Answer
$$\int^{8}_{4\sqrt2}\frac{24dy}{y\sqrt{y^2-16}}=\frac{\pi}{2}$$
Work Step by Step
$$A=\int^{8}_{4\sqrt2}\frac{24dy}{y\sqrt{y^2-16}}=\int^{8}_{4\sqrt2}\frac{24dy}{y\sqrt{y^2-4^2}}$$
We have $$\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\sec^{-1}\Big|\Big(\frac{x}{a}\Big)\Big|+C$$
Therefore, $$A=24\times\frac{1}{4}\sec^{-1}\Big|\frac{y}{4}\Big|\Big]^{8}_{4\sqrt2}$$ $$A=6\Big(\sec^{-1}\Big|\frac{8}{4}\Big|-\sec^{-1}\Big|\frac{4\sqrt2}{4}\Big|\Big)$$ $$A=6(\sec^{-1}2-\sec^{-1}\sqrt2)$$ $$A=6\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$ $$A=6\times\frac{\pi}{12}=\frac{\pi}{2}$$
