Consider one of the regions bounded only by functions $f(x)=\sin(x)$ and $g(x)=\cos(x)$ (say from $x=\frac\pi4$ to $x=\frac{5\pi}4$) Now if that region were to revolve around the $x$ axis, what would be the volume of the resulting shape.
Now, if you consider the region first, it intersects the $x$-axis. Which means that when it gets revolved, some parts are going to be encapsulated twice. To avoid that I tried to divide the region into four sub-regions. Since the shape is symmetric, it seems sufficient to calculate the volume of the first two regions then multiply the answer by $2$.
The first region is simple, it can be done by the washer method. The top part is $f(x)$ which is $\sin(x)$ and the bottom part is $g(x)$ which is $\cos(x)$. This gives us the volume by the following integral:
$$\pi\int_{\pi/4}^{\pi/2} \sin^2(x)-\cos^2(x)\ {dx} =\frac{\pi}{2}$$
Now the second region, since in the second region $\sin x$ is strictly bigger than $\cos(x)$. It seems to me that the revolution of $\cos x$ is going to be entirely encapsulated in the revolution of $\sin x$, so it can be ignored. This means that the significant region can be calculated by the disk method. This gives us the volume by the following integral:
$$\pi\int_{\pi/2}^{3\pi/4}{\sin^2(x)}\ {dx} =\frac{\pi^2}{8}+\frac{\pi}{4}$$
The total volume should be equal to double the sum of the previous two volumes, which should equal to: $$2\left(\frac{\pi}{2}+\frac{\pi^2}{8}+\frac{\pi}{4}\right)=\frac{\pi}{4}(6+\pi)$$
Is that the correct approach to that problem? The answer I got doesn't seem to agree with the answer I got from Wolfram Alpha.