Module MA1132 (Frolov), Advanced Calculus
Homework Sheet 8
Each set of homework questions is worth 100 marks
Due: at the beginning of the tutorial session Thursday/Friday, 24/25 March 2016
Name:
You may use Mathematica to sketch the integration regions and solids, and to check the
results of integration. Use polar coordinates to evaluate double integrals.
1. Consider the region R that is inside the curve r = 4 cos θ and outside the curve (lemniscate) r2 = −8 cos 2θ , where r and θ are the polar coordinates: x = r cos θ, y = r sin θ.
(a) What is the curve r = 4 cos θ?
(b) Sketch the region R.
(c) Find the area of R.
Solution:
(a) r = 4 cos θ is the circle of radius 2 centred at (2, 0). The curve r2 = −8 cos 2θ is the
lemniscate (x2 + y 2 )2 = 2a2 (y 2 − x2 ) with the parameter a = 2.
(b) The region R is shown below
(c) To find the area of R we need to find the range of θ. It is done by solving the equation
4 cos θ =
√
1
−8 cos 2θ =⇒ cos θ = ,
2
(1)
which gives θ = −π/3 and θ = π/3. Thus, the area A of R is
ZZ
Z
A=
π/3
dA =
R
−π/3
√
1
8π
(16 cos2 θ + 8 cos 2θ)dθ =
+ 4 3.
2
3
1
(2)
2. Find the volume V of the solid bounded by the planes x = 0, y = 0, z = 0, the cylinders
az = x2 , a > 0, x2 + y 2 = b2 , and located in the first octant x ≥ 0, y ≥ 0, z ≥ 0.
Solution: The solid, and its projection R onto the xy-plane are shown below (a = 2, b = 1)
Thus, the volume is
Z
Z
Z
1 2
1 π/2 b 2
b4 π/2
2
V =
x dxdy =
r cos θ r drdθ =
cos2 θdθ
a
a
4a
R
0
0
0
Z
b4 π/2 1
πb4
=
(1 + cos 2θ)dθ =
.
4a 0
2
16a
ZZ
(3)
3. Consider the solid inside the surface r2 + z 2 = 4 and outside the surface r = 2 sin θ.
(a) What is the surface r2 + z 2 = 4?
(b) What is the surface r = 2 sin θ?
(c) Sketch the solid. Sketch the intersection of the solid with the xy-plane
(d) Find the volume V of the solid
Solution:
(a) It is the sphere of radius 2 centred at the origin
(b) It is the circular cylinder of radius 1 with the axis of symmetry parallel to the z-axis
and going through the point (0, 1) on the xy-plane.
(c) The solid, and the intersection R of the solid with the xy-plane are shown below
2
(d) Since the solid is symmetric about the xy-plane its volume is
ZZ √
V =2
4 − r2 dA .
(4)
R
Then, since the solid also is symmetric about the yz-plane and the region R is symmetric
about the y-axis, the solid’s volume is
ZZ √
V =4
4 − r2 dA .
(5)
R/2
where R/2 is the half of R located in the right half-plane x ≥ 0. The region R/2 is a
simple polar region but it is better to represent it as the union of two simple polar regions
– the first one bounded by the rays θ = −π/2, θ = 0, and the second one by the rays
θ = 0, θ = π/2, see the picture above. Thus the volume is
√
ZZ
V =4
Z
4−
r2 dA
0
Z
2
√
r 4−
=4
Z
r2 drdθ
−π/2 0
π/2
R/2
π/2
Z
2
+4
0
√
r 4 − r2 drdθ
2 sin θ
Z
1
π 1
= 4 (− (4 − r2 )3/2 )|20 + 4
(− (4 − r2 )3/2 )|22 sin θ dθ
2 3
3
0
Z π/2
16π 32
16π 64
=
+
cos3 θdθ =
+
.
3
3 0
3
9
(6)
4. Find a parametric representation (different from the generalised stereographic projection)
of
(a) the elliptic cone
r
x2 y 2
+ 2.
a2
b
Solution: It is a generalisation of the parametrisation of a circular cone
z=
x = au cos θ ,
y = bu sin θ ,
3
z = u,
0 ≤ θ ≤ 2π , u ≥ 0 .
(7)
(b) the hyperboloid of one sheet
x2 y 2 z 2
+ 2 − 2 = 1.
a2
b
c
Solution: It is a generalisation of the parametrisation of a surface of revolution
√
√
x = a u2 + 1 cos θ , y = b u2 + 1 sin θ , z = cu , 0 ≤ θ ≤ 2π , −∞ ≤ u ≤ ∞ .
(8)
5. Read about the stereographic projection in Wikipedia and generalise it to parametrise
(a) the ellipsoid
x2 y 2 z 2
+ 2 + 2 = 1.
a2
b
c
Solution: The generalised stereographic projection from the “north” pole (the point with
coordinates (0, 0, c)) to the xy-plane is given by
x=a
2u
,
1 + u2 + v 2
y=b
2v
,
1 + u2 + v 2
z=c
u2 + v 2 − 1
,
1 + u2 + v 2
−∞ ≤ u, v ≤ ∞ .
(9)
(b) the hyperboloid of two sheets
x2 y 2 z 2
+ 2 − 2 = −1 .
a2
b
c
Solution: To find the projection we note that the replacement x → ix, y → iy converts
2
2
2
the equation of the ellipsoid xa2 + yb2 + zc2 = 1 into the equation of the hyperboloid
2
2
x2
+ yb2 − zc2 = −1. This replacement is equivalent to u → iu, v → iv. Thus, the
a2
generalised stereographic projection from the point with coordinates (0, 0, −c) to the
xy-plane is given by
x=a
2u
,
1 − u2 − v 2
y=b
2v
,
1 − u2 − v 2
z=c
u2 + v 2 + 1
,
1 − u2 − v 2
u2 + v 2 < 1 .
(10)
One can easily check that the formulae provide a solution of the equation of the
2
2
2
hyperboloid xa2 + yb2 − zc2 = −1. The stereographic projection covers only one (upper)
sheet of the hyperboloid because z > c. In the case a = b = c, one sheet of the
hyperboloid is called the hyperbolic or Lobachevsky plane, and the stereographic
projection is called the Poincare disc model of the Lobachevsky plane (it is disc
because u2 + v 2 < 1).
(c) the hyperboloid of one sheet
−
x2 y 2 z 2
+ 2 + 2 = 1.
a2
b
c
Solution: To find the projection we note that the replacement x → ix converts the equation of
2
2
2
2
2
2
the ellipsoid xa2 + yb2 + zc2 = 1 into the equation of the hyperboloid − xa2 + yb2 + zc2 = 1.
4
This replacement is equivalent to u → iu. Thus, the generalised stereographic
projection from the point with coordinates (0, 0, c) to the xy-plane is given by
2u
x=a
,
1 − u2 + v 2
2v
y=b
,
1 − u2 + v 2
−u2 + v 2 − 1
z=c
,
1 − u2 + v 2
−∞ ≤ u, v ≤ ∞ .
(11)
In the case a = b = c, the hyperboloid is called the two-dimensional de Sitter space,
and its four-dimensional generalisation plays an important role in cosmology.
6. Find the area of the portion of the elliptic paraboloid z = c −
2
2
cylinder xa2 + yb2 = c2 .
Solution: In the x, y coordinates the region R is the ellipse
elliptic paraboloid as
x = au ,
y = bv ,
x2
a2
z =c−a
+
y2
b2
x2
2a
−
y2
2b
that is inside the
= c2 . Let us parametrise the
u2
v2
−b .
2
2
(12)
The advantage of this parametrisation is that the region R in terms of the u and v
coordinates is just the disc u2 + v 2 ≤ c2 . We then compute
∂r
= a i − au k ,
∂u
∂r
= b j − bv k ,
∂v
|
√
√
∂r ∂r
×
| = a2 b2 + a2 b2 v 2 + a2 b2 u2 = ab 1 + u2 + v 2 .
∂u ∂v
(13)
Thus the surface area is
ZZ
ZZ
√
∂r ∂r
×
| dAuv = ab
S=
|
1 + u2 + v 2 dAuv
∂u
∂v
2
2
2
2
2
2
u +v ≤c
u +v ≤c
Z 2π Z c √
2abπ
2abπ
= ab
(1 + r2 )3/2 |c0 =
1 + r2 rdrdθ =
(1 + c2 )3/2 − 1) .
3
3
0
0
(14)
7. Find the area of the portion of the sphere x2 + y 2 + z 2 = a2 that is inside the cylinder
x 2 + y 2 = b2 .
Solution: The region R p
is the disc of radius b ≤ a centred at the origin, and using the upper
semi-sphere z = a2 − x2 − y 2 one gets the surface area
ZZ s
ZZ s
∂z
x2
∂z
y2
S=2
1 + ( )2 + ( )2 dA = 2
1+ 2
+
dA
∂x
∂y
a − x 2 − y 2 a2 − x 2 − y 2
R
R
ZZ
Z 2π Z b
(15)
a
1
p
√
=2
dA = 2a
rdrdθ
a2 − r 2
a2 − x 2 − y 2
0
0
R
√
= −4πa(a2 − r2 )1/2 |b0 = 4πa a − a2 − b2 ) .
If b = a one gets the area of a sphere of radius a: S = 4πa2 .
8. Find the area of the portion of the sphere x2 + y 2 + z 2 = a2 that is outside the cylinder
x2 + y 2 = ay.
Solution: The region R is shown below
5
To answer the question it is easier to find the portion of the sphere x2 + y 2 + z 2 = a2 that
is inside the cylinder x2 + y 2 = ay, and then to subtract it from the area of the sphere.
Thus, one gets
ZZ s
ZZ s
∂z 2
∂z 2
x2
y2
Sinside = 2
1 + ( ) + ( ) dA = 2
1+ 2
+
dA
∂x
∂y
a − x 2 − y 2 a2 − x 2 − y 2
R
R
Z π/2 Z a sin θ
ZZ
1
a
(16)
p
√
dA = 2a
rdrdθ
=2
a2 − r 2
a2 − x 2 − y 2
−π/2 0
R
Z π/2
Z π/2
2
2 1/2 a sin θ
(a − a cos θ) dθ = 2πa2 − 4a2 .
dθ (a − r ) |0
= 4a
= −4a
0
0
Thus
Soutside = 2πa2 + 4a2 .
(17)
9. Find the area of the portion of the cylinder x2 + y 2 = ay that is inside the sphere
x2 + y 2 + z 2 = a2 .
Solution: Due to the symmetry under the reflection x →p
−x it is sufficient to consider the portion
2
2
of the cylinder x +y = ay with x ≥ 0, i.e. x = ay − y 2 . The region R is the projection
of this portion onto the yz-plane, and its boundaries are found by solving the equation
2
ay − yp
= a2 − y 2 − z 2 . Thus, R is bounded by the lines y = 0, y = a, and the parabolas
z = ± a2 − ay. Thus, the area is equal to
ZZ s
ZZ s
∂x 2
∂x 2
(a − 2y)2
S=2
1 + ( ) + ( ) dA = 2
1+
dA
∂y
∂z
4(ay − y 2 )
R
R
(18)
Z a Z √a2 −ay s
Z a
a2
1
=2
dzdy = 2a3/2
√ dy = 4a2 .
√
2)
4(ay
−
y
y
2
0
− a −ay
0
6