# 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:

1. Draw $D$:

1. 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 $\frac{2\pi - \sqrt{3}}{6}$

• You answer is correct for your area but you have shaded the wrong area. Commented Nov 1, 2015 at 13:13
• @IanMiller Can you help me draw D? I still can not understand where is my mistake. Commented Nov 1, 2015 at 13:15
• Why G2 ? indeed a constraint is missing in definition of D. can y be negative ? Commented Nov 1, 2015 at 13:18

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.

• ok, here there is really a confinment by the 3 curves :-) (I would say it was bad not to define D with inequalities). Commented Nov 1, 2015 at 13:23
• Thx. But I still have wrong answer... Commented Nov 1, 2015 at 13:25
• Agree the use of the inequality is confusing. Commented Nov 1, 2015 at 13:26
• $2\pi$ minus your original answer gave me: $\frac{2\pi-\sqrt{3}}{6}$ which I thought was what you were after. Commented Nov 1, 2015 at 13:27
• @IanMiller Damn, in book there are no brackets and answer written in 2 lines. It confused me :) My answer is $\frac{2\pi -\sqrt{3}}{6}$. Suppose, it is right. Commented Nov 1, 2015 at 13:32