The area of the region A=(x,y):4x^2+y^2 8 and y^2 4x is:

Question

The area of the region $A={(x,y):4{x}^{2}+{y}^{2}\leqslant 8$ and ${y}^{2}\leqslant 4x}$ is:

Accepted Answer

$A=2{\int }_{1}^{0}2\sqrt{x}dx+2{\int }_{\sqrt{2}}^{1}\sqrt{8-4{x}^{2}}dx=\frac{8}{3}{({X}^{\frac{3}{2}})}_{1}^{0}+4{\int }_{\sqrt{2}}^{1}\sqrt{2-{x}^{2}}dx$

$\begin{gathered} \frac{8}{3}+4	imes \frac{1}{2}{[x\sqrt{2-{x}^{2}}+2{\sin }^{-1}(\frac{x}{\sqrt{2}})]}_{\sqrt{2}}^{1} \ \frac{8}{3}+2[2	imes \frac{\pi }{2}-1-2	imes \frac{\pi }{4}]=\frac{8}{3}+2\pi -2-\pi =\pi +\frac{2}{3} \end{gathered}$

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