Triple integration of a solid bounded by 2 curves $z=1-x^2-2y^2$ and $z=x^2$. 
I need to find the Vol bounded by $z=1-x^2-2y^2$ and $z=x^2$.

I have gotten
$$V=∫∫∫rdrdθdz$$
where the limits are $0$ to $2\pi$, $0$ to $\sqrt{1/2}$, and $0$ to $1$. 
The ans is $\pi$ but my ans is $\pi/2$. Can anyone tell me where my mistake is?

$ z = 1 - x^2 - 2y^2$ and $z = x^2 $ meet along the circle $x^2+y^2=\frac{1}{2} $
So, limits of $ x$ would be $ \frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$ and 
$y$ would be $ -\sqrt{\frac{1}{2}-x^2}<y<\sqrt{\frac{1}{2}-x^2}$ and
$z$  would be $x^2<z<1-x^2-2y^2$
So,
$V=\int_{x^2}^{1-x^2-2y^2} \int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}dxdydz$
$ \implies V=\int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}} \int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}(1-2x^2-2y^2)dxdy$
Now, convert to r-plane.
$A=\int \int rdrd{\theta }$   
Put   $x=r \cos \theta , y=r \sin \theta $
$V=\int_{0}^{\frac{1}{\sqrt {2}}} \int_{0}^{2\pi}(1-2r^2)rdrd{\theta}$
$ \implies V=4\int_{0}^{{\frac{1}{\sqrt {2}}}} \int_{0}^{\frac{\pi}{2}}(1-2r^2)rdrd{\theta}$
