This is relatively easy to do by the method of disks, where a
typical disk has radius x(k-x) and width dx. Thus the volume of the
solid of revolution is given by V = pi int_{x=0}^{x=k} x^2(k-x)^2 dx
= pi int_{x=0}^{x=k} x^2(k^2 - 2kx + x^2) dx
= pi [k^2(x^3/3) - 2k(x^4/4) + x^5/5]_{x=0}^{x=k}
= pi [k^5/3 - k^5/2 + k^5/5]
= pi(k^5/30)[10 - 15 + 6]
= pi(k^5/30).

second problem

you could probably do this by the method of disks, but the radii of the
outer and inner disks in this case are rather awkward functions
involving the (two) solutions of x(k-x) = y (in terms of x). So let's
use the method of shells instead, where the typical shell has radius x
and height x(k-x). Then the volume of the solid of revolution is given
by