Spinning a Circle, along a Circle!

We note that the radius of the circle at any $x$ is equal to $y$ , where $x^2+y^2 = 4$ . Therefore an volume element $\delta V = \pi y^2 \ \delta x$ and as $\delta x \to dx$ , we have:

$\begin{aligned} V & = \int_{-2}^2 \pi y^2 \ dx & \small \color{#3D99F6}{\text{As the solid is symmetrical on the origin}} \\ & = 4 \int_0^2 \pi y^2 \ dx & \small \color{#3D99F6}{\text{Let }x = 2 \cos \theta, \ y = 2 \sin \theta \implies dx = - 2 \sin \theta \ d \theta} \\ & = 32 \pi \int^\frac \pi 2 _0 \sin^3 \theta \ d \theta & \small \color{#3D99F6}{\text{Let }u = \cos \theta \implies du = - \sin \theta \ d \theta} \\ & = 32 \pi \int_0^1 (1-u^2) \ du \\ & = 32 \pi \left(u-\frac {u^3}3\right) \bigg|_0^1 \\ & = \frac {64\pi}3 \approx \boxed{67.0206} \end{aligned}$

This can be done with geometry without calculus!

C, r = Distance of centroid from diameter, radius of the circular path

$2 (2 \pi C)\left(\dfrac{\pi r^2}{2}\right) = 2 \left(2 \pi \dfrac{4\times 2}{3\pi}\right)\left(\dfrac{\pi (2^2)}{2}\right) = \textbf{67.0206}$