We can work this type of problem in general by requiring one of the roots of the parabolic boundary to be the origin, and the other on the positive $x$-axis.
So let the region be bounded by:
$$y=-kx(x-a)$$ and $$y=0$$, which is also the axis of rotation.
Note: $$0<k$$ and $$0<a$$.
Disk method:
The volume of an arbitrary disk is:
$$dV=\pi r^2\,dx$$
where:
$$r=y=-kx(x-a)$$
hence:
$$dV=\pi \left(-kx(x-a) \right)^2\,dx=\pi k^2\left(x^4-2ax^3+a^2x^2 \right)\,dx$$
Summing up the disks, we find:
$$V=\pi k^2\int_0^a x^4-2ax^3+a^2x^2\,dx$$
Applying the FTOC, we obtain:
$$V=\pi k^2\left[\frac{1}{5}x^5-\frac{a}{2}x^4+\frac{a^2}{3}x^3 \right]_0^a=\pi a^5k^2\left(\frac{1}{5}-\frac{1}{2}+\frac{1}{3} \right)=\frac{\pi a^5k^2}{30}$$
Shell method:
The volume of an arbitrary shell is:
$$dV=2\pi rh\,dy$$
where:
$$r=y$$
To find $h$, consider:
$$y=-kx(x-a)=-kx^2+akx$$
Arranged in standard quadratic form:
$$kx^2-akx+y=0$$
Application of the quadratic formula yields:
$$x=\frac{ak\pm\sqrt{a^2k^2-4ky}}{2k}$$
Hence:
$$h=\frac{ak+\sqrt{a^2k^2-4ky}}{2k}-\frac{ak-\sqrt{a^2k^2-4ky}}{2k}=\frac{\sqrt{a^2k^2-4ky}}{k}$$
And so we have:
$$dV=\frac{2\pi}{k}\left(y\sqrt{a^2k^2-4ky} \right)\,dy$$
To determine the upper limit of integration, we note that the axis of symmetry for the parabola is on $$x=\frac{a}{2}$$ and so the upper limit is:
$$y=-k\frac{a}{2}\left(\frac{a}{2}-a \right)=\frac{a^2k}{4}$$
And so the volume is given by:
$$V=\frac{2\pi}{k}\int_0^{\frac{a^2k}{4}} y\sqrt{a^2k^2-4ky}\,dy$$
At this point, we may want to develop a formula for the indefinite integral:
$$I=\int x\sqrt{a+bx}\,dx$$
Let's try the substitution:
$$u=a+bx\,\therefore\,du=b\,dx$$
Hence, the integral becomes:
$$I=\frac{1}{b^2}\int u^{\frac{3}{2}}-au^{\frac{1}{2}}\,du$$
Using the power rule, we find:
$$I=\frac{1}{b^2}\left(\frac{2}{5}u^{\frac{5}{2}}-\frac{2a}{3}u^{\frac{3}{2}} \right)+C=\frac{2}{15b^2}u^{\frac{3}{2}}\left(3u-5a \right)+C$$
Back-substituting for $u$, we obtain:
$$I=\frac{2}{15b^2}(a+bx)^{\frac{3}{2}}\left(3(a+bx)-5a \right)+C$$
$$I=\frac{2}{15b^2}(a+bx)^{\frac{3}{2}}\left(3bx-2a \right)+C$$
Now, applying this to our volume, we get through the application of the FTOC:
$$V=\frac{\pi}{60k^3}\left[\left(a^2k^2-4ky \right)^{\frac{3}{2}}(-12ky-2a^2k^2) \right]_0^{\frac{a^2k}{4}}=\frac{\pi}{60k^3}\left(0+2a^5k^5 \right)$$
And thus:
$$V=\frac{\pi a^5k^2}{30}$$
Applying this formula to the given problem, we then find:
$$V=\frac{\pi 2^51^2}{30}=\frac{16\pi}{15}$$