The problem is straightforward with the Disk-Washer method. To help me fully understand these rotation problems I like to work the problems using both Disk-Washer and Shell (some problems become too difficult to solve for the x or y variable, but most work at the Calculus 1 level).
The problem is: Find the volume of the solid that results when the region is rotated about the y-axis.
$y=3-2x, y=2, y=0, x=0$
Using $V=\pi \int_{0}^2 (\frac{3-y}{2})^2 dy$ I get the correct answer of $13\pi/6$
When evaluating how to do this via the Shell method I notice that I will need to break this region into 2 pieces - a rectangle and a triangle.
Since the line $y=2$ intersects the line $y=3-2x$ at $(\frac{1}{2},2)$ I chose the two shell integrals as follows:
$V=2 \pi\int_{0}^\frac{1}{2} (x)(2) dx + 2\pi\int_{\frac{1}{2}}^{\frac{3}{2}} (x-\frac{1}{2})(3-2x)dx $
This does NOT produce the same answer. I think the correct answer is produced if I use:
$2\pi\int_{\frac{1}{2}}^{\frac{3}{2}} (x)(3-2x)dx $ for the second integral. But I don't understand this. For my second integral, it would seem to me that my radius is NOT $x$ but $(x-\frac{1}{2})$.
Can you assist with doing this problem via the Shell Method?
In the second integral, the cylinder radius should be $x$, not $x-\tfrac{1}{2}$, as it is the distance of the position of the cylindrical shell from the rotation axis: the line $x=0$, not the line $x=\tfrac{1}{2}$. With integrand $x(3-2x)$, you obtain the result $\frac{13\pi}{6}$ as with disks.