I am looking to find the area enclosed by the following curves: $$(x+ \sqrt{x^2+y^2})^\frac{1}{2}=1,\;y=0, \; (*)\;x=\frac{1}{2}(y^\frac{2}{3}-y^\frac{4}{3})$$ in the quadrant $x,y\geq0$ using parabolic coordinates, $$x=\frac{u^2-v^2}{2},\;y=uv.$$
I know that the first curve reduces to $u=1$, the second gives $uv=0$, but I can't find how to incorporate $(*)$ to then find the integration limits. Any help would be great, thanks!
You can express your region as $$\frac{1}{2}(y^{2/3}-y^{4/3})<x<\frac{1-y^2}{2}$$$$0<y<1$$ After applying $x=\frac{1}{2}(u^2-v^2),y=uv$ and doing a bunch of algebra this becomes $$0<v<u^2$$ $$0<u<1$$ Therefore $$\int_0^1 \int_{\frac{1}{2}(y^{2/3}-y^{4/3})}^{\frac{1-y^2}{2}}\mathrm{d}x\mathrm{d}y=\int_0^1 \int_0^{u^2}\Bigg|\frac{\partial\Big(\frac{1}{2}(u^2-v^2),uv\Big)}{\partial(u,v)}\Bigg|\mathrm{d}v\mathrm{d}u=\int_0^1 \int _0^{u^2}(u^2+v^2)\mathrm{d}v\mathrm{d}u$$ The last integral evaluates to $\frac{26}{105}$.