Question

Find the area of the region common between the curve $x^2+y^2=4$ and $x^2+4y^2=9$.

Answer 1

REF. Image.

As shaded region is symmetric

about origin, getting area in

first quadrant,

For $P,4-x^2=\frac{9-x^2}{4}$

$\Rightarrow 16-4x^2=9-x^2$

$\Rightarrow 3x^2=7\Rightarrow x^2=7/3\Rightarrow x=\sqrt{7/3}$

$A=({\int }_0^{\sqrt{7/3}}\frac{\sqrt{9-x^2}}{2}+{\int }_{\sqrt{7/3}}^2\sqrt{4-x^2})dx$

$=\frac{1}{2}(\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}si{n}^{-1}(\frac{x}{3}\left)\right){\int }_0^{\sqrt{7/3}}$

$+\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}si{n}^{-1}\left(\frac{x}{2}\right){\int }_{\sqrt{7/3}}^2$

$=\frac{1}{2}(\sqrt{\frac{7}{3}}\times \frac{1}{2}\sqrt{9-\frac{7}{3}}+\frac{9}{2}si{n}^{-1}(\frac{\sqrt{7}}{3\sqrt{3}})=0)$

$+2si{n}^{-1}\left(1\right)-\frac{\sqrt{7/3}}{2}\sqrt{4-\frac{7}{3}}-2si{n}^{-1}\left(\frac{\sqrt{7}}{2\sqrt{3}}\right)$

$=\frac{\sqrt{7}}{4\sqrt{3}}\times \sqrt{\frac{20}{3}}+\frac{9}{4}si{n}^{-1}\left(\frac{\sqrt{7}}{3\sqrt{3}}\right)+\pi$

$-\frac{\sqrt{7}}{2\sqrt{3}}\times \frac{\sqrt{5}}{\sqrt{3}}-2si{n}^{-1}\left(\frac{\sqrt{7}}{2\sqrt{3}}\right)$

$=\frac{1}{4}si{n}^{-1}\left(\frac{\sqrt{7}}{3\sqrt{3}}\right)+\pi$