Hello bassman,
The volume we are being asked to compute is that of the frustum of a cone, and the formula for such a solid is derived using the calculus here:
http://mathhelpboards.com/math-notes-49/volumes-pyramids-6131.html
$$V=\frac{\pi h}{3}\left(R^2+Rr+r^2 \right)$$
In this problem, we have:
$$h=1,\,R=\frac{3}{2},\,r=1$$
Hence:
$$V=\frac{\pi}{3}\left(\frac{9}{4}+\frac{3}{2}+1 \right)=\frac{\pi}{3}\left(\frac{19}{4} \right)=\frac{19\pi}{12}$$
Thus, we know the answer given by the book is correct. So, let's work this in the manner you are expected to use.
Let's use the disk method. The volume of an arbitrary disk is:
$$dV=\pi r^2\,dx$$
where:
$$r=y=2-\frac{1}{2}x=\frac{4-x}{2}$$
and so we have:
$$dV=\pi\left(\frac{4-x}{2} \right)^2\,dx=\frac{\pi}{4}(4-x)^2\,dx$$
Adding the disks by integrating, we may state:
$$V=\frac{\pi}{4}\int_1^2(4-x)^2\,dx$$
Using the substitution:
$$u=4-x\,\therefore\,du=-dx$$
we may state:
$$V=\frac{\pi}{4}\int_2^3 u^2\,du=\frac{\pi}{12}\left[u^3 \right]_2^3=\frac{\pi}{12}\left(3^3-2^3 \right)=\frac{19\pi}{12}$$
