Answer
$\dfrac{8c \pi}{3}$
Work Step by Step
The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$
Now, $V=2\pi \int_{-1}^{1} (c-x) (1-x^2) \ dx\\= 2\pi \int_{-1}^{1} (c-x-cx^2+x^3) \ dx \\=2 \pi [cx-\dfrac{x^2}{2} -\dfrac{cx^3}{3}+\dfrac{x^4}{4}]_{-1}^{1} \\= 4\pi [\dfrac{2c}{3}] \\=\dfrac{8c \pi}{3}$
