Why the volume of this solid isn't equal?

Question

I calculated the Volume of a specific solid using 2 different methods, first I did Integration, and then pure geometry. Consider $$ x=2-\sqrt{y^2+z^2} $$ where : $$0< x < 1$$

I did integration and it gave $4 \pi/3 $ and then I simply calculated the volume of the solid by doing the difference between the cones' volumes, and it gave me $7\pi/3$ ... It's a silly doubt, but I am not understanding!

Basically, this is a cone divided in half, aligned on x-axis. Couldn't I simply do the difference of the bigger cone with the smaller cone?

Answer 1

Let, $x=t$ and integral

$V=∫_1^0 \pi(2-t)^2dt$ $=\pi[(2-t)^3/3]$ $=7\pi/3$

calculating by subtract cone

$V=4\pi*2/3-\pi/3=7\pi/3$

Answer 2

Let radius $r =\sqrt{y^2+z^2}$

For the truncated part that does not contain cone apex. The limits taken automatically decide truncation and its volume calculation.

$$Vol=∫_0^1 \pi r^2 dx = \pi∫_0^1 (2-x)^2 dx = 7 \pi/3. $$