Answer
$a)$ $\frac{6\pi}{5}$
$b)$ $\frac{4\pi}{5}$
$c)$ $2\pi$
$d)$ $2\pi$
Work Step by Step
a) the x axis
$V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(y)[12(y^{2}-y^{3}$)]$dy$
$V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$[(y^{3}-y^{4}$)]$dy$
$V$ = $24\pi$$(\frac{y^{4}}{4}-\frac{y^{5}}{5})$ $|_{{\,0}}^{{\,1}}$
$V$ = $24\pi$$[(\frac{1}{4}-\frac{1}{5})-(0-0)]$
$V$ = $\frac{6\pi}{5}$
b) the line y=1
$V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(1-y)[12(y^{2}-y^{3}$)]$dy$
$V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(y^{2}-2y^{3}+y^{4}$)$dy$
$V$ = $24\pi$$(\frac{y^{3}}{3}-\frac{y^{4}}{2}+\frac{y^{5}}{5})$ $|_{{\,0}}^{{\,1}}$
$V$ = $24\pi$$[(\frac{1}{3}-\frac{1}{2}+\frac{1}{5})-(0-0+0)]$
$V$ = $\frac{4\pi}{5}$
c) line y=$\frac{8}{5}$
$V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(\frac{8}{5}-y)[12(y^{2}-y^{3}$)]$dy$
$V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(\frac{8}{5}y^{2}-\frac{13}{5}y^{3}+y^{4}$)$dy$
$V$ = $24\pi$$(\frac{8}{15}y^{3}-\frac{13}{20}y^{4}+\frac{1}{5}y^{5}$)$dy$$|_{{\,0}}^{{\,1}}$
$V$ = $24\pi$$[(\frac{8}{15}-\frac{13}{20}+\frac{1}{5})-(0-0+0)]$
$V$ = $2\pi$
d) the line y=-$\frac{2}{5}$
$V$ = $\int_{{\,0}}^{{\,1}}$$2\pi(\frac{2}{5}+y)[12(y^{2}-y^{3}$)]$dy$
$V$ = $24\pi$$\int_{{\,0}}^{{\,1}}$$(\frac{2}{5}y^{2}+\frac{3}{5}y^{3}-y^{4}$)$dy$
$V$ = $24\pi$$(\frac{2}{15}y^{3}+\frac{3}{20}y^{4}-\frac{1}{5}y^{5}$)$dy$$|_{{\,0}}^{{\,1}}$
$V$ = $24\pi$$[(\frac{2}{15}+\frac{3}{20}-\frac{1}{5})-(0+0-0)$)]$dy$
$V$ = $2\pi$
