volume of $y^2-x^2=1$, $y=2$, about the $x$-axis

Question

The answer is $\frac{4\pi}{3}$, but shouldn't it be $\frac{8\pi}{3}$? I used the disk method and when I solved for the equivalent $x$ value of $y=2$ I got $x=\pm \sqrt 3$. So when setting up the integral I wrote it as $2 \int_0^{\sqrt 3}$. But to get $\frac{4\pi}{3}$ it's just $0$ to $\sqrt 3$, why is that?

Answer 1

The region has not been fully described. But from the post I assume we are rotating the region above the top half of the hyperbola, and below $y=2$, about, as the post specifies, the $x$-axis.

I get $4\pi\sqrt{3}$. Half the volume is $$\int_0^{\sqrt{3}}\pi\left(2^2-(x^2+1)\right)\,dx.$$ The integration is routine, we get $2\pi\sqrt{3}$. Then double.

Remark: If we rotate instead aout the $y$-axis, we are integrating $\pi(y^2-1)$ from $y=1$ to $y=2$. We get $\frac{4\pi}{3}$. There is no doubling. The solid is swept out twice, once by the right half and once by the left half.