Answer
$a)$ $\frac{5\pi}{3}$
$b)$ $\frac{4\pi}{3}$
$c)$ $2\pi$
$d)$ $\frac{2\pi}{3}$
Work Step by Step
from (1,2) and (2,2) we get $y = 2$
from (1,1) and (1,2) we get $x = 1$
from (1.1) and (2,2) can find the equation as
$m$ $(slope)$ = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
$m$ = $\frac{2-1}{2-1}$ = $1$
then
$1$ = $\frac{y-1}{x-1}$
$x-1$ = $y-1$
$x$ = $y$
so we have $y = 2$, $x = 1$ and $x$ = $y$
use shell method
a) the x axis
$V$ = $\int_{{\,1}}^{{\,2}}$$2\pi(y)(y-1)$$dy$
$V$ = $2\pi$$\int_{{\,1}}^{{\,2}}$$(y^{2}-y)$$dy$
$V$ = $2\pi$$(\frac{1}{3}y^{3}-\frac{1}{2}y^{2})$$|_{{\,1}}^{{\,2}}$
$V$ = $2\pi$$[(\frac{8}{3}-2)-(\frac{1}{3}-\frac{1}{2})]$
$V$ = $\frac{5\pi}{3}$
b) the y axis
$V$ = $\int_{{\,1}}^{{\,2}}$$2\pi(x)(2-x)$$dx$
$V$ = $2\pi$$\int_{{\,1}}^{{\,2}}$$(2x-x^{2})$$dx$
$V$ = $2\pi$$(x^{2}-\frac{1}{3}x^{3})$$|_{{\,1}}^{{\,2}}$
$V$ = $2\pi$$[(4-\frac{8}{3})-(1-\frac{1}{3})]$
$V$ = $\frac{4\pi}{3}$
c) the line x = $\frac{10}{3}$
$V$ = $\int_{{\,1}}^{{\,2}}$$2\pi(\frac{10}{3}-x)(2-x)$$dx$
$V$ = $2\pi$$\int_{{\,1}}^{{\,2}}$$(\frac{20}{3}-\frac{16}{3}x+x^{2})$$dx$
$V$ = $2\pi$$(\frac{20}{3}x-\frac{8}{3}x^{2}+\frac{1}{3}x^{3})$$|_{{\,1}}^{{\,2}}$
$V$ = $2\pi$$[(\frac{40}{3}-\frac{32}{3}+\frac{8}{3})-(\frac{20}{3}-\frac{8}{3}+\frac{1}{3}))]$
$V$ = $2\pi$
d) the line y = 1
$V$ = $\int_{{\,1}}^{{\,2}}$$2\pi(y-1)(y-1)$$dy$
$V$ = $2\pi$$\int_{{\,1}}^{{\,2}}$$(y^{2}-2y+1)$$dy$
$V$ = $2\pi$$(\frac{1}{3}y^{3}-y^{2}+y)$$|_{{\,1}}^{{\,2}}$
$V$ = $2\pi$$[(\frac{8}{3}-4+2)-(\frac{1}{3}-1+1)]$
$V$ = $\frac{2\pi}{3}$
