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