Chapter
6
~The area A of the region bounded by the
curves
y=f(x), y=g(x), and the lines x=a, x=b, where f and g are continuous
and f(x)
g(x) for all x in [a,b] is
A=
A=
or A=
Ex.: Find the area bounded by the curves:
Y= 20-x2 y = x2-12
20 - x2 = x2 –
12
A = [(32(4)-2(43 ))-(32(-4)-2(-43))]
32 - x2 = x2
A = [(128 - 128/3) - (-128 + 128/3)]
32 = 2x2
A = [(128 – 128/3 + 128 – 128/3)]
x2 =
16
Find a common denominator
x =
4
(a = -4, b =
4)
A = 384/3 –128/3 + 384/3 – 128/3
So if
A = -256/3 + 768/3 = 512/3
Then
Answer = A = 512/3
Ex:
Find the area enclosed by the line y
= x – 1
and parabola y2 = 2x + 6
Solve for x
Y = x – 1 y2
= 2x +
6
X = y + 1 x = ˝(y2)
– 3
Integrate between appropriate y-values
Y= -2 and y = 4
= [-1/2(y3/3)+(y2/2)+4y]
= [(-1/6(43) + 42/2
+
4(4)) – (-1/6(23 + 22/2 + 4(2)))]
= -1/6(64) + 8 + 16 – (4/3 + 2 – 8) =
18
Answer = A = 18 units2
Trigonometric Functions:
D/dx(sin x) = cos x
D/dx(cos x) = -sin x
D/dx(tan x) = sec2x
Ex: Find the area enclosed by y = sin x and y = -cos x , x = 0 x =
Solve for x
Sin x = -cos x
Sin x/cos x = -1
Tan x = -1
X = 3
/4
Integrate between appropriate x-values
(x = 0 to x = 3
/4),(x
= 3
/4
to x =
)
Section 2
Volume = V = AH
V =
AH
V =
r2h
V = LWH
Cross Sections:
For a solid S that isn’t a cylinder we first “cut” s into pieces and
approximate each piece by a cylinder. We estimate the volume of S by
adding
the
volumes of the cylinders. We then arrive at the exact
volume of S
though a limiting process in which the number of pieces becomes large.
We start by intersection S with a plane and obtaining a plane region
that is
called a cross-section. Let A(x) be the area of the cross-section of S
in a
plane Px
perpendicular
to the x-axis and passing through the point x, where
. The cross-sectional area A(x) will vary as x increases
from a to
*Definition of Volume
If the cross-sectional area of S in the plane Px, through x and
perpendicular
to the x-axis is A(x) where A is a continuous function, then the volume
of s
is
, where A(x) is area.
Cross sections perpendicular to the x-axis are squares
solve for
ŕ Solution
Disc
Ex: Find the volume of the solid obtained by rotating about the x-axis
the
region under the curve
from 0 to 1.
Find A(x) =
(
)
from 0 to 1
dx
xdx
(1/2)-
(0/2)
Solution: V(x) =
/2
Washer
Ex: Find the volume of the resulting solid from the region enclosed by
the
curves y = x and y = x
is rotated about x-axis, 0 to 1
ccc
cccc
=
(
)
V =
Shell
Ex:
find the volume of the solid obtained by rotating the region enclosed
by the
curves y = x and
y = x2 rotated about y =2.
[
Solution: 
/15
Chapter 6 Links:
http://www.math.uwo.ca/courses/Online_calc_notes/050/unit10/Unit10.pdf
http://www.math.uci.edu/~jlambers/math2b/fall03/
http://tutorial.math.lamar.edu/AllBrowsers/2413/AreaBetweenCurves.asp
Recommendations for
AP Calculus students:
Do your work and don’t sleep in class. You
need to
pay attention on most days because this isn’t an easy class. Do
your
homework at takes
notes and you
might be alright. You can survive
for a
little while doing the minimum, but it will eventually get to
you. It’s
hard to keep focus at the
end of the year, but you have to
do it so you
get the
points.
To succeed in
AP Calc,
students need to be organized, give about 1 hour a day for homework,
pay
attention, and go in for extra help if needed. I
would recommend
getting
a lot of sleep at night, eating a good breakfast, and having some spare
batteries for your calculator. You might lose all
respect your
teacher ever
had for you if you fall asleep too much in class, and from there on,
your grade
goes down pretty fast.
If you pay
attention
and do your work you will do fine. If not, you will hate this
class. Make sure you pay attention on limits. If you don’t
pay
attention to limits it will come
back to haunt you. Everything
eventually
comes back. You will use stuff in later chapters so it is
important to
pay
attention.