Answer
\begin{align}
\int_{0}^{3}\sqrt {6x-x^{2}} \ dx = 2.25\pi
\end{align}
Work Step by Step
The given integral is
\begin{align}
\int_{0}^{3}\sqrt {6x-x^{2}} \ dx
\end{align}
Let us solve it by completing the square and applying substitutions/geometric formulas:
\begin{alignat}{2}
&\int_{0}^{3}\sqrt {6x-x^{2}}dx=\int_{0}^{3}\sqrt {-(x^{2}-6x+9)+9} \ dx=\\ & = \int_{0}^{3}\sqrt {-(x-3)^{2}+9} \ dx
\end{alignat}
Let u = x - 3 $\Rrightarrow$ du = dx
\begin{alignat}{1}
\int_{-3}^{0}\sqrt {-u^{2}+9} \ du
\end{alignat}
Now, use trigonometric substitution u = 3 $\sin t$ $\Rrightarrow$ du = 3 $\cos t$ dt. Thus,
\begin{alignat}{2}
& \int_{-\frac{\pi}{2}}^{0}\sqrt {9-9(\sin t)^2} \times 3\cos t \ dt = \int_{-\frac{\pi}{2}}^{0} 9 (\cos t)^{2} \ dt = \\ &=4.5 \times \int_{\frac{\pi}{2}}^{0} (1 + \cos (2t)) \ dt = 4.5 \times (\frac{\pi}{2} + 0) = 2.25\pi
\end{alignat}
