



There
is a very interesting relationship



between
the equation for the surface area of a tube ... 





A = 2 p r h






and
the equation for the volume of a cylinder ... 





V = p
r^{2}h






Let's
rearrange the order of the terms 


to
make the relation more obvious. 





A = p
h × 2r



V = p
h × r^{2}






See
it? 





Sure
you do! 


It's
... 











in
general ... 











Wow!
The equation for the volume of a cylinder 


(of
radius "r") is the integral of the area of a tube as
"r". 


the
radius, goes from zero to some "r" value. 





We
can use this idea to find the volume of 


round
3 dimensional objects. 


What
we need to know is ... 





1)
The distance from the center of the object 


to
the edge (the radius). 





2)
The height of the object. 





A
way to think of this is a cylinder, 


made
up of very thin tubes (sometimes called shells) 


sitting
inside each other like a weird onion. 











If
we were to look at this from the side 


and
set the thing on a coordinate axis 


we
get something like ... 











The
radius, "r", of each of these tubes is a certain distance 


in
the X direction 


so
now instead of using "r" in the equation, 


now
we could also use X. 


The
height (h) of the cylinder is some Y distance, 


so
instead of using h in the equation, 


we
could also use Y. 


Using
X's and Y's the equation becomes ... 











Here
Y is just some constant height, 


so
we can write ... 











This
is the volume of the entire cylinder. 





What
happens if we have a cylinder 


that
puts on antigravity shoes and floats up off the X axis? 











Hey!
No problem! 


We
just find out where the top of the cylinder is (call it Y_{2}) 


and
where the bottom of the cylinder is (call it Y_{1}) 


and
instead of Y in the equation we have (Y_{2}  Y_{1}). 











Now
we have ... 











(Y_{1}
and Y_{2} are still just constants) 





OK,
let's get really wild! 


Suppose
the top and bottom of this round thing 


aren't
straight lines like Y = 3 or something? 


What
if they are functions? 


What
if the top is Y = X ^{2} + 4 


and
the bottom is Y = 2X ^{2} 











This
forms a round thing that looks sort of like a shallow bowl 


(but
it wouldn't sit on a table too well) 





To
find the volume of this shape, substitute X
^{2} + 4 for Y_{2}. 


It
is the equation of the upper surface boundary line. 


Substitute
2X ^{2} for Y_{1}. 


It
is the equation of the lower surface boundary line. 





We
use the place where the two equation lines cross on the right 


as
the upper limit of integration. 


Let's
find it. 





Y = X^{2} + 4 Y =
2X^{2}



2X^{2} =
X^{2} + 4



X^{2} =
4



X = 2






So
the equation is ... 











THAT'S
THE EQUATION FOR THE VOLUME OF MATERIAL 


OUR
FUNNY CUP IS MADE OUT OF! 





Don't
get confused and think that it's the equation 


for
the volume of stuff you can put IN the cup, 


that's
some other equation. 


So
let's figure out the volume of the material 


that
the cup is made of ... 











We
move the constants (p
and 2) left, 


and
multiply everything else out. 











Integrate
and solve ... 











In
general, if we have any shape like this ... 











The
equation for the volume of material is ... 











Since
2p is a constant, we can make this ... 

















We
can also make a shape 


out
of two lines that don't touch the Y axis. 


Suppose
we have ... 











NO
BIG DEAL! 





Just
use "b" as your upper integration limit, 


and
"a" (not zero) as the lower limit. 











This
gives us a shape that looks something like 


a
donut or an onion ring. 











One
thing to mention, is that the view shown 


is
actually slightly looking down on the shape. 


If
we looked at it straight from the side, 


the
view would be pretty boring. 











What
if we wanted to make an object like a donut 


that
is NOT centered around the Y axis? 


What
if we wanted it centered around the line X = 2? 











It's
about the same shape, but much bigger around. 


The
radius has been increased. 





HOW
MUCH? 





Well
we moved the center of the donut 2 units, 


so
I'd say the radius is 2 units bigger. 





Before,
when the donut was centered on the Y axis 


we
used the value of X for the radius distance. 


Now
it looks like we use X + 2! 





Big
Time Example: 





Say
we want to find the volume of some object 


that
is bounded on the top by the line ... 





F(X) = 
(X  2)^{2} + 2






and
on the bottom by the line ... 





G(X) = (X
 2)^{2}






And
we want to rotate it around the line X = 3. 











First
we need to find the places that the two lines cross, 


also
known as the upper and lower limits of integration (X = a and X = b)
... 





(X 
2)^{2}
= 
(X 
2)^{2} + 2 


X^{2}  4X +
4 = 
X^{2} + 4X  2 


2X^{2}  8X
+ 6 = 
0 


X^{2}  4X +
3 = 
0 


(X  3)(X
 1) = 
0 


X = 3
or X = 1 


(b = 3,
a
= 1) 





We
can see from the picture that the radius is X + 3, 


so
we use X + 3 for "r" instead of just X. 











First
a bunch of algebra ... 











Now
the integration! 











So
the volume of this "donut" is 83.8 cubic inches. 


There
is a "quick and dirty" way to estimate these puppies ... 


When
you are working on a problem that takes this many steps to solve, 


it
is easy to make a simple math error and get a very wrong answer. 


It
would be great if we could find a way to estimate the answer. 


Then
when we finish the big calculation, 


we
can see if the answer we got was reasonable. 





OK,
How do we do it? 





Say
we have an object with a cross section of: 











Who
knows exactly what the area of this thing is, 


but
the square that surrounds it has an area of 4. 


So
let's guess that the donut has an area of about 3. 











The
radius of the center of the object is 5, 


so
if we cut this donut and straightened it out 


into
a cylinder it would have a height equal 


to
the distance around the object. 


(circumference
= 2pr, and r = 5) 


The
volume of this cylinder is equal to 


the
base times the height. 





V =

bh 


V =

3 × 10 p 


V = 
30 p 


V = 
94.2 





Our
guess was 94.2 


and
the actual area was 83.8. 





copyright 2005 Bruce Kirkpatrick 
