R is the region in the first quadrant that is bounded on the left by the y-axis, on the right by the curve $x = \tan(y)$, and above by the line $y = \pi/4$; l is the line $x = 1$.
I came up with the equation so far of $$\pi\int_0^{\pi/4}(1)^2-(1-\tan(y))^2 dy$$
However, I'm not sure if it is right
On
Assuming that the axis of revolution is the line $x=1$, then you are integrating along the y-axis and using the disc method. So the radius of each disc is
$r = |1 - \tan y| = 1 - \tan y$ since $0 \le y \le \dfrac{\pi}{4}$
so the element of volume is
$dV = \pi r^2 dh = \pi (1 - \tan y)^2 dy$
and the total volume is $\boxed{V = \displaystyle\pi\int_0\limits^{\pi/4}{(1-\tan y)^2 dy}}$
You don't have to include the line $x=1$.The formula is $\pi \int_a^b(R(x))^2dx$. In your case, $R(y)=tan(y)$ so the integral should be $\pi \int_0^\frac{\pi}{4}tan(y)^2dy = \pi(tan(\frac{\pi}{4})-\frac{\pi}{4}-tan(0)+0)=\pi(tan(\frac{\pi}{4})-\frac{\pi}{4})$.