AM1
Mathematical Analysis 1
Oct. 2011 – Feb. 2012
Exercises Lecture 4
Date: October 25
Exercise 4.1. Use integration to compute the volume of a right circular cone generated by revolving
the ordinate set of f (x) = γx on the interval [0, b]. See figure 4.1.
Solution:
The formula for the volume of a solid S obtained from the revolution around the x-axis of the
ordinate set of the function f , which is square-integrable over the interval [a, b] is
b
Z
π f 2 (x) dx
v(S) =
(4.1)
a
Therefore the volume of the cone is
b
Z
γ 2 x2 dx = πγ 2
v(C) = π
a
b3
3
(4.2)
The basis of the cone is a circle having radius r = f (b) = γb; therefore γ = rb . Inserting the value
of gamma into (4.2) gives
1
v(C) = πr2 b
(4.2’)
3
which is the standard formula of the cone having radius r and height b.
Exercise 4.2. (Paraboloid of revolution) Compute the volume of the solid P generated by revolving
√
the ordinate set of the function f (x) = x over the interval [0, 1]. See figure 4.2
Solution:
Using (4.1) we get
1
Z
v(P ) =
π x dx =
0
π
2
(4.3)
Exercise 4.3. Compute the volume of the solid P generated by revolving the ordinate set of the
√
function f (x) =4 x over the interval [0, 1]. See figure 4.3
Solution:
Using (4.1) we get
Z
v(P ) =
1
π
√
0
4-1
x dx =
2π
3
(4.4)
4-2
Figure 4.1: The cone obtained revolving the ordinate set of γx.
Exercise 4.4. Compute the volume of the solid S generated by revolving the ordinate set of the
function f (x) = x2 over the interval [−1, 2]. See figure 4.4
Solution:
Using (4.1) we get
Z
2
π x4 dx =
v(S) =
−1
33
π
5
(4.5)
Exercise 4.5. Compute the volume of the solid S generated by revolving the ordinate set of the
function f (x) = sin(x) over the interval [0, π]. See figure 4.6
Solution:
Using (4.1) we get
Z
v(P ) =
0
π
π sin2 (x) dx =
π2
2
(4.6)
Exercise 4.6. Compute the volume of the solid S generated by revolving the ordinate set of the
function f (x) = cos(x) over the interval [0, π2 ]. See figure 4.6
4-3
Figure 4.2: The solid obtained revolving the ordinate set of
√
x.
Solution:
Using (4.1) we get
Z
v(P ) =
π
2
π cos2 (x) dx =
0
π2
4
(4.7)
Exercise 4.7. Compute the volume of the solid S generated by revolving the ordinate set of the
function f (x) = sin(x) + cos(x) over the interval [0, π]. See figure 4.7
Solution:
Using (4.1) we get
Z
π
π (sin(x) + cos(x))2 dx
Z π
2 sin(x) cos(x) dx
=π π+
Z 0π
= π2 + π
sin(2x) dx
v(P ) =
(4.8)
0
(4.9)
(4.10)
0
= π2
(4.11)
4-4
Figure 4.3: The solid obtained revolving the ordinate set of
4 √x.
Figure 4.4: The solid obtained revolving the ordinate set of x2 .
4-5
Figure 4.5: The solid obtained revolving the ordinate set of sin(x).
Figure 4.6: The solid obtained revolving the ordinate set of cos(x).
4-6
Figure 4.7: The solid obtained revolving the ordinate set of sin(x) + cos(x).
4-7
Exercise 4.8. A napkin-ring is obtained by drilling a cylindrical hole symmetrically through the
center of a solid sphere. The radius of the spere is R, that of the cylinder is r4. Compute the
volume of the napkin-ring. See figure 4.10.
Solution:
The napkin-ring can be thought of beings formed by rotating the ordinate set of the red curve in
figure 4.8
7,5
5
2,5
r
-10
-7,5
-5
-R -h
-2,5
0
2,5
h
R
5
7,5
-2,5
Figure 4.8: Rotating the ordinate set of the red curve gives the napkin-ring in figure 4.9
√
The height 2h of the cylinder to drill is 2 R2 − r2 . Therefore we can compute the volume of the
napkin-ring by subtracting from the volume of the whole sphere ( 34 πR3 ) the volume of the cylinder
(πr2 2h) and twice the volume of the sperical caps (drawn in blue and green in figure 4.9).
The sperical caps, in turn, are obtained by rotating the ordinate set of the blue and green curve
from figure 4.8. Therefore the volume of the single spherical cap C is
Z
R
R2 − x2 dx
(4.12)
R 3 h3
3
2
+
=π R −R h−
3
3
(4.13)
v(C) = π
h
4-8
So the volume of the napkin-ring is
4
R3 h3
+
v(R) = πR3 − 2πr2 h − 2π R3 − R2 h −
3
3
3
1
= 2π (R2 − r2 )h − h3
3
4 3
= πh
3
(4.14)
(4.15)
(4.16)
Figure 4.9: Napkin-ring obtained by rotating figure 4.8
Exercise 4.9. (Ellipsoid of revolution). Compute the volume of the solid E generated by rotating
2
the ellipsis x4 + y 2 ≤ 1 around the x-axis. See figure 4.11
Solution:
2
To obtain the ellipsoid from the ellipsis x4 + y 2 = 1, we have to revolve around the x-axis the
q
2
ordinate set of the function f (x) = 1 − x4 over its domain, that is [−2, 2].
Therefore the volume of the ellipsoid is
2
x2
8
v(E) = π
1−
dx = π
4
3
−2
Z
2
(4.17)
4-9
Figure 4.10: Napkin-ring obtained by drilling the sphere
Figure 4.11: Ellipsoid of revolution
4-10
Exercise 4.10. (One-sheet Hyperboloid of revolution). Compute the volume of the solid H1 generated by rotating the set {(x, y) s.t. − γ ≤ x ≤ γ, 0 ≤ y 2 − x2 ≤ 1} around the x-axis. See
figure 4.12
Solution:
The hyperboloid can be obtained from the hyperbola
y 2 − x2 = 1. We have to revolve around the
√
x-axis the ordinate set of the function f (x) = 1 + x2 over the interval [−γ, γ].
Therefore the volume of the hyperboloid is
Z
γ
v(H1 ) = π
1+x
−γ
2 2
γ3
dx = 2π γ +
3
(4.18)
Figure 4.12: One-sheet hyperboloid of revolution
Exercise 4.11. (Two-sheets Hyperboloid of revolution). Compute the volume of the solid H2 obtained by revolving around the x-axis the ordinate set of the graph of the hyperbola x2 − y 2 = 1 over
the interval [−γ, γ]. See figure 4.13.
Solution:
√
The function we have to consider in rotating the ordinate set is f (x) = x2 − 1. This function
has domain (−∞, −1] ∪ [1, , +∞). Then, if γ > 1 the integral to compute for the volume of the
4-11
hyperboloid is
Z
−1
Z
2
x − 1 dx +
2
= 2π
x − 1 dx
1
3
4
γ
= π + 2π
−γ
3
3
v(H2 ) = π
−γ
Z γ
γ
x − 1 dx
2
(4.19)
1
(4.20)
(4.21)
Figure 4.13: Two-sheets hyperboloid of revolution
Exercise 4.12. A solid has a circular base of radius r = 2. Each cross section cut by a plane
perpendicular to a given diameter δ is an equilateral triangle, see figure 4.14. Compute the volume
of the solid.
Solution:
Recall the following theorem, that we discussed during our classes:
Theorem 4.1. Let R be a Cavalieri solid in A with a cross-sectional area function aR which is
integrable on an interval [a, b] and is zero outside it. Then the volume of R is equal the integral of
the cross-sectional area
Z
b
v(R) =
aR (u) du
a
(4.22)
4-12
Next, recall the formula for the area of an equilateral triangle of side s
√
3 2
A(s) =
s
4
(4.23)
Take a u-axis lying on the diameter δ. The side of the triangle cut by a plane perpendicular to δ
at the point u0 , with −r ≤ u0 ≤ u0 , is the lenght
p of the chord orthogonal to δ at the point u0 . By
Pitagora’s theorem such a chord has length 2 r2 − u20 .
This means that the cross-sectional area function is
√
aR (u) = 3(r2 − u2 )
(4.24)
Thus the volume of the solid is
√ Z r 2
3
r − x2 dx
−r
√
4 3 3
=
r
3
v(R) =
Since r = 2, v(R) =
√
32 3
3
Figure 4.14: Each cross section represented is an equilateral triangle
(4.25)
(4.26)