Given $f(x)=\sin{(x-1)}, [0, 1 + \frac{\pi}{2}]$, find the volume of revolution with respect to the $y$-axis, given the limits $y=0$, $y=1$, and $x=0$.
If I use the disk method to find the volume: $\pi\int^{1}_{0}{(1+\arcsin{y})^2}dy$, the answer will be $\approx 8.19$.
However, if I do it using the shells method: $2\pi\int^{1+\frac{\pi}{2}}_{1} {x\sin(x-1)}dx$, the answer would be $4\pi \approx 12.6.$
What's my mistake, or why is there a discrepancy?
You are calculating the volumes of two different solids.
Consider $y=\sin x$ on $[0,\pi/2]$, and calculates the volume of the solid obtainded by rotating about the $y$-axis the region bounded by $x=0$, $y=1$ and the curve on $[0,\pi/2]$. Using your way to do the calculation, you would get $$ \int_0^{\pi/2}2\pi x\sin x\ dx=2\pi $$ and $$\int_0^1\pi\arcsin^2 y\ dy=\frac{1}{4}\pi^3-2\pi $$ But actually, $$ \color{blue}{\pi\cdot \left(\frac{\pi}{2}\right)^2\cdot 1-}\int_0^{\pi/2}2\pi x\sin x\ dx=\int_0^1\pi\arcsin^2 y\ dy=\frac{1}{4}\pi^3-2\pi $$