Why is the volume of the solid of revolution between two graphs $\int_{0}^{b}\pi[f^2(x)-g^2(x)]dx$ instead of $\int_{0}^{b}\pi[f(x)-g(x)]^2dx$?

Question

Let $f$ and $g$ be two non-negative functions which are integrable over the interval $[0,b]$ and satisfy $f \geq g$ on $[0,b]$. I am told that the volume of the solid of revolution created by rotating the region between the graphs of $f(x)$ and $g(x)$ about the x-axis is given as $\int_{0}^{b} \pi f^2(x)dx-\int_{0}^{b}\pi g^2(x)dx$, or $\int_{0}^{b}\pi[f^2(x)-g^2(x)]dx$.

I think I understand how to see this integral calculation visually. The solid of revolution created by the region under the smaller graph "carves out" a hollow in the solid of revolution created by the region under the bigger graph, and the volume of the carved out solid is what we obtain through the integral calculation. More specifically, one can think of summing up the volumes of infinitesimally thin annuli over the interval, with the inner and outer radii of the annulus at some $x $ on the interval given by $g(x)$ and $f(x)$ respectively.

However, I don't understand why the volume is not instead defined as $\int_{0}^{b}\pi[f(x)-g(x)]^2dx$, as this seems to follow naturally from how the area between between the two graphs would be found. So in this scenario, you find the volume of a solid of revolution created by summing up the areas of infinitesimally thin circles of radius $f(x)-g(x)$ at some given $x$ in the interval. It seems as if you could define the solid of revolution this way instead. Why is the volume not defined through this calculation? Is the other calculation more useful?

Am I missing something obvious here? Is the way I see the calculation wrong? If not, is there a good reason why the volume is defined the way it is?

Answer 1

For some intuition, consider revolving the region between the graphs of $y=y_0$ and $y=y_1$ on $[a,b]$. For definiteness let's say $y_1>y_0>0$. Now the region in space that you are creating is the space between two cylinders whose axes are the $x$ axis. The inner cylinder has radius $y_0$ (the distance between the bottom of the segment and the axis of revolution) and the outer cylinder has radius $y_1$ (the distance between the top of the segment and the axis of revolution). So the cross sectional area is $\pi y_1^2 - \pi y_0^2$ (the area between two concentric circles). You would have $\pi (y_1-y_0)^2$ if you revolved around $y=y_0$ instead of $y=0$, because then there would be no separation between the axis and the segment.

What you described, which is commonly called the washer method, is just the generalization of this to the case when the radii depend on $x$.

Answer 2

Consider a region bounded by the lines $x=0$ and $x=b$ and the curves $y=f(x)$ and $y=g(x)$, with $c\le g(x)\le f(x)$. Suppose we revolve this region about a line $y=c$ and use the washer method to find the volume. Consult the sketch below.

The sketch shows one such washer, whose volume is

$$\pi (f(x)^2 - g(x)^2) \Delta x$$

The volume of a cylinder is $\pi R^2 h$, where the radius is $R$ and the height is $h$. From this cylinder we cut out another cylinder with a smaller radius $r$. The resulting washer has volume $\pi R^2h-\pi r^2h = \pi (R^2-r^2)h$, and we just replace symbols with $f,g,\Delta x$ accordingly, make $\Delta x\to0$, and take infinitely many washers to get

$$V = \pi \int_0^b (f(x)^2-g(x)^2) \, dx$$

(assuming we revolve about the $x$-axis so $c=0$)

On the other hand, integrating with $(f(x)-g(x))^2$ amounts to applying the disk method on a single function $h(x)=f(x)-g(x)$, for which the volume would be

$$V' = \pi \int_0^b (f(x)-g(x))^2 \, dx$$