Answer
$\text {(a) The volume is}$
\begin{align}
V = 2\pi \int_0^1 (1-x)x \ dx
\end{align}
$\text {(b) The volume is}$
\begin{align}
V = 2\pi \int_0^1 (1-y)(1+y) \ dy
\end{align}
Work Step by Step
$\text {It is given that}$
\begin{align}
y = x; y = 0 \ \ and \ \ x = 1
\end{align}
$\text {(a) Revolved about x = 1}$
$\text {Meaning that the cross section generates a cylindircal surface of height}$
$\text {x and radius (1-x). Thus, the volume is}$
\begin{align}
V = 2\pi \int_0^1 (1-x)x \ dx
\end{align}
$\text {(b) Revolved about y=-1}$
$\text {Meaning that the cross section generates a cylindircal surface of height}$
$\text {(1-y) and radius (1+y). Thus, the volume is}$
\begin{align}
V = 2\pi \int_0^1 (1-y)(1+y) \ dy
\end{align}
