<s>[INST] Calculate area by using double integral Area $D$ confined by lines:
$x=\sqrt{4-{y}^{2}},\: y=\sqrt{3x}, \: x\geq 0$
Need to calculate $I = \iint\limits_D \, dx\,dy$
My steps:


*

*Draw $D$:





*Set boundaries:


$I = {I}_{1} + {I}_{2}$
${I}_{2} = \frac{1}{4}\left(\pi {r}^{2} \right)=\pi$
${I}_{1} =  \int_{0}^{\sqrt{3}} dy \int_{\frac{1}{3}{y}^{2}}^{\sqrt{4-{y}^{2}}}dx = \int_{0}^{\sqrt{3}}\sqrt{4-{y}^{2}}-\frac{1}{3}{y}^{2} \: dy = \left[2 \arcsin\frac{y}{2} + \frac{y}{2}\sqrt{4-{y}^{2}}  \right]_{0}^{\sqrt{3}} - \frac{1}{3}\left[\frac{{y}^{3}}{3} \right]_{0}^{\sqrt{3}} = \frac{\sqrt{3}}{6}+\frac{2\pi }{3}$
So $I = \frac{\sqrt{3}}{6}+\frac{2\pi }{3} + \pi = \frac{5\pi }{3} + \frac{\sqrt{3}}{6}$
I have check ${I}_{1}$ via WolframAlpha.
But answer should be $\frac{2\pi - \sqrt{3}}{6}$
 [/INST] You selected the wrong area. You want the bit bound by all 3 lines mentioned.
You can get that by doing $2\pi$ take away your answer. Or recalculate the easier integral.

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