MATH11007 SOLUTION 21: POLAR COORDINATES
(1) (a)
(b)
(c)
(d)
>
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plots[polarplot]([1+sin(theta),theta,theta=0..2*Pi]);
plots[polarplot]([1+2*cos(theta),theta,theta=0..2*Pi]);
plots[polarplot]([cos(3*theta),theta,theta=0..2*Pi]);
plots[polarplot]([sqrt(cos(2*theta)),
theta,theta=0..2*Pi]);
(2) To find the curvature κ, we write the equation of the curve in the polar
form r = R(θ) and use the formula
2
R + 2(R0 )2 − RR00 .
κ=
3
[R2 + (R0 )2 ] 2
(a) This is the equation of an expanding spiral; the curvature
e−θ
κ= √
2
decreases exponentially as θ increases.
(b)
κ = 2,
showing that this is the equation of a circle of radius 1/2.
(c) A lemniscate; the curvature is
3p
κ=
cos(2θ) .
2
(d) Another circle; the curvature is 2/5.
(3) There are two equally natural ways of doing these calculations: one is to
use the chain rule, i.e.
∂
∂
∂
= cos θ
+ sin θ
∂r
∂x
∂y
∂
∂
∂
= −r sin θ
+ r cos θ
.
∂θ
∂x
∂y
and
The other is to set x = r cos θ and y = r sin θ in the formula for f and then
calculate the partial derivatives directly. The solution will illustrate both
approaches.
(a) fx = y and fy = x. By the chain rule
ur = cos θy + sin θx = 2r sin θ cos θ = r sin(2θ)
and
uθ = −r sin θy + r cos θx = r2 cos(2θ) .
c
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2
MATH11007 SOLUTION 21: POLAR COORDINATES
(b) fx = fy = cos(x + y). By the chain rule,
ur = [cos θ + sin θ] cos(x + y) = [cos θ + sin θ] cos (r cos θ + r sin θ)
and
uθ = [−r sin θ + r cos θ] cos(x + y) = r (cos θ − sin θ) cos (r cos θ + r sin θ) .
√
(c) Here u = 1 + r2 . Hence
r
and uθ = 0 .
ur = √
1 + r2
(d) For r 6= 0, u = 2 ln r and so
2
r
ur =
and uθ = 0 .
(4) We start from the formulae
∂
sin θ ∂
∂
= cos θ
−
∂x
∂r
r ∂θ
and
∂
∂
cos θ ∂
= sin θ
+
.
∂y
∂r
r ∂θ
Then
sin θ ∂
∂
sin θ ∂
∂2
∂
−
cos θ
−
= cos θ
∂x2
∂r
r ∂θ
∂r
r ∂θ
2
∂
sin θ ∂
sin θ ∂ 2
= cos θ cos θ 2 + 2
−
∂r
r ∂θ
r ∂r∂θ
2
sin θ
∂
∂
cos θ ∂
sin θ ∂ 2
−
− sin θ
+ cos θ
−
−
r
∂r
∂θ∂r
r ∂θ
r ∂θ2
2 sin θ cos θ ∂
2 sin θ cos θ ∂ 2
sin2 θ ∂
sin2 θ ∂ 2
∂2
+
−
+
+
.
∂r2
r2
∂θ
r
∂θ∂r
r ∂r
r2 ∂θ2
A similar calculation yields
= cos2 θ
∂2
∂2
2 sin θ cos θ ∂
2
=
sin
θ
−
2
2
∂y
∂r
r2
∂θ
2 sin θ cos θ ∂ 2
cos2 θ ∂
cos2 θ ∂ 2
+
+
.
r
∂θ∂r
r ∂r
r2 ∂θ2
So, adding the two, we obtain
+
1 ∂2
∂2
∂2
∂2
1 ∂
+
+
=
+
.
∂x2
∂y 2
∂r2
r ∂r r2 ∂θ2
(5) R is the section of an annulus in the right-half plane. We have
ZZ
2
Z
π/2
√
Z
x dxdy =
R
2
2
2
Z
π/2
r cos θ rdrdθ =
−π/2
1
−π/2
√
r4
2
2
cos θ dθ
4
1
Z π/2
3
3π
=
cos2 θ dθ =
.
4 −π/2
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MATH11007 SOLUTION 21: POLAR COORDINATES
3
(6) This is the portion of the annulus that lies above the line y = x.
ZZ
R
Z
xy
dxdy =
x2 + y 2
5π/4
π/4
√
Z
2
1
√
cos θ sin θ 2
cos θ sin θ rdrdθ =
r dθ
2
1
π/4
Z 5π/4
sin2 θ 5π/4
1
= 0.
cos θ sin θ dθ =
=
2 π/4
4 π/4
Z
5π/4
(7) In each case, we first write the integral in the form
ZZ
f (x, y) dxdy
R
and then switch to polar coordinates.
(a) Here R is the circle of radius a centered at the origin. Hence the
integral is
2π
Z
Z
0
a
rdrdθ = πa2 .
0
(b) Here R is the portion of the same circle contained in the first quadrant.
Hence the integral is
Z
π/2
Z
a
π/2
Z
2
r rdrdθ =
0
0
0
a4
πa4
dθ =
.
4
8
(c) Here R is the portion of the previous region that lies below the line
y = x. Hence the integral is
π/4
Z
Z
a
π/4
Z
r cos θ rdrdθ =
0
0
0
a3
a3
cos θ dθ = √ .
3
3 2
(8) It is a good idea to use Maple’s plots[polarplot] command to visualise
the region before attempting to write down the integral. The region is to
the right of the y-axis and we have
Z
π/2
√
Z
Area =
cos θ
Z
π/2
rdrdθ =
−π/2
0
−π/2
√
r2 cos θ
dθ
2 0
Z π/2
cos θ
sin θ π/2
=
dθ =
= 1.
2
2 −π/2
−π/2
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MATH11007 SOLUTION 21: POLAR COORDINATES
Also
√
√
−1
cos θ
dθ
(1 − r2 )3/2 3
0
−π/2 0
−π/2
"
23 #
Z
Z
i
3
θ
1 π/2 h
1 π/2
θ
1 − (1 − cos θ) 2 dθ =
1 − 1 − cos2 + sin2
dθ
3 −π/2
3 −π/2
2
2
"
"
32 #
32 #
Z
Z
1 π/2
2 π/2
2 θ
2 θ
1 − 2 sin
1 − 2 sin
=
dθ =
dθ
3 −π/2
2
3 0
2
Z
Z
√
√
2 π/2
θ
2 π/2
θ
θ
1 − 8 sin3
1 − 8 sin
=
dθ =
1 − cos2
dθ
3 0
2
3 0
2
2
π
√
2
θ 2
θ
2
θ − 8 −2 cos + cos3
3
2 3
2
0
(
" √
√ #
)
√
2 π √
2 2 8
2
=
− 8 −2
+
+ 8 −2 +
≈ 0.75 .
3 2
2
3 8
3
Z
π/2
Z
cos θ
Volume =
p
1 − r2 rdrdθ =
Z
π/2
(9) Let R be the disk of radius 1 centered at the origin. The volume removed
is twice
ZZ p
Z 2π Z 1 p
Z 2π −1
1
4 − x2 − y 2 dxdy =
4 − r2 rdrdθ =
(4 − r2 )3/2 dθ
3
0
R
0
0
0
Z 2π h
i
√
3
3
2π
1
4 2 − 3 2 dθ =
8−3 3 .
3 0
3
(10) We have
Z
2π
Z
I(a, b) =
0
a
b


b − a
r−n rdrdθ = 2π ln ab

 b2−n −a2−n
2−n
So the limit only exists if n = 1.
if n = 1
if n = 2 .
if n > 2