Evaluate the volume generated by revolving the region bounded by the curve $x^2-y^2=9$ and the lines $y=4$ and $y=-4$ about (1) the $x$-axis and (2) the $y$-axis.
For 'about the $x$-axis', I've tried the cylindrical method and got $254.154$ (correct to 3 dp) and I'm not too sure if I'm correct. These were my steps: $$V = 2 \pi \int_{-4}^4 (4 + y)(5-\sqrt{9+y^2})dy$$ $$=2\pi \left(80-36\ln \left(3\right)\right)$$ $$=254.154$$ I was thinking that I don't need to take $-\sqrt{9+y^2}$ and just took $\sqrt{9+y^2}$. I'm thinking that it's redundant, since if it rotates it'll rotate 180° to become a cylindrical shape.
Any form of help is much appreciated. Thanks!
Your attempt at shell integration of the solid revolved about the $x$-axis involves the wrong integrand (a correct derivation is below). Here's a diagram of the rotated region:
We can rearrange $x^2-y^2=9$ to yield $x=\sqrt{9+y^2}$.
For rotation about the $x$-axis, the $yz$ plane splits the solid into two equal parts. Then shell integration gives $$V=2\cdot2\pi\int_0^4xy\,dy$$ $$=2\cdot2\pi\int_0^4y\sqrt{9+y^2}\,dy$$ $$=4\pi\left[\frac13(9+y^2)^{3/2}\right]_0^4$$ $$=4\pi\left(\frac{125}3-\frac{27}3\right)=\frac{392\pi}3=410.501$$ For rotation about the $y$-axis, disc integration gives $$V=\pi\int_{-4}^4x^2\,dy$$ $$=\pi\int_{-4}^4\left(\sqrt{9+y^2}\right)^2\,dy$$ $$=\pi\int_{-4}^4(9+y^2)\,dy$$ $$=\pi\left[9y+\frac{y^3}3\right]_{-4}^4$$ $$=\pi\left(36+\frac{64}3+36+\frac{64}3\right)=\frac{344\pi}3=360.236$$