



In
the method we just used,



we
found things that were centered on the Y axis, 


or
on a line parallel to the Y axis. 


We
do have ways of using that method to find 


the
volume of things centered on the X axis, 


but
there is another method that is easier to use 


on
things centered on the X axis. 





Suppose
we have the equation 





F(X)
= X^{2}






and
we want to find the volume of the object 


that
is built by taking the area under F(X)
= X ^{2}
from X = 0 to X = 2. 











and
spinning the X axis ... 











This
thing kind of looks like a funnel, but it's solid. 


Hey,
what kind of an equation will find the area of that? 


If
we look at it from the end on the right, 


it
will look like a circle ... 











In
fact, we could slice this "funnel" at lots of X values 


and
create zillions of circular disks ... 











The
equation for the area of a circle is: 





Area
= pr^{2}






The
question is, what is the radius of any of these circles? 





The
answer is, the equation (F(X))
value at the X value 


where
the slice is made. 


Our
original equation was F(X)
= X ^{2},
so r = X ^{2}. 


We
substitute the function X
^{2} into the
equation A = pr
^{2} for
"r". 


We
get ... 





Area
= p(X^{2})^{2}






Our
area went from X = 0 to X = 2, 


if
we find the integral of this term from X = 0 to X = 2 


We
will have the volume of our funnel. 











Your
kidding? 





Nope,
it works ... every time! 











Example: 





Suppose
that we have a sphere with a radius of 5 inches 


(something
about the size of a basketball). 











and
we fill it with water 


until
the water is 3 inches deep at the center. 











At
this point, what is the volume of the water? 





Think
of the water as being made up of 


lots
of incredibly thin disks (like very thin coins) 


stacked
on top of each other. 


If
we found the sum of the volume of these "coins" 


we
would know the volume of the water. 











The
formula for the volume of a coin 


(it's
just a cylinder that never got very tall) 


is: 





Volume = pr^{2}h






If
we can make these coins thin enough 


we
will be able to use the dh for the height. 


The
volume equation will be: 











We
will be integrating from the bottom of the sphere, 


to
the top of the water (3 inches up from the bottom), 


the
trick is going to be figuring out a way 


to
express the radius of these coins as a function of height. 





That
is, we need to substitute something with h in it for r 


so
we have h and dh in the equation. 


How
do we do it? 





First
look at the "basketball", sitting on an XY coordinate
system. 


The
water is 3 inches deep so it's surface is at Y = 2. 











We
said that this ball has a radius of 5 inches, 


so
let's draw that in. 


A
line from the center of the circle to the edge 


where
it meets the top of the water. 











Look
at: 


the
5" radius line, 


the
negative piece of the Y axis above the water, 


and
the surface of the water. 





Those
three things form a right triangle ... 











The
Y value is one leg, 


the
radius of the water surface is the other leg, 


and
5 is the hypotenuse. 





The
Y distance is the water height (h), 


so
we can use h instead of Y in our equations. 


So
as old Pythagoras says: 





h^{2}
+ r^{2} = 5^{2}






or,
solving for r ^{2}
... 





r^{2}
= 25  h^{2}






The
thing to remember in what we just did, 


is
that "r" is the radius of the water disk 


not
the radius of the ball (that was 5). 





Back
at the beginning of this problem, 


we
were looking to substitute something into the equation 


to
give us h's instead of r's. 


Guess
what we found. 


In
fact, we had an r ^{2}
in the original equation, 


so
things worked out real nice! 





We
can use this 25  h ^{2}
term in our volume equation for r
^{2}. 


So
... 











We
are going to find the integral of this equation 


from
h  5 to h = 2 


(Remember,
in this equation h is used for Y). 











copyright 2005 Bruce Kirkpatrick 
