Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 5
Due: at the end of the tutorial session Tuesday/Thursday, 23/25 February 2016
Name and student number:
1. Consider the portion of the surface (y − 1)2 + z 2 = 8 that is above the rectangle
R = {(x, y) : −1 ≤ x ≤ 1 , −1 ≤ y ≤ 3}.
(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:
a) It is a cylinder
b) The portion of the cylinder and its projection are shown below
3
2
y
1
0
-1
3
2
2
1
1
0
-1.0
-0.5
-1.0
0.5
-0.5
1.0
0.0
x
0.5
-1
1.0
c) The area is given by the formula
s
s
2 2
2 2
ZZ
Z 1Z 3
∂z
∂z
∂z
∂z
S=
1+
+
dA =
1+
+
dy dx ,
∂x
∂y
∂x
∂y
R
−1 −1
where
z=
p
8 − (y − 1)2
Computing the derivatives we get
s
√
2 2 s
∂z
∂z
(y − 1)2
2 2
1+
+
= 1+
=p
.
∂x
∂y
8 − (y − 1)2
8 − (y − 1)2
1
Thus, we get
1
Z
Z
√
2 2
3
S=
−1
−1
p
dy dx = 4
8 − (y − 1)2
√
2 2
p
dy ,
8 − y2
2
Z
0
To compute the integral we do the substitution
p
√
√
√
y = 2 2 sin t , dy = 2 2 cos t dt ,
8 − (y − 1)2 = 2 2 cos t ,
0≤t≤
π
4
and get
√
√
2 2 dt = 2 2 π .
π/4
Z
S=4
0
√
2. Consider the solid bounded by the surface z = y and the planes
2x + y = 4 , x = 0, and z = 0.
(a) Sketch the projection of the portion onto the xy-plane.
(b) Use a triple integral to find the volume of the solid.
Solution:
(a) The solid, G, and its projection, R, onto the xy-plane are shown below
4
2.0
3
1.5
2
1.0
0.5
1
0.0
0.0
0.5
1.0
1.5
4
3
2
1
2.0 0
0.5
(b) Thus, the volume is equal to
ZZZ
V
ZZ Z
=
√
y
Z
=
0
2
2
Z
dz dA =
dV =
G
Z
R
0
0
4−2x
√
y dy dx
0
2
2
21
64
3/2
5/2 (4 − 2x) dx = −
(4 − 2x) =
.
3
35
15
0
2
1.0
1.5
2.0
3. Consider the lamina with density δ(x, y) = 4y + π2 bounded by
y = sin x2 , y = 0 , x = 0, and x = 2π.
(a) Sketch the lamina .
(b) Find the mass and centre of gravity of the lamina.
Solution:
(a) The lamina, R, is shown below
1.0
0.8
0.6
0.4
0.2
1
2
3
4
5
6
(b) Its mass is equal to
Z
ZZ
2π
sin
Z
δ(x, y) dA =
M =
Z
=
0
R
2π
0
0
x
2
π
(4y + ) dy dx =
2
Z
2π
(2 sin2
0
x π
x
+ sin ) dx
2 2
2
π
x
x
(1 − cos x + sin ) dx = (x − sin x − π cos )|2π
= 4π .
2
2
2 0
By the symmetry of the lamina the x-coordinate of its centre of gravity is equal to
xcg = π .
The y-coordinate is given by
ycg
Z 2π Z sin x
Z 2π
2
1
π
1
4
x π
x
y δ(x, y) dA =
y(4y + ) dy dx =
( sin3 + sin2 ) dx
4π 0
2
4π 0 3
2 4
2
0
ZR 2π
Z 2π
2
x
x
1
= −
(1 − cos2 ) d(cos ) +
(1 − cos x) dx
3π 0
2
2
32 0
2π
2
x 1
π
2
2
π
8
π
3 x = − (cos − cos ) +
= − (−2 + ) +
=
+
.
3π
2 3
2 0
16
3π
3
16
9π 16
1
=
M
ZZ
Thus, the coordinates of the centre of gravity of the lamina are
xcg = π
ycg =
8
π
+
≈ 0.479292 .
9π 16
3