Discussion 24 solution – Thursday, April 14th
topic: polar coordinates
The trick to working with the polar coordinate change of variables is that we don’t worry about the substitution r(x, y) and
θ(x, y) to determine D∗ . We are supposed to recognize D∗ from previous work in calculus. For example, the region D in
xy-space that describes the the first quadrant portion of the disk x2 + y 2 ≤ 1 is the polar rectangle 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2.
1. Describe the polar rectangle D∗ (the region in rθ-space) that describes the following D in xy-space under the polar coordinates change of variable x(r, θ) := r cos θ and y(r, θ) := r sin θ.
(a) Let D be the disk x2 + y 2 ≤ 4.
Solution :
Replacing x = r cos θ and y = r sin θ, we get
r2 cos2 θ + r2 sin2 θ ≤ 4.
This is the same as r2 ≤ 4. Since there is no restriction on θ and r is always
greater than or equal to zero, we conclude
0 ≤ r ≤ 2,
0 ≤ θ ≤ 2π.
(b) Let D be the region bounded by the arcs of the circles x2 + y 2 = a2 , x2 + y 2 = b2 (where 0 < a < b), the lines y = x,
y = −x with y ≥ 0.
Solution :
We obtain the following
r2 = a2 , r2 = b2 , r sin θ = r cos θ, r sin θ = −r cos θ, and r sin θ ≥ 0.
We get a ≤ r ≤ b first. We can now observe that y ≥ 0 implies 0 ≤ θ ≤ π.
Also y = x corresponds to θ =
π
4
and y = −x corresponds to θ =
Therefore
a ≤ r ≤ b,
π
3π
≤θ≤
.
4
4
√
(c) Let D be the region in the first quadrant bounded by x2 + y 2 = 4, y = x/ 3 and y = 0.
Solution :
3π
4 .
The first quadrant corresponds to 0 ≤ θ ≤ π2 . x2 + y 2 = 4 becomes r2 = 4.
√
Finally y = x/ 3 corresponds to θ = π6 (because tan π6 = √13 ) and y = 0
corresponds to θ = 0. Hence
0 ≤ r ≤ 2,
ZZ
p
2. Evaluate
0≤θ≤
π
.
6
x2 + y 2 + 1 dA, where D is the portion of the unit disk in the first quadrant.
D
Solution :
We change to polar coordinate and evaluate. The Jacobian for polar coordinate transform is r.
ZZ p
Z π/2 Z 1 p
x2 + y 2 + 1 dA =
r2 + 1 · r drdθ
D
Z0 π/2 0
1
1 2
(r + 1)3/2 0 dθ
=
Z0 π/2 3√
8−1
=
dθ
3
0
π √
= ( 8 − 1)
6
√
xy dA where D is the region in the first quadrant bounded by x2 + y 2 = 4, y = x/ 3 and y = 0.
ZZ
3. Evaluate the integral
D
Solution :
We change to polar coordinate and evaluate. Note that the region is the
same as in 1(c).
ZZ
Z
π/6 Z 2
(r cos θ)(r sin θ) · r drdθ
xy dA =
D
Z0
π/6
Z0
=
Z0 π/6
=
Z0 π/6
2
r3 cos θ sin θ drdθ
0
2
r4
cos θ sin θ0 dθ
4
4 cos θ sin θ dθ
π/6
= 2 sin2 θ0
=
0
= 1/2
topic: general change of variables
4. Let T (u, v) := (u2 − v 2 , 2uv).
(a) Show that the image of the square D∗ = {(u, v)|0 ≤ u ≤ 1, 0 ≤ v ≤ 1} under T is the region in xy-space bound by
the x-axis and by the parabolas y 2 = 4 − 4x and y 2 = 4 + 4x. (Hint: parametrize the four boundary lines of D∗ and draw the
image of this curves under T .)
Solution :
Part 1 : Consider the curve C1 (t) = (t, 0), 0 ≤ t ≤ 1. C1 (t) is the line
segment of the bottom of the square. T (C1 (t)) describes a curve in xy-space
which is going to be part of the boundary of the image of D∗ .
T (C1 (t)) = (t2 − 02 , 2 · t · 0) = (t2 , 0).
As t increases from 0 to 1, T (C1 (t)) goes from (0,0) to (1,0) along the curve
described by y = 0.
Part 2 : Consider the curve C2 (t) = (0, t), 0 ≤ t ≤ 1. C2 (t) is the line
segment of the leftside of the square.
T (C2 (t)) = (02 − t2 , 2 · 0 · t) = (−t2 , 0).
As t increases from 0 to 1, T (C2 (t)) goes from (0,0) to (-1,0) along the curve
described by y = 0.
Part 3 : Consider the curve C3 (t) = (t, 1), 0 ≤ t ≤ 1. C3 (t) is the line
segment of the top of the square.
T (C3 (t)) = (t2 − 12 , 2 · t · 1) = (t2 − 1, 2t).
As t increases from 0 to 1, T (C3 (t)) goes from (-1,0) to (0,2) along the curve
described by x = t2 − 1 = (y/2)2 − 1 (which is y 2 = 4 + 4x).
Part 4 : Consider the curve C4 (t) = (1, t), 0 ≤ t ≤ 1. C4 (t) is the line
segment of the rightside of the square.
T (C4 (t)) = (12 − t2 , 2 · 1 · t) = (1 − t2 , 2t).
As t increases from 0 to 1, T (C4 (t)) goes from (1,0) to (0,2) along the curve
described by x = 1 − t2 = 1 − (y/2)2 (which is y 2 = 4 − 4x).
We combine our results. We found out that the image of D∗ under T is
bounded by the three curves y = 0, y 2 = 4 − 4x, and y 2 = 4 + 4x. Also the
corners of the square goes to the points (0,0), (1,0), (0,2), and (-1,0).
(b) Compute the area of the region D in xy-space bound by the x-axis and by the parabolas y 2 = 4 − 4x and y 2 = 4 + 4x.
(Hint: Note D = T (D∗ ). I wouldn’t calculate this directly in xy-space. Use a Jacobian and the Change of Variables Theorem.)
Solution :
We compute the
|
Jacobian first.
∂x ∂x 2u −2v | = 4u2 + 4v 2
∂u ∂v | = | ∂y ∂y 2v
2u
∂u ∂v
The area is obtained by integrating 1.
ZZ
Z 1Z 1
dA =
1 · (4u2 + 4v 2 ) dudv
D
Z0 1 Z0 1
4u2 + 4v 2 dudv
=
Z0 1 0
4
=
+ 4v 2 dv
0 3
4 4 8
=
+ =
3 3 3
5. aZchange
of variables dictated by integrand
Z
(x+y)/(x−y)
Let
e
dA where D is the trapezoidal region with vertices (1, 0), (2, 0), (0, −2) and (0, −1).
D
(a) Take 10 seconds and convince yourself that you do NOT know a useful antiderivative to f (x, y) := e
or y.
Solution :
x+y
x−y
in either x
(10 seconds of silence)
(b) Do you know an antiderivative to f (u, v) := eu/v in either u or v? Define a change of variables u(x, y) and v(x, y) which
changes the integrand to eu/v .
Solution :
We know how to integrate eu/v with respect to u.
Define u(x, y) = x + y and v(x, y) = x − y.
(c) Find the image of D∗ under the substitution u(x, y) and v(x, y). (Remark: The image will also be a trapezoidal region.
Sadly, not all D∗ are rectangles.)
Solution :
D is bounded by the lines 1 = x − y, x − y = 2, y = 0, and x = 0.
Notice that v = x − y is bounded between 1 and 2. y = 0 corresponds to
(u − v)/2 = 0 and x = 0 corresponds to (u + v)/2 = 0. Therefore the image
of D∗ under the substitution is the trapezoidal region bounded by the lines
v = 1, v = 2, v = u, and v = −u.
(d) Find the inverse transformation T (u, v) := (x(u, v), y(u, v)) and determine the Jacobian
∂(x, y)
.
∂(u, v)
Solution :
T (u, v) =
u+v u−v
1 −1 1 1
1
,
. The Jacobian is | ·
− · |= .
2
2
2 2
2 2
2
ZZ
x+y
e x−y dA. (Hint: Even though D∗ is not a rectangle, it is a fairly
(e) Use the change of variables theorem and calculate
D
easy region to describe as a single double integral in the order of dudv. Why would you not want to do dvdu?)
Solution :
We would not want dvdu because in 5(b) we already decided the order we
can evaluate.
ZZ
e
x+y
x−y
ZZ
Z 2DZ∗
D
u
ev ·
dA =
v
1
dA
2
1 u
e v dudv
2
Z1 2 −v
v u v
e v −v dv
=
2
Z1 2
v
(e − e−1 ) dv
=
1 2
4−1
3e − 3e−1
−1
=
(e − e ) =
4
4
=