION:
L(u,v)= (2u,v)
(b) Returning to the problem of finding T taking S to R, come up with formulas for T (u,0),
T (u,1), T (0,v), and T (1,v). Hint: For three of these, use your answer in part (a).
SOLUTION:
T (u,0)= (2u,0) T (u,1)= (2u,1)
T (1,v)= (2,v) T (0,v)= (v − v2,v)
(c) Now extend your answer in (b) to the needed transformation T . Hint: Try “filling in” be-
tween T (0,v) and T (1,v) with a straight line.
SOLUTION:
T (u,v)= (2u+ v(1− v)(1−u),v)
(d) Compute the area of R in two ways, once using T to change coordinates and once directly.
SOLUTION:
To change coordinates we compute the Jacobian
J (T )= det
(
2− v(1− v) (1−2v)(1−u)
0 1
)
= 2− v(1− v)
So we have the area of R given byÏ
R
dx dy =
∫ 1
0
∫ 1
0
2− v(1− v)dudv = 11/6
Computing directly we have the area of R given by∫ 1
0
2− (y − y2)dy = 11/6