[Calc II]: Shell Method - Dealing with offsets So I'm told to use the shell method for a triangle bounded by $(1,1)$, $(1,2)$ and $(2,1)$. The function values, I've found, are then $y = 3 - x$, $y = 1$ and $x = 1$. I'm rotating it around the $y$-axis. I'm having trouble finding my values for $h(x)$ and $ρ(x)$ though. The two offsets are really screwing with me. Any help understanding how to deal with offsets using the shell method?
I know that the $a$ and $b$ for the definite integral should be $1$ and $2$, respectively. I've tried setting $h(x) = 3 - x$ and $ρ(x) = x$, but it does not give me the correct answer of $4\pi/5$.

Since we are rotating around the $y$-axis, the method of shells says that the volume is 
$$ V = \int_a^b A(x)\, dx $$
where $A(x)$ is the cross-sectional area given by
$$ A(x) = 2\pi\left(\textrm{radius}\right)\left(\textrm{height}\right). $$
Since we are rotating around $x=0$ and the triangle is on the first quadrant, the radius is $\rho(x) = x$. Now, vertically (in the $y$-direction) the triangle is bounded by $y_\textrm{lower}=1$ and $y_\textrm{upper}=3-x$. Consequently, the height is 
$$ h(x) = y_\textrm{upper} - y_\textrm{lower} = 3-x-1 = 2-x. $$
Since the integration variable is $x$, the limits $a, b$ is determine by looking at the $x$-direction; it should be clear that $a = 1$ and $b = 2$. Hence, 
\begin{align*}
V & = \int_1^2 2\pi x(2-x)\, dx \\
& = 2\pi\int_1^2 2x - x^2\, dx \\
& = 2\pi\left(x^2 - \frac{x^3}{3}\right)\bigg|_1^2 \\
& = 2\pi\left[\left(2^2 - \frac{2^3}{3}\right) - \left(1^2 - \frac{1^3}{3}\right)\right] \\
& = 2\pi\left[\left(4 - \frac{8}{3}\right) - \left(1 - \frac{1}{3}\right)\right] \\
& = 2\pi\left[\frac{4}{3} - \frac{2}{3}\right] \\
& = \frac{4\pi}{3}.
\end{align*}
