# Find the volume of the solid bounded by $z = x^2 + y^2$ $\frac{x^4}{4}+y^2=1$ and the plane $xy$

Find the volume of the solid bounded above by $z = x^2 + y^2$ below by the plane $xy$ and laterally by $\frac{x^4}{4}+y^2=1$

So far I have tried to use polar coordinates I started from$$\int\int_R x^2+y^2 dx dy$$

$$\int_0^{2\pi}\int_?^{\frac{1}{\sqrt{\frac{cos^2\theta}{4}+sen^2\theta}}} r(r) dr d\theta$$

But I am not sure how to set the r limits or if there is an easier way to calculate this volume

$V=\left\{(x,y,z)\in\Bbb R^3:0\le z\le x^2+y^2,\,\frac{x^4}{4}+y^2=1\right\}$ $$V=\iiint_V\mathrm dV=\iint_S\int_0^{x^2+y^2}\,\mathrm dz\,\mathrm dS=\iint_S(x^2+y^2)\,\mathrm d S$$ where $S=\left\{(x,y)\in\Bbb R^2:\,\frac{x^4}{4}+y^2=1\right\}$.
By symmetry $$V=4\int_0^1\int_0^{[4(1-y^2)]^{1/4}} (x^2+y^2)\,\mathrm dx\,\mathrm dy= 4\int_0^1 \left(\frac{[4(1-y^2)]^{3/4}}{3}+{[4(1-y^2)]^{1/4}}y^2\right)\mathrm dy$$ Putting $y^2=u$, $y=\sqrt u$, $\mathrm dy=\frac{1}{2\sqrt u}\mathrm du$, we have \begin{align} V&=\frac{4\sqrt 2}{3}\int_0^1 u^{-1/2}(1-u)^{3/4}\mathrm du+{2\sqrt2}\int_0^1 u^{1/2}(1-u)^{1/4}\mathrm du\\ &=\frac{4\sqrt 2}{3}B\left(\frac12,\frac74\right)+2\sqrt 2B\left(\frac32,\frac54\right) \end{align} where $B(x,y)=\int_0^1 u^{x-1}(1-u)^{y-1}\,\mathrm d u$ is the Euler's Beta function.