So I have:
Let $R \in \mathbb{R^+}$
and $$\int_{-R}^{0} dx \int_{- \sqrt{R-\frac{x^2}{R} }}^{(x+R)^2)} dy +\int_{0}^{R} dx \int_{- \sqrt{R-\frac{x^2}{R} }}^{\sqrt{R^4-R^2x^2}} dy $$
So I have to first draw the area of integrating and then change the order of integration. I don't know if I have to just exchange $dy$ and $dx$ parts with boundaries, or how is it suppose to be.
So I notice that since they both are similar, I thought of just trying to make one integral, but i don't know if it is possible. At least the integration by $x$ is possibly the same, just not sure about the $y$ part. Lower bounds are the same, but the upper, seem similar but they aren't exactly the same either. Or should I just write: $$x\in (-R,0) \cup (0,R),$$ $$ y\in \left( - \sqrt{R-\frac{x^2}{R} },(x+R)^2 \right) \cup \left(- \sqrt{R-\frac{x^2}{R}} ,\sqrt{R^4-R^2x^2} \right)$$
Is it possible to change it into: $$x\in (-R,R),$$ $$ y\in\left(- \sqrt{R-\frac{x^2}{R} }, max\left[\left(\left(x+R\right)^2\right),\left(\sqrt{R^4-R^2x^2} \right)\right]\right)$$ ?
Any help or thoughts on it would be appreicated.
Thank you in advance.
Note that, if $x\in[-R,0]$, $$-\sqrt{R-\frac{x^2}{R}}\le y\le(x+R)^2$$ and if $x\in[0,R]$, $$-\sqrt{R-\frac{x^2}{R}}\le y\le\sqrt{R^4-R^2x^2}$$ After changing the order, the area becomes that, if $y\in[-\sqrt{R},0]$, $$ -\sqrt{R^2-Ry^2}\le x\le\sqrt{R^2-Ry^2}$$ and if $y\in[0,R^2]$, $$-R-\sqrt y\le x\le\sqrt{R^2-\frac{y^2}{R^2}}. $$ So \begin{eqnarray} &&\int_{-R}^{0} dx \int_{- \sqrt{R-\frac{x^2}{R} }}^{(x+R)^2)} dy +\int_{0}^{R} dx \int_{- \sqrt{R-\frac{x^2}{R} }}^{\sqrt{R^4-R^2x^2}} dy\\ &=&\int_{-\sqrt R}^0dy\int_{-\sqrt{R^2-Ry^2}}^{\sqrt{R^2-Ry^2}}dx+\int_0^{R^2}dy\int_{-R-\sqrt y}^{\sqrt{R^2-\frac{y^2}{R^2}}}dx\\ &=&\frac12R^{3/2}\pi+\frac14R^3\pi+\frac13R^2. \end{eqnarray}