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

    \[V(x)=\int \pi  (f(x))^2 dx\]


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

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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