Plotting the diagram of $x^2+y^2\le 21 \& y^2\le 4x$ for $x\ge 1$ we get,

Now from above diagram required Area $\mathrm{\Delta }=$$2{\int }_1^{3}2\sqrt{x}dx+2{\int }_3^{\sqrt{21}}\sqrt{21-x^2}dx$

$\Rightarrow \mathrm{\Delta }=\frac{8}{3}\left(3\sqrt{3}-1\right)+21{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)-6\sqrt{3}$

So, $\frac{1}{2}\left(\mathrm{\Delta }-21{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)$

$=\frac{1}{2}\left(\frac{8}{3}\left(3\sqrt{3}-1\right)+21{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)-6\sqrt{3}-21{\sin }^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)$

$=\frac{1}{2}\left(\frac{8}{3}\left(3\sqrt{3}-1\right)-6\sqrt{3}\right)$

$=\frac{2\sqrt{3}-\frac{8}{3}}{2}$

$=\sqrt{3}-\frac{4}{3}$