Volume of Revolution Discussion (iv) -

In this discussion, we are tasked with finding the rate of change of the volume function. By volume function, we refer for now to a function that maps a value on the positive $x$ axis to the volume of the solid formed when the function is rotated 360° around the $x$ axis, between $x=0$ and our chosen point.

With this function, the volume of the cylinder below, from $x=0$ to $x=10$ is $V(10)$ .

$V(10)= \pi \times 3^2 \times 10 = 90\pi$ .

Also, the volume from $x=0$ to $x=4$ would be $V(4)$ .

$V(4)=\pi \times 3^2 \times 4 = 36\pi$ .

Therefore the volume from $x=4$ to $x=10$ is simply $V(10)-V(4)$

The volume of the cylinder in figure 1 is therefore $90 \pi - 36\pi = 54 \pi$ .

Figure 1.

Now let’s look at a more complicated curve. We are going to examine the section of the curve between $x=3$ and $x=3.25$ . Following the argument above, the volume of that brown solid is $V(3.25)-V(3)$ .

Figure 2.

Let’s extract that rotated portion of the curve to see it more closely (Figure 3). We see that the function that forms it is increasing between $x=3$ and $x=3.25$ .

Figure 3.

In Figure 4, we draw a cylinder that is smaller in volume that the actual solid we are studying. It is drawn with a radius that is the minimum value of $f(x)$ between $x=3$ and $x=3.25$ , that is, $f(3)$ . The length of the cylinder is 0.25. Using the familiar volume of cylinder formula, the volume of that cylinder is $\pi \times (f(3))^2 \times 0.25$ .

Figure 4.

Now on figure 5 we see that a larger cylinder has been formed using the maximum value of $f(x)$ on the interval, which is $f(3.25)$ . The volume of this cylinder is $\pi \times (f(3.25))^2 \times 0.25$ .

Figure 5.

The volume of the brown solid we are attempting to calculate lies between the volume of those two cylinders. We can write:

$\pi \times (f(3))^2 \times 0.25\leq V(3.25)-V(3) \leq \pi \times (f(3.25))^2 \times 0.25$

Let’s generalise to two points reasonably close to each other. In place of $3$ lets use $x$ and in place of $3.25$ let’s use $x+h$ – for now let’s suppose that $h$ is positive. Let’s for the sake of argument suppose that our curve is increasing, therefore $f(x)$ is the minimum value of $f$ on this interval and $f(x+h)$ is the maximum value of $f$ on the interval.

Now, the volume of the brown solid can be written more generally as:

$\pi (f(x))^2 h\leq V(x+h)-V(x) \leq \pi (f(x+h))^2 h$

Now our formula is beginning to look familiar. Let’s divide this whole expression by the positive number $h$ .

$\pi (f(x))^2 \leq \frac{V(x+h)-V(x)}{h} \leq \pi (f(x+h))^2$

Finally, let’s take the limit of each part of the expression as $h$ tends to zero.

Left hand side:

$\lim_{h \rightarrow 0} \pi (f(x))^2=\pi (f(x))^2$

Right hand side:

$\lim_{h \rightarrow 0} \pi (f(x+h))^2=\pi (f(x))^2$

Now it’s interesting, as the left and the right hand side of our inequality

$\pi (f(x))^2 \leq \frac{V(x+h)-V(x)}{h} \leq \pi (f(x+h))^2$

tend to the same quantity. The expression in the middle must agree! Read about the pinching theorem (or squeeze theorem) here.

The middle:

$\lim_{h \rightarrow 0}\frac{V(x+h)-V(x)}{h}=V'(x)$

In other words, the rate of change of the volume function is equal to $\pi (f(x))^2$ .

Now we have

$V'(x)=\pi (f(x))^2$

it is a small step to claim that

$V(x)=\int \pi (f(x))^2 dx$

and

$V(b)-V(a)=\int_a^b \pi (f(x))^2 dx$

which takes care of the constant of integration. (In this case, simply the solution to V(0)=0).

Another way to think of this, perhaps intuitively, is to multiply the expression $\pi \times (f(x))^2$ by 1 so that it recognisably matches the formula for a simple cylinder. We recognise this expression to be that of the volume of a cylinder with length one and radius the $y$ value of the curve.

You may have noticed that our original simple example had a constant rate of change. In the example of the cylinder generated by the function $f(x)=3$ , for every one unit increase on the $x$ axis the volume increased by $\pi \times 3^2 \times 1$ . So at any point on the more complicated curves $f(x)$ , should the volume suddenly take on a constant rate of change, the curve would need to have the form $f(x)=$ constant, and the solid would become at cylindrical at this point. As we know, we are not obliged to live out a full minute to get a one minute heart rate – we imagine the solid to become cylindrical just for an instant.

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