Answer
$\text{(a) V = $\frac{2\pi}{3}$}$
$\text{(b) V = $\frac{16}{3}$}$
$\text{(c) V = $\frac{4\sqrt 3}{3}$}$
Work Step by Step
$\text{It is given that}$
\begin{align}
x^2 + y^2 = 1
\end{align}
$\text{(a) Semicircles}$
$\text{The cross section area:}$
\begin{align}
& A(x) = \frac{1}{2} \pi r^2 \ \ where \ \ r = y = \sqrt {1-x^2} \\
& A(x) = \frac{1}{2} \pi (1-x^2)
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_{-1}^{1} \frac{1}{2} \pi (1-x^2) \ dx = \frac{1}{2} \pi \left[x-\frac{x^3}{3} \right]_{-1}^1 = \frac{1}{2} \pi \times \frac{4}{3} = \frac{2\pi}{3}
\end{align}
$\text{(b) Squares}$
$\text{The cross section area:}$
\begin{align}
& A(x) = a^2 \ \ where \ \ a = 2y = 2 \sqrt {1-x^2} \\
& A(x) = 4(1-x^2)
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_{-1}^{1} 4(1-x^2) \ dx = 4 \left[x-\frac{x^3}{3} \right]_{-1}^1 = 4 \times \frac{4}{3} = \frac{16}{3}
\end{align}
$\text{(c) Equilateral triangles}$
$\text{The cross section area:}$
\begin{align}
& A(x) = \frac{\sqrt 3}{4} a^2 \ \ where \ \ a = 2y = 2 \sqrt {1-x^2} \\
& A(x) = \sqrt 3 (1-x^2)
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_{-1}^{1} \sqrt 3 (1-x^2) \ dx = \sqrt 3 \left[x-\frac{x^3}{3} \right]_{-1}^1 = \sqrt 3 \times \frac{4}{3} = \frac{4\sqrt 3}{3}
\end{align}
