Split an absolute integral at the sign change

$x\cos2x=0$ gives $x=0,\pi/4,3\pi/4$. The intercept points are $(0,0)$, $(\pi/4,0)$ and $(3\pi/4,0)$, with the origin also the y-intercept. The curve is positive on $(0,\pi/4)$, negative on $(\pi/4,3\pi/4)$ and positive on $(3\pi/4,\pi]$, ending at $(\pi,\pi)$. The origin is an included domain endpoint, rather than a two-sided crossing.

An antiderivative is $F(x)=\tfrac12x\sin2x+\tfrac14\cos2x$. Hence $\int_{\pi/4}^{\pi}|x\cos2x|dx=-[F]_{\pi/4}^{3\pi/4}+[F]_{3\pi/4}^{\pi}=\pi/2+(1/4+3\pi/8)=7\pi/8+1/4$. The modulus bars in the rendered question are essential; the native extraction drops them.

The stated region adjacent to $x=\pi$ starts at $x=3\pi/4$. Thus $V=\pi\int_{3\pi/4}^{\pi}(x\cos2x)^2dx\approx10.46525977=10.5$ cubic units (3 s.f.). The negative lobe from (iii) is not part of this region.