Integration application to volume

Question

I'm trying to find out the volume of: Area enclosed by $y=\arctan x, \ y=0, \ x=1$ rotated about y-axis.
I tried to use both the disc and cylindrical method, and I can get it with the cylindrical shell method, but not with the disc method. Here's my cylindrical shell solution:
$$ 2\pi rh\delta t=2\pi x\arctan x\delta x \\ \int^1_0 2\pi x\arctan xdx=\pi(\frac{\pi}{2}-1) $$ And here's my attempt at a disc solution:
Inner shell$$ \pi r^2\delta h=\pi x^2\delta y=\pi(1-\tan y)^2\delta y \\ \int^{\frac{\pi}{4}}_0{\pi(1-tany)^2)} $$Outer shell$$ \pi r^2 \delta h=\pi1^2\delta h \\ \pi\int^{\frac{\pi}{4}}_0{1}dy $$However,$$ \pi\int^{\frac{\pi}{4}}_0{1}dy-\int^{\frac{\pi}{4}}_0{\pi(1-tany)^2)}\neq\pi(\frac{\pi}{2}-1) $$Could someone help me pick out where I have gone wrong? Thank you very much for your help!

Answer 1

Yes your working for the cylindrical shell method is correct. In disk method, you have a mistake in the volume integral for the inner shell.

It should be $\displaystyle \int_0^{\pi/4} \pi x^2 \ dy = \int_0^{\pi/4} \pi \ \color {blue} {(\tan y)}^2 \ dy = \frac{\pi}{4}(4-\pi) \ $.

So the volume using disk method $ = \displaystyle \frac{\pi^2}{4} - \frac{\pi}{4}(4-\pi) = \pi (\frac{\pi}{2}-1)$