Section 6.1 Volumes By Slicing and Rotation About an Axis
Generalized Cylinder
A cylinder is a solid that is generated when a plane region is translated along a line
or axis that is perpendicular to the region.
If a cylindrical solid is generated y translating a region of area A through a distant h,
then h is called the height of the cylinder, and the volume V of the cylinder is
defined to be
V=Ah =[area of a cross section] X [height]
Volumes By Slicing
•
To solve this problem, we begin by dividing the interval [a, b] into n
subintervals, thereby dividing the solid into n slabs.
•
If we assume that the width of the kth subinterval is xk, then the volume of
the kth slab can be approximated by the volume A(xk)xk of a right cylinder
of width (height) xk and cross-sectional area A(xk), where xk is a point in
the kth subinterval.
•
Adding these approximations yields the following Riemann sum that
approximates the volume V:
n
V   A( xk )xk
k 1
•
Taking the limit as n increases and the width of the subintervals approach
zero yields the definite integral
V  lim
max xk 0
n
 A( x )x  
k 1
k
k
b
a
A( x)dx
Volume Formula
The Volume of a solid can be obtained by integrating the cross-sectional area from
one end of the solid to the other.
How to Calculate the Volume
To apply the formula in the definition to calculate the volume of a solid, take the
following steps:
Example
A pyramid 3m high has a square base that is 3 m on a side. The cross
section of the pyramid perpendicular to the altitude x m down from
the vertex is a square x m on a side. Find the volume of the pyramid.
Solids of Revolution: The Disk Method
Examples
Example: The region between the curve y  x , 0  x 4, and the x-axis is
revolved about the x-axis to generate a solid. Find its volume.
Example
Example: Find the volume of the solid generated by revolving the region bounded
by y  x and the lines y=1, x=4 about the line y=1.
To find the volume of a solid generated by revolving a region between
the y-axis and a curve x=R(y), c  y d, about the y-axis, we use the
same method with x replaced by y. In this case, the circular crosssection is
A(y)= [radius]2 = [R(y)]2
Example
Example: Find the volume of the solid generated by revolving the region between
the y-axis and the curve x=2/y, 1  y 4, about the y-axis.
Example
Example: Find the volume of the solid generated by revolving the region between
the parabola x=y2+1 and the line x=3 about the line x=3.
Solids of Revolution: The Washer Method
Example: The region bounded by the curve y=x2+1 and the line y=-x+3 is revolved
about the x-axis to generate a solid. Find the volume of the solid.
Example
Example: The region bounded by the parabola y=x2 and the line y=2x in the first
quadrant is revolved about the y-axis to generate a solid. Find the volume of the
solid.
Section 6.2 Volumes by Cylindrical Shells
The method of slicing in section 6.1 is sometimes awkward to apply. To overcome
This difficulty, we use the same integral definition for volume, but obtain the area by
slicing through the solid in a different way.
Volume of Cylindrical Shells
A cylindrical shell is a solid enclosed by two concentric right circular cylinders.
The volume V of a cylindrical shell with inner radius r1, outer radius r2, and height
h can be written as
V=2 [1/2(r1+r2) ] h (r2-r1)
So V=2 [ average radius ] [height] [thickness]
Method of Cylindrical Shells
•
The idea is to divide the interval [a, b] into n subintervals, thereby
subdividing the region R into n strips, R1, R2,, …, Rn.
•
When the region R is revolved about the y-axis, these strips generate
“tube-like” solids S1, S2, …, Sn that are nested one inside the other and
together comprise the entire solid S.
•
Thus the volume V of the solid can be obtained by adding together the
volumes of the tubes; that is
V=V(S1)+V(S2)+…+V(Sn).
Method of Cylindrical Shells
Method of Cylindrical Shells
Suppose that the kth strip extends from xk-1 to xk and that the width of the strip
is xk. If we let xk* be the midpoint of the interval [xk-1, xk], and if we
construct a rectangle of height f(xk*) over the interval, then revolving this
rectangle about the y-axis produces a cylindrical shell of average radius xk*,
height f(xk*), and thickness xk.
Then the volume Vk of this cylindrical shell is
Vk=2xk*f(xk*) xk
n
V   2 xk* f ( xk* )xk
k 1
Hence, we have
V  lim
max xk 0
n
 2 x
k 1
*
k
b
f ( x )xk   2 xf ( x)dx
*
k
a
Volume by Cylindrical Shells about the y-axis
Let f be continuous and nonnegative on [a, b] (0a<b), and let R be the region
that is bounded above by y=f(x), below by the x-axis, and on the sides by
the lines x=a and x=b. Then the volume V of the solid of revolution that is
generated by revolving the region R about the y-axis is given by
b
V   2 xf ( x)dx
a
Generally
Summary
Example
Example: The region bounded by the curve y  x , the x-axis, and the line x=4 is
revolved about the y-axis to generate a solid. Find the volume of the solid.
Example
So far, we have used vertical axes of revolution. For horizontal axes, we replace
the x’s with y’s.
Example: The region bounded by the curve y  x , the x-axis, and the line x=4 is
revolved about the x-axis to generate a solid. Find the volume of the solid by the
shell method.
Examples
Example. Use cylindrical shells to find the volume of the solid generated when
the region enclosed between y  x , x  1, x  4, and the x-axis is revolved
about the y-axis.
Solution:
6.3 Length of a Plane Curve
Arc Length Formula for Parametric Curves
Example: Find the circumference of a circle of radius 2 from the parametric equations
x  2cos t , y  2sin t
Solution:
(0  t  2 )
Example
Example: Find the length of the astroid.
x  cos3 t, y  sin3 t,
(0  t  2 )
y=f(x) is a smooth curve on [a, b] if f ’ is continuous on [a, b].
Where convenient, (3) can also be expressed as
Example Find the arc length of the curve y  x
3/2
Solution:
from (1, 1) to (2,2 2)
Dealing with Discontinuities in dy/dx
At a point on a curve where dy/dx fails to exist, dx/dy may exist and we may be able to
Find the curve’s length by expressing x as a function of y and applying the following:
Example
Example: Find the length of the curve y=(x/2)2/3 from x=0 to x=2.