MA1S12: SOLUTIONS TO TUTORIAL 4
1. Let R be the region bounded by the graph of
x3
,
2
f (x) = 1 +
x ∈ [0, 2]
Find the volume of the solid generated by revolving R about the
y-axis.
Solution: We use the method of cylindrical shells.
V olume =
Z
2
2π x f (x) dx
0
=
Z
=
Z
2
2π x (1 +
0
2
2π (x +
0
x3
) dx
2
x4
) dx
2
5 2
x2 x
+
2
10 0
16
= 2π 2 +
− 2π [0]
5
52π
=
5
= 2π
Region R bounded by the graph of f HxL
5
4
3
2
R
1
0
0.0
0.5
1.0
1
1.5
2.0
2
MA1S12: SOLUTIONS TO TUTORIAL 4
2. Compute the length of the curve
3
1
y = (x2 + 2) 2 ,
3
x ∈ [0, 3]
Solution: The length of the curve is
Z
Length =
3
p
1 + f 0 (x)2 dx
0
where
f (x) =
3
1 2
(x + 2) 2
3
We have
1
f (x) =
3
0
1
3
(x2 + 2) 2 (2x)
2
1
= x(x2 + 2) 2
=⇒ 1 + f 0 (x)2 = 1 + x2 (x2 + 2)
= x4 + 2x2 + 1
= (x2 + 1)2
=⇒
p
1 + f 0 (x)2 = x2 + 1
Now
Length =
Z
=
3
(x2 + 1) dx
0
x3
+x
3
3
0
= [9 + 3] − [0]
= 12
MA1S12: SOLUTIONS TO TUTORIAL 4
3
The curve in question 2
12
10
8
y = 1  3 Hx ^ 2 + 2L ^ H3  2L
6
4
2
0.5
1.0
1.5
2.0
2.5
3.0
3. Find the surface area of the surface of revolution obtained by revolving the line segment
x ∈ [1, 2]
y = 3x,
about the x-axis.
Solution: Writing f (x) = 3x we have
Surf ace area =
Z
=
Z
2
p
2π f (x) 1 + f 0 (x)2 dx
1
2
√
2π 3x 1 + 32 dx
1
2 2
√
x
= 6 10π
2 1
√
1
= 6 10π 2 −
2
√
= 9 10π
4. Let R be the triangular region in the first quadrant which is bounded
by the line
y =1−
x
2
Compute the volume of the cone obtained by revolving R about the
x-axis.
4
MA1S12: SOLUTIONS TO TUTORIAL 4
The line segment in question 3
6
5
y =3x
4
3
2
1
0.5
1.0
1.5
2.0
Surface of revolution generated by revolving line about x-axis
5
0
-5
-5
0
5
1.5
1.0
2.0
Solution: The region R is the interior of a triangle with vertices
(0, 0), (2, 0),
(0, 1)
By revolving R about the x-axis we obtain a cone.
At each point x ∈ [0, 2] the cross section of the cone is a disk with
radius 1 −
x
2
and area
x 2
A(x) = π 1 −
2
MA1S12: SOLUTIONS TO TUTORIAL 4
5
Region in first quadrant bounded by the line y=1-x2
1.0
0.8
y = 1-x2
0.6
0.4
R
0.2
0.0
0.0
0.5
1.0
1.5
2.0
Surface of revolution generated by revolving R about the x-axis
-0.5
0.0
1.0
0.5
-1.0
1.0
0.5
0.0
-0.5
-1.0
0.0
0.5
1.0
1.5
2.0
The volume of the cone is
Z
2
A(x) dx =
0
=
=
=
=
Z
2
x 2
dx
π 1−
2
0
Z 2
x2
π
1−x+
dx
4
0
2
x2 x3
π x−
+
2
12 0
2
π 2−2+
− π [0]
3
2π
3