Q:

AP Calculus Volume Problem Explaination

Let R and S be the regions in the first quadrant shown in the
figure above. The region R is bounded by the x-axis and the
graphs of $y= 2-x^3$ and $y=\tan(x)$ . The region S is bounded by
the y-axis and the graphs of $y=2-x^3$ and $y=\tan(x)$.
Find the volume of the solid generated when S is revolved about the x-axis?
I attempted the problem using the disk method, and got:
$$V=\pi\int_\limits{0}^{1.266}[(2-x^3)-\tan(x)]^2dx$$
Apparently the problem requires the washer method, and I do not understand why. Could somebody explain why the answer is: 
$$V=\pi\int_\limits{0}^{1.266}[(2-x^3)^2-(\tan(x))^2]dx$$
and why it requires the washer method?

A:

Let me resort to a different example here, trying to make you see why the washer method should be used instead of your disk method. Consider $f(x)=5$ and $g(x)=4$ on interval $x$,say $[0,1]$. What happens if we use the disk method? For that consider $(f(x)-g(x))$ for the integral, which is essentially $y=1$ rotated about the x-axis on $[0,1]$. You can envision a fairly small volume, right?. But know consider the "rectangular" strip created between $y=5$ and $y=4$ from $[0,1]$ and rotate THAT about the x-axis. How can that volume be the same? It is much greater. It is the latter that is meant by the question  

