Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 8
To be solved during the tutorial session Thursday/Friday, 17/18 March 2016
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 in the first quadrant bounded by the curves r = 2a sin θ, and
θ = φ ≤ π/2, where r and θ are the polar coordinates: x = r cos θ, y = r sin θ.
(a) What is the curve r = 2a sin θ?
(b) What is the curve θ = φ?
(c) Sketch the region R.
(d) Find the area of R.
Solution:
(a) r = 2a sin θ is the circle of radius a centred at (0, a)
(b) θ = φ is the ray making the angle φ with the x-axis
(c) The region R is shown below
2.0
1.5
1.0
0.5
-1.0
0.5
-0.5
1.0
(d) The area A of R is
ZZ
Z
A=
dA =
R
0
φ
1 2 2
4a sin θdθ = a2
2
Z
0
φ
(1 − cos 2θ)dθ = a2 (φ −
1
sin 2φ) .
2
(1)
2. Find the volume V of the solid bounded by the planes x = 0, y = 0, z = 0, the cylinder
x2 + y 2 = a2 and the hyperbolic paraboloid z = xy in the first octant.
Solution: The solid, and its projection R onto the xy-plane are shown below (a = 2)
1
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
Thus, the volume is
ZZ
Z
V =
π/2
Z
a
xydA =
R
0
0
1
r sin θ cos θdrdθ = a4
8
3
Z
0
π/2
1
sin 2θdθ = a4 .
8
(2)
3. Find a parametric representation (different from the generalised stereographic projection)
of
(a) the ellipsoid
x2 y 2 z 2
+ 2 + 2 = 1.
a2
b
c
Solution: It is a generalisation of the sphere parametrisation
x = a cos θ sin φ ,
y = b sin θ sin φ ,
z = c cos φ ,
0 ≤ θ ≤ 2π , 0 ≤ φ ≤ π . (3)
(b) the hyperboloid of two sheets
x2 y 2 z 2
+
− 2 = −1 .
a2 a2
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| ≥ 1 .
(4)
4. Consider the portion of the surface (y − 2)2 + (z + 1)2 = 12 that is above the rectangle
√
√
3
R = {(x, y) : −
≤ x ≤ 3 , −1 ≤ y ≤ 5}.
2
(a) What is the surface?
(b) Sketch the projection of the portion onto the xy-plane.
(c) Use double integration to find the area of the portion.
Show the details of your work.
Solution:
2
5
4
3
2
1
-0.5
0.5
1.0
1.5
-1
a) It is a circular cylinder of radius
and is parallel to the x-axis.
√
12. Its centre line goes through the point (0, 2, −1)
b) The portion of the cylinder and its projection are shown below
c) The area is given by the formula
s
s
2 2
2 2
Z √3 Z 5
ZZ
∂z
∂z
∂z
∂z
+
dA = √
+
dy dx ,
S=
1+
1+
∂x
∂y
∂x
∂y
R
− 3/2 −1
where
z = −1 +
p
12 − (y − 2)2
Computing the derivatives we get
s
√
2 2 s
∂z
∂z
(y − 2)2
12
+
= 1+
1+
=p
.
2
∂x
∂y
12 − (y − 2)
12 − (y − 2)2
Thus, we get
√
Z
S=
3
√
− 3/2
Z
√
12
5
−1
p
Z
12 − (y − 2)2
5
1
dy dx = 9
p
−1
Z
3
= 18
0
12 − (y − 2)2
1
p
dy .
12 − y 2
Z
3
dy = 9
−3
1
p
dy
12 − y 2
To compute the integral we do the substitution
y=
√
√
12 sin t = 2 3 sin t ,
dy =
√
12 cos t dt ,
p
√
12 − y 2 = 12 cos t ,
0≤t≤
π
3
and get
Z
π/3
dt = 6 π .
S = 18
0
5. Find the area of the portion of the hyperbolic paraboloid z = xy that is in the first octant
x, y, z ≥ 0 and inside the cylinder x2 + y 2 = a2 .
3
Solution: The region R is a quarter of the disc of radius a centred at the origin. Thus the surface
area is
ZZ s
ZZ p
Z π/2 Z a √
∂z 2
∂z 2
2
2
S=
1 + ( ) + ( ) dA =
1 + x + y dA =
1 + r2 rdrdθ
∂x
∂y
0
0
R
R
π
π
= (1 + r2 )3/2 |a0 =
(1 + a2 )3/2 − 1) .
6
6
(5)
4