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Calculus with Algebra and Trigonometry IILecture 16
Volume of solids of revolution: shells
Mar 19, 2015
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 1 / 19
Finding volumes using shells
To find the volume generated when the area between the graph of afuncion y = f (x) is rotated about the y axis.
A representative slice of the area when rotated about the x axis generatesa cylindrical shell with a height equal to the length of the slice and aradius equal to the distance to the axis of rotation.
The volume is the sum of all these cylinders
Volume =
∫ b
a2π(radius)(height) dx =
∫ b
a2πx(f (x)) dx
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 2 / 19
Example 1
Find the volume generated when the region, bounded byy = x(x − 2)(x − 4); 0 ≤ x ≤ 2 and the x axis, is rotated about the y axis
Use the shell method. The radius is x and the height is x(x − 2)(x − 4)
Volume = 2π
∫ 2
0x2(x2 − 6x + 8) dx = 2π
[x5
5− 3x4
2+
8x3
3
]20
=112π
15
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 3 / 19
Example 2
Find the volume generated when the region bounded by y = 1√1+x2
, x = 1
and the coordinate axes is rotated about the y axis.
Again use shells. The radius is x and the height is 1√1+x2
.
Volume = 2π
∫ 1
0
x√1 + x2
dx
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 4 / 19
To evaluate the integral use u substitution. Let
u = 1 + x2 ⇒ du = 2x dx ⇒ dx =du
2x
x = 0 ⇒ u = 1 x = 1 ⇒ u = 2
Volume = 2π
∫ 1
0
x√1 + x2
dx
= 2π
∫ 2
1
x√u
(du
2x
)= 2π
∫ 2
1
1
2u−1/2 du
= 2π[u1/2
]21
= 2π(√
2− 1)
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 5 / 19
Example 3
Find the volume generated when the region bounded byy = sin x ; 0 ≤ x ≤ π and the x axis is rotated about the y axis.
Again use shells. The radius is x and the height is sin x .
Volume = 2π
∫ π
0x sin x dx
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 6 / 19
To evaluate the integral use integration by parts. Let
f (x) = x g ′(x) = sin x ⇒ f ′(x) = 0 g(x) = − cos x
Volume = 2π
∫ π
0x sin x dx
= 2π
([x(− cos x)]π0 −
∫ π
0(− cos x) dx
)= 2π
(π +
∫ π
0cos x dx
)= 2π (π + [sin x ]π0 ) = 2π2
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 7 / 19
Example 4
Find the volume generated when the region bounded by y = x2 andy = x + 2 rotated about the line x = −2.
First the intersection points
x2 = x + 2 ⇒ x2 − x − 2 = 0 ⇒ x = 2,−1
The radius is x + 2 and the height is x + 2− x2.Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 8 / 19
The volume is given by
Volume = 2π
∫ 2
−1(x + 2)(x + 2− x2) dx
= 2π
∫ 2
−1(4 + 4x − x2 − x3) dx
= 2π
[4x + 2x2 − x3
3− x4
4
]2−1
= 2π
((8 + 8− 8
3− 4
)−(−4 + 2 +
8
3− 1
4
))=
107π
6
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 9 / 19
Example 5
Find the volume generated when the region bounded by y = x ln x , x = e,and the x axis, is rotated about the line x = 1.
Use shells: radius = x − 1 and the height = x ln x
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 10 / 19
Volume = 2π
∫ e
1(x − 1)x ln x dx
Use integration by parts
f (x) = ln x g ′(x) = x(x − 1) ⇒ f ′(x) =1
xg(x) =
x3
3− x2
2
Volume = 2π
∫ e
1(x − 1)x ln x dx
=
[(x3
3− x2
2
)ln x
]e1
−∫ e
1
(x3
3− x2
2
)1
xdx
=e3
3− e2
2−[x3
9− x2
4
]e1
=2e3
9− e2
4− 5
36
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 11 / 19
Example 6
Find the volume generated when the region between the parabolsx = 3y2 − 2, x = y2 and the x axis, is rotated about the x axis
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 12 / 19
Slice horizontally and use shells. The radius = y and the height = 2− 2y2.
To find the upper limit for the y integration we need to find theintersection point
3y2 − 2 = y2 ⇒ y = 1
Volume = 2π
∫ 1
0y(2− 2y2) dy
= 2π
[y2 − y2
2
]10
= π
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 13 / 19
Example 7
Find the volume of a torus. A torus is obtained when the circle(x − R)2 + y2 = r2 is rotated about the y axis
Using shells the radius i= x , the height = 2√r2 − (x − R)2, and
R − r ≤ x ≤ R + r .
Volume = 2π
∫ R+r
R−rx(2√r2 − (x − R)2) dx
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 14 / 19
Firstly let u = x − R then
Volume = 4π
∫ r
−r(u + R)
√r2 − u2 du
= 4π
(∫ r
−ru√
r2 − u2 du +
∫ r
−rR√r2 − u2 du
)The first integral is zero. Let u = −v then∫ r
−ru√r2 − u2 du =
∫ −rr
(−v)√r2 − v2 (−dv) =
∫ −rr
v√
r2 − v2 dv
and ∫ r
−r
√r2 − u2 du = area of a semicicle of radius r =
πr2
2
thusVolume = 2πRr2
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 15 / 19
Example 8
Find the volume remaining when a cylindrical hole with radius 3 is drilledthrough a sphere of radius 5.
The volume will be the same as rotating the reginn between the line x = 3and the circle x2 + y2 = 25 about the y axis.
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 16 / 19
Using the shell method radius = x and height = 2√
25− x2
Volume = 2π
∫ 5
32x√
25− x2 dx
Use u substitution. Let
u = 25− x2 ⇒ du = −2x dx ⇒ dx = −du
2x
x = 3 ⇒ u = 16 x = 5 ⇒ u = 0
Volume = 2π
∫ 0
162x√u
(−du
2x
)= 2π
∫ 16
0
√u du
= 2π
[2
3u3/2
]160
=256π
3
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 17 / 19
Volumes with the cross sections as a function of x
Suppose we have a body whose cross section at x is a known function of x,A(x) then its volume will be
Volume =
∫ b
aA(x) dx
For example to find the volume of a pyramid with height, h, and its base asquare of side length b.
A vertical section though the pyramid will be an isosceles triangle withheight h and base b. We can represent it as the region between the twolines y = ± b
2hx with 0 ≤ x ≤ h. So the cross section at position x is a
square of side length bxh .
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 18 / 19
The volume is then
Volume =
∫ h
0
(bx
h
)2
dx
=b2
h2
∫ h
0x2 dx =
b2
h2
[x3
3
]h0
=1
3b2h
Calculus with Algebra and Trigonometry II Lecture 16Volume of solids of revolution: shellsMar 19, 2015 19 / 19
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