Find the centroid of the solid generated by rotating

Find the centroid of the solid generated by rotating about the $y$-axis the area bounded by $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ and the lines $y=0$ and $y=b$ ($b>0$).

Step 1: Cross-sectional radius

\[r(y)=\frac ab\sqrt{b^2+y^2} \implies \pi r(y)^2=\frac{\pi a^2}{b^2}(b^2+y^2)\]

Step 2: Volume

\[V=\frac{\pi a^2}{b^2}\int_0^b(b^2+y^2)\,dy=\frac{\pi a^2}{b^2}\left(b^3+\frac{b^3}3\right)=\frac{4\pi a^2b}3\]

Step 3: Moment about the $xz$-plane

\[M_{xz}=\frac{\pi a^2}{b^2}\int_0^by(b^2+y^2)\,dy=\frac{\pi a^2}{b^2}\left(\frac{b^4}2+\frac{b^4}4\right)=\frac{3\pi a^2b^2}4\]

Step 4: Centroid

\[\bar y=\frac{M_{xz}}V=\frac{3\pi a^2b^2/4}{4\pi a^2b/3}=\frac{9b}{16}\]

\[\boxed{\left(0,\frac{9b}{16},0\right)}\]

Step 5: Sanity check

Since $\bar y=\frac{9b}{16}\approx0.5625b>\frac b2$, the centroid sits above the geometric midpoint, consistent with the hyperboloid widening ($r^2\propto b^2+y^2$) as $y$ increases — more volume is concentrated toward $y=b$.