Section 7.2 Volumes
2010 Kiryl Tsishchanka
Volumes
DEFINITION OF A DEFINITE INTEGRAL: If f is a function defined on [a, b], the definite integral
of f from a to b is a number
Z b
n
X
f (x)dx = lim
f (x∗i )∆xi
max ∆xi →0
a
i=1
provided that this limit exists.
DEFINITION OF VOLUME: Let S be a solid that lies between x = a and x = b. If the cross-section
area of S in the plane Px , through x and perpendicular to the x-axis, is A(x), where A(x) is an integrable
function, then the volume of S is
Z b
n
X
∗
V = lim
A(xi )∆xi =
A(x)dx
max ∆xi →0
a
i=1
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region under the
curve y = ex from 0 to 1.
3
2.5
2
2
1
y 1.5
0
0
1
0.2
0.4
0.6
0.8
1
-1
0.5
-2
0
0
0.2
0.4
0.6
0.8
1
x
Solution: We have
2
x 2
2x
A(x) = πR = π(e ) = πe
=⇒
V =
Z
1
A(x)dx =
0
1
Z
1
2x
πe dx = π
0
Z
0
1
1
e dx = π e2x
2
2x
1
0
=
π 2
(e − 1)
2
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE:
√ Find the volume of the solid obtaining by rotating about the x-axis the region under the
curve y = x from 0 to 1.
Solution: We have
2
√
2
A(x) = πR = π( x) = πx
=⇒
V =
Z
1
A(x)dx =
0
Z
0
1
πxdx = π
Z
0
1
x2
xdx = π
2
1
0
=
π
2
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region bounded by
y = x2 − 4x + 5, x = 1 and x = 4.
2
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region bounded by
y = x2 − 4x + 5, x = 1 and x = 4.
Solution: We have
A(x) = π(x2 − 4x + 5)2 = π(x4 − 8x3 + 26x2 − 40x + 25)
therefore
V
=
Z
4
A(x)dx =
1
=π
Z
4
4
1
π(x4 − 8x3 + 26x2 − 40x + 25)dx
4
26x3
78π
x5
4
2
− 2x +
− 20x + 25x =
(x − 8x + 26x − 40x + 25)dx = π
5
3
5
1
4
1
Z
3
2
1
EXAMPLE: Show that the volume of a cone of height h and radius r is V = πr 2 h.
3
3
Section 7.2 Volumes
2010 Kiryl Tsishchanka
1
EXAMPLE: Show that the volume of a cone of height h and radius r is V = πr 2 h.
3
Solution: We have
A(x) = πR2 = π
therefore
V
=
Z
h
A(x)dx =
0
Z
h
0
r 2 x3
=π 2·
h
3
h
=π
0
rx 2
h
=π
r2
r 2 x2
π 2 dx = π 2
h
h
Z
r 2 x2
h2
h
x2 dx
0
r 2 h3
1
·
= πr 2 h
2
h
3
3
4
EXAMPLE: Show that the volume of a sphere of radius r is V = πr 3 .
3
Solution: We have
A(x) = πR2 = π
therefore
Z r
Z
V =
A(x)dx =
−r
√
r
−r
2
2
π(r − x )dx = π
r 2 − x2
Z
2
= π(r 2 − x2 )
r
−r
(r 2 − x2 )dx
r
x3
r3
(−r)3
2
2
2
=π r x−
=π r ·r−
− π r · (−r) −
3 −r
3
3
4
= πr 3
3
EXAMPLE: Find the volume of the solid obtaining by rotating about the y-axis the region bounded by
y = x3 , y = 8, and x = 0.
4
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of the solid obtaining by rotating about the y-axis the region bounded by
y = x3 , y = 8, and x = 0.
Solution: We have
√
A(y) = πR2 = π( 3 y)2 = πy 2/3
therefore
V =
Z
8
A(y)dy =
0
Z
0
8
πy
2/3
dy = π
Z
8
y
2/3
0
y 2/3+1
dy = π
2/3 + 1
8
0
3
= π y 5/3
5
8
0
=
96π
5
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region bounded by
y = x and y = x2 .
5
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region bounded by
y = x and y = x2 .
Solution: We have
2
2
2
2
A(x) = πRbig
− πRsmall
= π(Rbig
− Rsmall
) = π(x2 − (x2 )2 ) = π(x2 − x4 )
Note that x = x2 if x = 0, 1. Therefore
1
3
Z 1
Z 1
Z 1
1 1
2π
x5
x
2
4
2
4
=π
=
−
−
V =
A(x)dx =
π(x − x )dx = π
(x − x )dx = π
3
5 0
3 5
15
0
0
0
EXAMPLE: Find the volume of the solid obtaining by rotating about the x-axis the region bounded by
y = x3 and y = x2 .
Solution: We have
2
2
2
2
A(x) = πRbig
− πRsmall
= π(Rbig
− Rsmall
) = π((x2 )2 − (x3 )2 ) = π(x4 − x6 )
Note that x3 = x2 if x = 0, 1. Therefore
1
5
Z 1
Z 1
Z 1
2π
x7
1 1
x
4
6
4
6
V =
=
A(x)dx =
=π
−
−
π(x − x )dx = π
(x − x )dx = π
5
7 0
5 7
35
0
0
0
EXAMPLE: Find the volume of the solid obtaining by rotating about the y-axis the region bounded by
√
x
y = 3 x and y = , where x, y ≥ 0.
4
6
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of the solid obtaining by rotating about the y-axis the region bounded by
√
x
y = 3 x and y = , where x, y ≥ 0.
4
Solution: We have
y=
√
3
x =⇒ x = y 3
y=
x
4
=⇒ x = 4y
therefore
2
2
2
2
A(y) = πRbig
− πRsmall
= π(Rbig
− Rsmall
) = π((4y)2 − (y 3)2 ) = π(16y 2 − y 6 )
Note that y 3 = 4y if y = 0, ±2. We exclude the negative root, since y ≥ 0. We get
V =
Z
0
2
A(y)dy =
Z
0
2
2
6
π(16y − y )dy = π
Z
0
2
16y 3 y 7
−
(16y − y )dy = π
3
7
2
6
2
0
=
512π
21
EXAMPLE: Find the volume of the solid obtaining by rotating about the line y = 2 the region bounded
by y = x and y = x2 .
7
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of the solid obtaining by rotating about the line y = 2 the region bounded
by y = x and y = x2 .
Solution: We have
A(x) = π(2 − x2 )2 − π(2 − x)2
Note that x = x2 if x = 0, 1. Therefore
1
5
Z 1
Z 1
Z 1
8π
x
x3
x2
2 2
2
4
2
V =
A(x)dx =
[π(2 − x ) − π(2 − x) ]dx = π
(x − 5x + 4x)dx = π
=
−5 +4
5
3
2 0
15
0
0
0
EXAMPLE: Find the volume of the solid obtaining by rotating about the line y = 4 the region bounded
by y = x2 − 2x and y = x.
Solution: We have
2
2
2
2
A(x) = πRbig
− πRsmall
= π(Rbig
− Rsmall
) = π((4 − (x2 − 2x))2 − (4 − x)2 ) = π(x4 − 4x3 − 5x2 + 24x)
Note that x2 − 2x = x if x = 0, 3. Therefore
Z 3
Z 3
V =
A(x)dx =
π(x4 − 4x3 − 5x2 + 24x)dx
0
=π
Z
0
3
3
5x3
153π
x5
4
−x −
+ 12x =
(x − 4x − 5x + 24x)dx = π
5
3
5
0
4
0
3
2
EXAMPLE:
√ Find the volume of the solid obtaining by rotating about the line x = −1 the region bounded
by y = 2 x − 1 and y = x − 1.
8
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE:
√ Find the volume of the solid obtaining by rotating about the line x = −1 the region bounded
by y = 2 x − 1 and y = x − 1.
Solution: We have
√
y2
+1
y = 2 x − 1 =⇒ x =
4
y =x−1
therefore
A(y) =
Note that
2
πRbig
−
2
πRsmall
=
2
π(Rbig
−
2
Rsmall
)
=⇒ x = y + 1
2
= π (y + 1 + 1) −
y2
+1+1
4
2 !
y4
= π 4y −
16
y2
+ 1 = y + 1 if y = 0, 4. Hence
4
Z 4
Z 4 y4
dy
V =
A(y)dy =
π 4y −
16
0
0
=π
Z
4
0
4
y5
96π
y4
2
dy = π 2y −
=
4y −
16
80 0
5
EXAMPLE: Find the volume of a pyramid whose base is a square with sides of length L and whose height
is h.
9
Section 7.2 Volumes
2010 Kiryl Tsishchanka
EXAMPLE: Find the volume of a pyramid whose base is a square with sides of length L and whose height
is h.
Solution: We have
y
L
s
=
=⇒ s = y
L
h
h
therefore
A(y) = s2 =
hence
V =
Z
0
h
A(y)dy =
Z
h
0
L2
L2 2
y
dy
=
h2
h2
10
L2 2
y
h2
Z
0
h
L2 y 3
y dy = 2 ·
h
3
2
h
0
1
= L2 h
3