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A Stack of Chips
Disk Volume Integrals


 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) = X2

 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 = pr2

 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(X2)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!

 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)

 Volume = pr2h

 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:

 h2 + r2 = 52

 or, solving for r 2 ...

 r2 = 25 - h2

 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

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