Answer
$$\displaystyle{\int_0^{\frac{2}{3}}\sqrt{4-9{x}^2}dx=\frac{\pi }{3}}\\
$$
Work Step by Step
$\displaystyle{I=\int_0^{\frac{2}{3}}\sqrt{4-9x^2}dx}\\
\displaystyle{I=\int_0^{\frac{2}{3}}\sqrt{4-{\left(3x\right)}^2}dx}\\$
$\displaystyle \left[\begin{array}{ll} 3x=2\sin\theta & 9 x^2=4\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{2}{3}\cos\theta & dx=\frac{2}{3}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_0^{\frac{\pi}{2}}\sqrt{4-4\sin^2\theta}\times\frac{2}{3}\cos\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi}{2}}\sqrt{4\left(1-\sin^2\theta\right)}\times\frac{2}{3}\cos\theta\ d\theta}\\
\displaystyle{I=\int_0^{\frac{\pi}{2}}2\cos\theta\times\frac{2}{3}\cos\theta\ d\theta}\\
\displaystyle{I=\frac{4}{3}\int_0^{\frac{\pi}{2}}\cos^2\theta\ d\theta}\\
\displaystyle{I=\frac{2}{3}\int_0^{\frac{\pi}{2}}\cos2\theta+1\ d\theta}\\
\displaystyle{I=\frac{2}{3}\left[\frac{1}{2}\sin2\theta+ x\right]_0^{\frac{\pi}{2}}}\\
\displaystyle{I=\frac{1}{3}\sin\left(2\pi\times\frac{1}{2}\right)\ +\frac{\pi }{3}-0-0}\\
\displaystyle{I=\frac{1}{3}\sin\left(\pi\right)\ +\frac{\pi }{3}}\\
\displaystyle{I=\frac{\pi }{3}}\\
$
