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

Question

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

Accepted Answer

By symmetry about the $x$-axis:

$$A = 2\int_0^1 2\sqrt{x}\,dx + 2\int_1^{\sqrt2}\sqrt{8-4x^2}\,dx = \frac{8}{3} + 4\int_1^{\sqrt2}\sqrt{2-x^2}\,dx$$

$$= \frac{8}{3} + 4\cdot\frac{1}{2}\left[x\sqrt{2-x^2}+2\sin^{-1}\frac{x}{\sqrt2}\right]_1^{\sqrt2} = \frac{8}{3} + 2\left[2\cdot\frac{\pi}{2}-1-2\cdot\frac{\pi}{4}\right] = \pi + \frac{2}{3}$$

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