If the region $D$ is revolved about the $z$-axis in $ℝ^3$, then the volume of the resulting solid is

Question

Consider the region $D$ in the $yz$ plane bounded by the line $y=\frac{1}{2}$ and the curve $y^2+z^2=1$, where $y\geq 0$. If the region $D$ is revolved about the $z$-axis in $ℝ^3$, then the volume of the resulting solid is

But Answer given was $\pi\frac{\sqrt 3}{2}$. Is my method of calculation correct?

Answer 1

How about cylindrical shells? $2\pi\int_0^{1/2}y(\sqrt{1-y^2})\operatorname dy=2π[-(1-y^2)^{3/2}(1/3)]_0^{1/2}=2/3π(1-3\sqrt{3}/8)=π(8-3\sqrt3)/12$.

My answer agrees with yours.