Transcript Chapter 7. Applications of the Definite integral in

Chapter 7. Applications of the
Definite integral in Geometry,
Science, and Engineering
By
Jiwoo Lee
Edited by
Wonhee Lee
Area Between Two Curves
• If f and g are continuous functions
on the interval [a,b] and if f(x) > g(x)
for all x in [a,b] then the area of the
region bounded by y=f(x), below by
g(x), on the left by the line x=a, and
on the right by the line x=b is
• ∫ba[f(x)-g(x)]dx
7.1.2 Area Formula
Step 1
Determine which
function is on top
Step 2
Solve
Olive green area=
∫ba[f(x)-g(x)]dx
Beige area= ∫ba[g(x)]dx
Tip1
g(x) f(x)
• Sometimes it is
easier to solve by
integrating with
respect to y rather
than x
• ∫dc[f(x)-g(x)]dx
Tip2
When finding the area enclosed by two
functions, let the two functions equal
each other and solve for the
intersecting points to find a and b.
Tip3
If the two functions
switch top and
bottom, then the
regions must be
subdivided at those
points to find total
area
Solve
The area between the parabolas
X=y2-5y and x=3y-y2
Solution:
1. Intersections at (0,0) and (-4,4)
2. Determine upper function by either
plugging in points or graphing
Upper function is
3y-y2 , therefore
∫403y-y2 –(y2-5y)dy
=64/3
Solve
the area enclosed by the two functions
y=x3-2x and y=(abs(x))1/2
Solution:
1. Intersection at x= -1, 1.666
2. Functions Switch top and bottom at
x=0 so the integral must be divided
from -1 to 0 and 0 to 1.666
∫0-1x3-2x-(-x)1/2dx+ ∫1.6660(x)1/2-(x3-2x)dx
=.083+2.283=2.367
7.2 Volumes by Slicing; Disks and
Washers
The volume of a solid can be
obtained by integrating the
cross-sectional area from one
end of the solid to the other.
Volume Formula
• Let S be a solid bounded by
two parallel planes
perpendicular to the x-axis at
x=a and x=b. If, for each x in [a,
b], the cross-sectional area of S
perpendicular to the x-axis is
A(x), then the volume of the
solid is
• ∫ba[A(x)]dx
• ∫ba[A(x)]dx
Example
The base of a solid is the region
bounded by y=e-x ,the x-axis, the yaxis, and the line x=1. Each cross
section perpendicular to the x-axis is a
square. The volume of the Solid is...
V= ∫10 (e-x)2dx = (1-1/e2)/2
If each cross section is a
circle...
V= ∫ba ∏ (A(x))2dx
A special case, known as method of
disks, often used to find areas of
functions rotated around axis or lines.
If there are two functions rotated, then
subtract the lower region from the
upper region, A method known as
method of washers
∫ba(∏(f(x))2dx- ∏(g(x))2dx]
f(x)
g(x)
Rotated around a line
A(X)
height = A(x)-a
Therefore, V= ∫ba ∏ (A(x)-a)2dx
A(X)
If “a” is below
height
the x-axis, then “a” would be added to the
area between the two
functions rotated around a line
G(x)
A(x)
∫ba(∏(G(x)-a)2dx- ∏(A(x)-a)2dx]
Set up but do not solve for the
area using washers method
1.y=3x-x2 and y=x rotated around the xaxis
2. y=x2 and y=4 rotated around the x-axis
1
V= ∏ ∫20[(3x-x2)2-x2dx]
2
V=2∏∫20[(4+1) 2-(x2+1)2dx
=2∏∫20[24-x4-2x2]dx
Volume by Cylindrical Shells
Another method to determine the
volume of a solid
•
f(x)
•
•
•
a
a
b
b
•
f(x)
•
•
•
a
a
b
b
•
f(x)
•
•
a
b
When the section is flattened...
width =dx
height
= f(x)
length= 2∏x
Area of Cross section= 2∏xf(x)dx
Volume of f(x) rotated around the
y-axis = ∫ba2∏xf(x)dx
•
•
f(x)
a
b
Reminder
Shells Method:
∫ba[2∏x(fx)dx]
Washers Method:
∫ba [∏(f(x))2dx- ∏(g(x))2dx]
When shells method is used and includes dx,
then the function is rotated around the y axis
When washers method is used and uses dx,
then the function is rotated around the x axis
Solve
• the region bounded by y=3x-x2 and y=x
rotated about the y-axis
V= ∫ba2∏(2x2-x3)dx =
8∏/3
Length of a Plane Curve
• If f(x) is a smooth curve on the
interval [a,b] then the arc
length L of this curve over [a,b]
is defined as
• ∫ba√ 1 + [f’(x)]2 dx
Length of a Plane Curve
• If no segment of the curve
represented by the parametric
equations is traced more than
once as t increases from a to b,
and if dx/dt and dy/dt are
continuous functions for a<t<b,
then the arc length is given by
• ∫ba√ (dx/dt) 2 +(dy/dt) 2 dx
Area of a Surface of Revolution
• If f is smooth, nonnegative
function on [a,b] then the
surface area S of the surface of
revolution that is generated by
revolving the portion of the
curve y = f(x) between x=a and
x=b about the x-axis is defined
as
• ∫ba2∏f(x) √1 + [f’(x)]2 dx
Set up the integral for
The length the curve y2=x3 cut off by the
line x=4
• Solution:
• 2y dy/dx = 3 x2 , dy/dx = (3√x)/2
• 2∫40√ 1 + 9x/4 dx
Set up the integral for
The surface area of y=2x3 rotated around
the x-axis from 2 to 7
• Solution: y’ =6x2
• ∫722∏2x3√1 + [6x2]2dx
Work
• If a constant force of
magnitude F is applied in the
direction of motion of an object,
and if that object moves a
distance d, then we define the
work W performed by the
force on the object to be
• W=Fd
Work
• Suppose that an object moves in the
positive direction along a coordinate
line over the interval [a,b] while
subjected to a variable force F(x) that
is applied in the direction of motion.
Then we define work W performed
by the force on the object to be
• W= ∫baF(x)dx
Solve
• A square box with a side length of 7
feet is filled with an unknown chemical.
How much work is required to pump the
chemical to a connecting pipe on top of
the box?
• Hint: The weight density of the chemical
is found to be 90lb/ft3
Square box
with a side
length of 7
Density of
chemical
found to be
90lb/ft3
Solution
• Volume of each slice of chemical =
7*7*dy
• Increment of force= 90*49dy
=4410dy
• Distance lifted =7-y
• Work=force * distance
= 4410 ∫70(7-y)dy
=108045
Thank you for your undivided
attention