Math 212 Spring 2008: Solutions: HW #9
Instructor: S. Cautis
1. section 6.1 #6
We can take a linear map
maps (2, −1) to (1, 0) and (−1, 3) to
T which
a b
(0, 1). So if T looks like
we must have
c d
2a − b = 1, 2c − d = 0
and
−a + 3b = 0, −c + 3d = 1
Solving we get d = 2c so c = 1/5 and d = 2/5 while a = 3b so b = 1/5 and
a = 3/5. So we get
3/5 1/5
1/5 2/5
2. section 6.1, #10
The general parallelogram in R2 can be described as the set of points
p + λv + µw
where λ, µ ∈ (0, 1). If we apply a linear map T we get
T (p + λv + µw) = T (p) + λT (v) + µT (w)
where λ, µ ∈ (0, 1). This is also a parallelogram.
3. section 6.2, #2
In this case we are given D so the first step is to figure out D∗ .
Now D is a triangle with vertices (0, 0), (1, 0) and (1, 1). Since our map is
linear D∗ will also be a triangle – we just need to figure out it’s vertices.
(0, 0) is the vertex which maps to (0, 0). Denote by (a, b) the vertex which
maps to (1, 0) then a + b = 1and a − b = 0 which means a = b = 1/2.
Similarly denote by (c, d) the point which maps to (1, 1) then c + d = 1
and c − d = 1 so c = 1 and d = 0. Hence D∗ is the triangle with vertices
(0, 0), (1/2, 1/2), (1, 0).
1 1
The jacobian of our map linear map is det
= −2. So we need to
1 −1
compute
Z
Z
(u + v + u − v)| − 2|dudv =
4ududv
D∗
D∗
We break up D into two smaller triangles D1 and D2 . The first has
vertices (0, 0), (1/2, 0) and (1/2, 1/2) so the integral is
Z 1/2 Z u
Z 1/2
4udvdu =
4u2 du = 4/3(1/8) = 1/6
∗
0
0
0
The second integral is
Z 1 Z 1−u
Z
4udvdu =
1/2
0
1
1/2
4u(1 − u)du = (2 − 1/2) − 4/3(7/8) = 1/3
So the total integral is 1/6 + 1/3 = 1/2.
We can check this directly
Z 1Z x
Z
Z 1
(x+y)dxdy =
(x+y)dydx =
(x2 +x2 /2)dx = 1/3+1/6 = 1/2
D
0
0
0
which agrees with our previous answer.
4. section 6.2, #6
D∗ is the quarter of the unit disk lying in the first quadrant. Let’s figure
out where the three sides of D∗ go to.
The first side is (t, 0) where 0 ≤ t ≤ 1 and this maps to (t2 , 0) which is the
interval ([0, 1], 0). Similarly (0, t) maps to the interval ([−1, 0], 0). The
third side is can be parametrized as (cos t, sin t) where 0 ≤ t ≤ π/2. This
maps to (cos2 t − sin2 t, 2 cos t sin t) = (cos 2t, sin 2t). Here I’ve used trig
identities. You don’t need to remember them off the top of your head but
you should be aware that they exist and be able to look them up.
So as t varies the image of the third side is the semicircle of unit length
between (1, 0) and (−1, 0), centered at the origin.
So D is the half of the unit disk lying in the upper half plane. Hence
Z
dxdy = area(D) = π/2.
D
5. section 6.2, #14
We can parametrize the unit disk in polar coordinates (r, θ) as the region
where r ∈ [0, 1] and θ ∈ [0, 2π]. The Jacobian of this coordinate change is
r (we did this in class and if you don’t know how to do it then you should
review it – it’s an elementary exercise).
Since x = r cos θ and y = r sin θ we get
Z 1 Z 2π
(1 + r2 cos2 θ + r2 sin2 θ)3/2 rdθdr
0
Z
1
=
1
=
Z
0
0
Z
2π
(1 + r2 )3/2 rdθdr
0
2π(1 + r2 )3/2 rdr
0
= 2π[(1 + r2 )5/2 · 2/5 · 1/2]10
= 2π/5 · (25/2 − 1)
6. section 6.2, #15
In polar coordinates the lamniscate becomes r4 = 2a2 (r2 cos2 θ − r2 sin2 θ)
which can be simplified to r2 = 2a2 cos(2θ).
Since r2 ≥ 0 and a2 ≥ 0 we need to look where cos(2θ) ≥ 0. This is the
locus θ ∈ [−π/4, π/4] ∪ [3π/4, 5π/4]. The area of the first is given by
Z π/4 Z a√2 cos(2θ)
Z π/4
rdrdθ =
a2 cos(2θ)dθ
−π/4
0
−π/r
=
π/4
a [sin(2θ)/2]−π/4 = a2
2
The area of the second region is computed in the same way to be a2 . So
the total area is 2a2 .
7. section 6.2, #18
a) These are spherical coordinates in disguise. Take u = 1 and fix w. Then
varying v the image is a circle at height sin(w) of radius cos(w) (a slice of
the sphere). As we vary w now we get all the slices of the sphere.
b) Notice that any two points (u, v, w) and (u, v, w + 2π) map to the same
point. So T is not one-to-one.
8. section 6.2, #24
Drawing a figure helps! The region D is an eigths of the unit disk. Changing to polar coordinates D is the region 0 ≤ r ≤ 1 and π/4 ≤ θ ≤ π/2.
Hence
Z
Z 1 Z π/2
x2 dxdy =
r2 cos2 θdθdr
D
0
=
π/4
Z
1
Z
1
0
=
0
=
π/2
r2 [cos(θ) sin(θ)/2 + θ/2]π/4
r2 dr[π/4 − (1/4 + π/8)]
1/3(π/8 − 1/4)
9. section 6.2 #32
a) x2 + y 2 = 1 is the unit disk while x2 + y 2 = 2y can be rewritten as
x2 + (y − 1)2 = 1 which is the unit disk with center (0, 1). The region
we want is inside the first disk but not the second disk and in the first
quadrant.
b) This region is bounded by u = 1, v = 0, 2y = u − v ≥ 0. What
happened to the condition x ≥ 0? It became superfluous. Before x ≥ 0
cut our region into half. However, the map (u, v) takes the points (−x, y)
and (x, y) to the same point and hence maps these two halves onto the
same region – namely the triangle D with vertices (0, 0), (1, 0), (1, 1).
c) The Jacobian is
det
2x
2y
= −4x
2x 2y − 2
The general change of coordinates formula says that
Z
Z
f (u, v)dudv =
f (x2 + y 2 , x2 + y 2 − 2y)|4x|dxdy
D
R
So if we can find a function f (u, v) such that
f (x2 + y 2 , x2 + y 2 − 2y) · |4x| = xey
we are in business. Since our integral is over x ≥ 0 we can remove the
absolute value and divide by x to get
f (x2 + y 2 , x2 + y 2 − 2y) = ey
Now u − v = 2y which means that e(u−v)/2 = ey and so we take f (u, v) =
e(u−v)/2 .
So the integral we are left with is
Z
Z
(u−v)/2
e
dudv =
1
0
D
=
1
Z
1
0
u
e(u−v)/2 dvdu
0
Z
0
=
Z
[−2e(u−v)/2 ]u0 du
(−2 + 2 · eu/2 )du
=
[−2u + 4eu/2 ]10
=
−2 + 4e1/2 − 4 = −6 + 4e1/2
The tricky part in this question is not to get confused about what the
change of coordinates formula tells you and what you need to find out.