Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
12.11.2
Volumes (6.2-6.3)
Volume of a a simple body like circular\elliptical cylinder or rectangular
parallelepiped (a box) is given by V = Ah, where A is the area of the basis and h is the
height. For a more complicated body B we cut it into small pieces and approximate it
by cylinders and adding the volumes. The volume of B is obtained by limiting
process, when the pieces become infinitely small.
x=b
x=a
a
S ( xi )
S ( xi +1 )
xi
xi +1
b
We intersect\slicing B with a plane and obtaining a plane region that is called crosssections of B. Let A ( xi ) be the area of cross section in a plane Px . The volume of
i
cylinder is A ( xi* ) Δx and therefore the volume of the body is
b
( )
n
V = ∑ A x Δx = ∫ A ( x ) dx
*
j
j=1
Ex 5.
r
x
a
Volume of sphere
r
⎛
x3 ⎞
V = ∫ A ( x ) dx = ∫ π y dx = ∫ π ( r − x ) dx = 2π ∫ r − x dx = 2π ⎜ r 2 x − ⎟ =
3 ⎠0
⎝
−r
−r
−r
0
r
y
r
r
r
2
2
2
2
2
⎛
r3 ⎞
2 4
⎛ 1⎞
= 2π ⎜ r 3 − ⎟ = 2π r 3 ⎜ 1− ⎟ = 2π r 3 ⋅ = π r 3
⎝ 3⎠
3⎠
3 3
⎝
Ex 6.
Find volume of body enclosed between the axes and 1− x 3
rotated around a) x-axis and b) y-axis.
When we rotate around x-axis, we “slice” into discs and so the volume is given by
b
V = π ∫ f 2 (x) dx since the area is π r 2 = π f ( x ) , thus:
2
a
1
(
Vx = π ∫ 1− x
0
3
)
2
1
1
⎛
⎛ 1 1⎞ 9
2
1 ⎞
dx = π ∫ 1− 2x + x dx = π ⎜ x − x 4 + x 7 ⎟ = π ⎜ 1− + ⎟ = π
4
7 ⎠0
⎝
⎝ 2 7 ⎠ 14
0
3
6
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
In order to find the volume by on can “switch the axes” or in other words integrate by
y, i.e. to find the area enclosed between axes and x = 3 1− y . Thus
0
Vy = π ∫
1
(
3
)
0
0
1− y dy = π ∫ (1− y ) dy = −π ∫ t
2
2/3
1
1
1
2/3
1
dt = π ∫ t
2/3
0
t 5/3
3
dt = π
= π
5/ 3 0 5
12.11.2.1 Volumes by Cylindrical Shells (6.3) The other method to solve the last volume can be used for more
complicated shapes is called “Method of Cylindrical Shells”.
xi
xi +1
Consider cylindrical shell with inner radius x1 and outer radius x2 ,
the volume of the shell is the difference between the volumes of outer and inner
cylinder. Thus
V = V2 − V1 = π x22 h − π x12 h = π ( x22 − x12 ) h = π ( x2 + x1 ) ( x2 − x1 ) h =
= 2π
x2 + x1
( x2 − x1 ) h = 2π ravg ( Δr ) h = 2π ravg ⋅ h ⋅ ( Δr ) = circumference ⋅ height ⋅ thikness
2
When we get body obtained by rotating around y-axis, instead of slicing cross
sections we divide\peel the body into cylindrical shells, such that f(x) is a height, the
volume of every shell is given by Vj = 2π x j ⋅ f ( x j ) ⋅ Δx where x j =
( )
n
x j + x j+1
and therefore
2
V = lim ∑ 2π x j f x j Δx = ∫ 2π xf ( x ) dx
n→∞
Ex 7.
j=1
b
a
Find volume of body enclosed between the axes and 1− x 3
rotated around y-axis using method of cylindrical shells.
1
⎛ x 2 x5 ⎞
⎛ 1 1⎞ 3
V y = 2π ∫ x 1− x dx = 2π ∫ x − x dx = 2π ⎜ − ⎟ = 2π ⎜ − ⎟ = π
⎝ 2 5⎠ 5
⎝ 2 5 ⎠0
0
0
1
(
3
)
1
4
12.11.3
Arc Length (6.4)
When we speak about length of a curve we mean the distance between the end points
of the graph after they are stretched like a graph was a string or thread. However we
can’t really do it this way, therefore we need to a better method.
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
We approximate the curve by straight lines, i.e. we divide the curve into small pieces
each one approximated by a line. The length of the curve y = f ( x ) is approximated by
a sum of lengths of these lines.
Such lines can be formulated as
lj ( x) =
( ) ( )
( )
( )
f x j − f x j−1
x j f x j−1 − x j−1 f x j
x − x j−1
x − xj
f xj +
f x j−1 =
x+
x j − x j−1
x j−1 − x j
x j − x j−1
x j − x j−1
( )
( )
The length of each line is given by (Pythagoras formula):
(x
lj =
) ( ( ) ( ))
2
j
− x j−1 + l j x j − l j x j−1
(
= x j − x j−1
)
( ) ( )
2
(x
=
) ( ( ) ( ))
2
j
− x j−1 + f x j − f x j−1
2
=
2
⎛ f x j − f x j−1 ⎞
2
1+ ⎜
⎟ ≈ 1+ ( f ' ( x )) Δx
x j − x j−1
⎝
⎠
Thus the length of the curve is given by
n
n
b
L = lim ∑ l j = lim ∑ 1+ ( f ' ( x )) Δx = ∫ 1+ ( f ' ( x )) dx
n→∞
n→∞
j=1
2
j=1
2
a
b
Similarly, when the curve is given by x = f ( y ) we have L = ∫
a
2
⎛ dx ⎞
1+ ⎜ ⎟ dy .
⎝ dy ⎠
When the curve is given by parameterization x = f ( t ), y = g ( t ) we have f ' ( t j ) ≈
Δx ≈ f ' ( t ) Δt and similarly Δy ≈ g' ( t ) Δt , thus
lj =
(x
) (
2
j
− x j−1 + y j − y j−1
2
b
)
2
( f '(t ) Δt ) + ( g'(t ) Δt )
2
≈
2
=
Δx j
, i.e.
Δt
( f '(t )) + ( g'(t )) Δt
2
2
2
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt
dt
dt
and therefore L = ∫
a
Consider a polar curve r = f (θ ) , in polar coordinates we have
x = r cosθ = f (θ ) cosθ , y = r sin θ = f (θ ) sin θ and therefore
dx
= f ' (θ ) cosθ − f (θ ) sin θ and
dθ
dy
= f ' (θ ) sin θ + f (θ ) cosθ . Thus
dθ
2
b
2
b
⎛ dx ⎞ ⎛ dy ⎞
L = ∫ ⎜ ⎟ + ⎜ ⎟ dθ = ∫
⎝ dθ ⎠ ⎝ dθ ⎠
a
a
b
=∫
a
( f '(θ ))
2
b
( f '(θ ) cosθ − f (θ ) sinθ ) + ( f '(θ ) sinθ + f (θ ) cosθ )
2
+ f 2 (θ ) dθ = ∫ r 2 + ( f ' (θ )) dθ
a
2
2
dθ
Course: Accelerated Engineering Calculus I
Instructor: Michael Medvinsky
Compute arc length for curve described by ( x, y ) = ( t,t 2 ) on [ 0,1]
Ex 8.
2
b
2
1
2
1
2
⎛ dx ⎞ ⎛ dy ⎞
L = ∫ ⎜ ⎟ + ⎜ ⎟ dt = ∫ 1+ ( 2t ) dt =
1+ u 2 du =
∫
u=2t
⎝
⎠
⎝
⎠
dt
dt
20
a
0
du=2dt
= u 1+ u
Ex 9.
2
2
0
2
− ∫u
0
2u
2 1+ u 2
du = ... =
(
(
1
u 1+ u 2 + ln u + 1+ u 2
2
)) = 12 (2
2
0
(
5 + ln 2 + 5
))
Compute arc length of r = 3cosθ + 7 sinθ
b
b
L = ∫ r 2 + ( f ' (θ )) dθ = ∫
2
a
a
12.11.4
(
3cosθ + 7 sin θ
) (
2
+ −3sin θ + 7 cosθ
)
2
b
dθ = ∫ 9 + 7 dθ = 4 ( b − a )
a
Average value of function
Consider an average
*
*
*
y1 + y2 ++ yn f ( x1 ) + f ( x2 ) ++ f ( xn ) f ( x1* )+ f ( x2* )++ f ( xn* )
Avg =
=
=
=
b−a
n
n
Δx
=
( )
b
1 n
1
f x j Δx =
f ( x ) dx
∑
b − a j=1
b − a ∫a
Ex 1.
An experiment measurement showed that the investigated phenomena
could be described as f ( x ) = x 3 − 1 on time interval [1, 3] . Find the average value
of this function.
3
3
⎞
1
1 ⎛ x4
1 ⎛ 81
1 ⎞ 81− 9 72
3
Avg =
x
−
1dx
=
− x ⎟ = ⎜ − 3 − + 1⎟ =
=
=9
∫
⎜
3−1 1
2⎝ 4
4 ⎠
8
8
⎠1 2 ⎝ 4
Note that x 3 − 1 = 9 ⇒ 10 = x 3 ⇒ x = 3 10 , i.e. f ( −2 ) = favg . It is no coincidence.
Theorem: If f(x) is continues on [a,b], the there is exists a c in [a,b] such that:
f (c) =
b
1
f ( x ) dx , in other words we have
b − a ∫a
b
∫ f ( x ) dx = f ( c )(b − a ) .
a