\begin{align*}
&A = \pi r^2 \\
{\therefore}\
&V = \int_{-3}^2 \pi (6 - x - x^2)^2\ dx \\
&=\pi \int_{-3}^2 (36-6x-6x+x^2+x^3-6x^2+x^3+x^4)\ dx \\
&=\pi \int_{-3}^2 (36-12x-11x^2+2x^3+x^4)\ dx \\
&=\pi \left[36x - 6x^2 - \frac{11x^3}{3} + \frac{x^4}{2} + \frac{x^5}{5} \right]_{-3}^2 \\
&=\pi \left[{\frac{625}{6} \right] \\
&= \frac{625 \pi}{6} units^3 \\
\end{align*}
\section{Question 3}
\textbf{(a)} Solve the differential equation:\\
\begin{align*}
&\frac{dy}{dx} = \frac{6x}{y^2(3-x^2)}\\
&{\therefore}\ \frac{dy}{dx} y^2 = \frac{6x}{3-x^2} \\
&{\therefore}\ dy \times y^2 = \frac{6x \times dx}{3-x^2} \\
&{\therefore}\ \int y^2 \time dy = \int \frac{6x}{3-x^2) dx \\
\text{Let $u=3-x^2$} \\
{\therefore}\ \frac{du}{dx} = -2x \\
&{\therefore}\ \left[\frac{y^3}{3} = 6 \int \frac{x}{u} \frac{du}{dx}\ dx \\
&=6 \int \frac{1}{u} \frac{-1}{2}\ du \\
&=-3 \ln(|3-x^2|) + c \\
&{\therefore\} y = \sqrt[3]{-9 \ln(|3-x^2|)} + c \\
\end{align*}
