The curve $$x^2-y^2=16$$ is a hyperbola, which is of the form $$x^2-y^2=a^2$$
The hyperbola is marked in red lines. The line x=4 is marked in blue and x=8 is marked in green. The area of the bounded figure by the curve and the two given lines is the area of OABCO.
By symmetry, area of OABCO= 2$$\times\ $$ area of OBAO.
So, required area, A= 2$$_4^8\int\left(y\right)dx\ $$ , where y= $$\pm\ \sqrt{x^2-16\ }$$.
But in the first quadrant, y=$$\sqrt{x^2-16\ }$$
So, A= $$2\left(_4^8\int\left(\sqrt{x^2-16\ }\right)dx\right)\ $$
We know that, $$\int\ \sqrt{x^2-a^2\ }dx=\left[\ \frac{\ x}{2}\sqrt{x^2-a^2}-\ \frac{\ a^2}{2}\log\left(x+\sqrt{x^2-a^2}\right)\right]\ $$
So, A= $$2\left[\ \frac{\ x}{2}\sqrt{x^2-4^2}-\ \frac{\ 4^2}{2}\log\left(x+\sqrt{x^2-4^2}\right)\ \right]_4^8$$
=> A= $$2\left[\frac{\ 8}{2}\sqrt{64-16}-\ \frac{\ 16}{2}\log\left(8+\sqrt{64-16}\right)\right]-2\left[\frac{4\ }{2}\sqrt{16-16}-\ \frac{\ 16}{2}\log\left(4+\sqrt{16-16}\right)\right]\ \ $$
=>A= $$2\left[\ 4\sqrt{48}-\ \ 8\log\left(8+\sqrt{48}\right)\right]-2\left[2\sqrt{0}-\ \ 8\log\left(4+\sqrt{0}\right)\right]\ \ $$
=>A=$$2\left[\ 16\sqrt{3}-\ \ 8\log\left(8+4\sqrt{3}\right)\right]-2\left[-\ \ 8\log4\right]\ \ $$
=>A= $$32\sqrt{3}-\ \ 16\log\left(8+4\sqrt{3}\right)+16\log4\ \ $$
=>A= $$32\sqrt{3}-\ \ 16\log\left(\ \frac{\ 8+4\sqrt{\ 3}}{4}\right)$$
.'. A= $$32\sqrt{3}-\ \ 16\log\left(2+\sqrt{3}\right)$$