$$\displaystyle\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{48}{\sqrt{9-4z^2}} dz$$ is equal to

We need to evaluate $$\displaystyle\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \dfrac{48}{\sqrt{9 - 4z^2}}\,dz\;.$$

We simplify the integrand by writing $$\dfrac{48}{\sqrt{9 - 4z^2}} = \dfrac{48}{\sqrt{4\bigl(\tfrac{9}{4} - z^2\bigr)}} = \dfrac{48}{2\sqrt{\bigl(\tfrac{3}{2}\bigr)^2 - z^2}} = \dfrac{24}{\sqrt{\bigl(\tfrac{3}{2}\bigr)^2 - z^2}}\;.$$

Using the standard formula $$\int \dfrac{dz}{\sqrt{a^2 - z^2}} = \arcsin\!\bigl(\tfrac{z}{a}\bigr) + C$$ with $$a = \tfrac{3}{2}$$ gives $$\int \dfrac{24\,dz}{\sqrt{\bigl(\tfrac{3}{2}\bigr)^2 - z^2}} = 24\,\arcsin\!\bigl(\tfrac{2z}{3}\bigr) + C\;.$$

For the definite integral we evaluate at the limits. When $$z = \tfrac{3\sqrt{3}}{4}$$ one finds $$\arcsin\!\bigl(\tfrac{2 \cdot \tfrac{3\sqrt{3}}{4}}{3}\bigr) = \arcsin\!\bigl(\tfrac{\sqrt{3}}{2}\bigr) = \tfrac{\pi}{3}\;.$$ When $$z = \tfrac{3\sqrt{2}}{4}$$ one finds $$\arcsin\!\bigl(\tfrac{2 \cdot \tfrac{3\sqrt{2}}{4}}{3}\bigr) = \arcsin\!\bigl(\tfrac{\sqrt{2}}{2}\bigr) = \tfrac{\pi}{4}\;.$$

Subtracting these values gives $$24\bigl(\tfrac{\pi}{3} - \tfrac{\pi}{4}\bigr) = 24 \cdot \tfrac{\pi}{12} = 2\pi\;.$$ Hence the value of the integral is $$2\pi$$, which corresponds to Option D.