Let’s find out why. Here’s the outline of the discussion:
- Examine the simple case, .
- Agree that the function exists.
- Agree that if is smooth and continuous, then is smooth and continuous.
- Conclude that which leads to
Examine a simple case
Let’s take , from to .
By rotating the straight, horizontal line around the axis, we generate a cylinder. We already know that the volume of a cylinder can be found using the formula:
Where is the radius of the cross sectional circle, and is the length or height of the cylinder.
We also remember that a function maps a value from one set (domain) to another set (range). Let’s create a function that maps the length of the cylinder to the volume of the cylinder. Our domain is the length of the cylinder, which in our case is the measurement on the axis. If we keep one end of the cylinder parked at , then the length of the cylinder is just .
The cylinder length . The cylinder radius . Using the volume of cylinder formula given above, our volume function is:
We see that this agrees with when as follows:
Let’s check out the volume of the cylinder when and , using our function :
Now let’s check out the volume of the cylinder when and .
Lastly, let’s check out the volume of the cylinder when and . We can calculate this by taking the longer cylinder (from zero to ten) and subtracting the shorter cylinder (from zero to 4).
The calculation above is exactly the calculation
Let’s move on to understand why this calculation works for any smooth, continuous function , not just the simple straight line that generates the familiar cylinder.
Next step: Agree that the function exists for other smooth continuous functions. (under construction)
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