Find the volume of the solid of revolution generated by rotating about $y = a$ the region bounded by the loop of the given relation
$$2ay^2=x (a−x)^2$$
for $0≤x≤a$ and $a>0$.
This seems like a place for the washer method. I would do the volume from the farther edge minus the volume from the closer edge. But, how do I figure out those volumes - it is a loop?
The loop circumference consists of two segments, one above and the other below the $x$-axis, given respectively below by rearranging the loop equation $2ay^2=x(x-a)^2$,
$$y_1(x)=\sqrt{\frac{x(a-x)^2}{{2a}}},\>\>\>\>\>y_2(x)=-\sqrt{\frac{x(a-x)^2}{{2a}}}$$
The segment above the $x$-axis is closer to the revoling axis $y=a$ than the segment below the $x$-axis.
The volume resulting from the loop revolving around the axis $y=a$ is a ring-like solid with the center line same as $y=a$, yet with variable widths in the $x$-direction. It can be viewed as a stack of washers along the $x$-direction over $0\le x\le a$.
For the washer at $x$ (the blue line in the diagram), its the inner and outer radii are respectively $r_1$ and $r_2$, given by
$$r_1= a - y_1(x),\>\>\>\>\> r_2= a - y_2(x)$$
Then, the area of the washer at $x$ is
$$A(x) = \pi(r_2^2-r_1^2) =\pi \left(a+\sqrt{\frac{x(a-x)^2}{{2a}}}\right)^2-\pi\left(a-\sqrt{\frac{x(a-x)^2}{{2a}}}\right)^2=2\pi\sqrt{2ax}(a-x) $$
Then, the volume is obtained from integrating over the washer areas over $0\le x\le a$,
$$V = \int_0^a A(x)dx = 2\pi\sqrt{2a}\int_0^a \sqrt{x}(a-x)dx$$ $$=2\pi\sqrt{2a}\cdot\frac{4a^{5/2}}{15} =\frac{8\sqrt2}{15}\pi a^3$$