I have to find the volume of the region bounded the two paraboloids: $$z=x^{2}+y^{2}$$ and $$z=4x^{2}+4y^{2}$$ and also limited by the cylinder $$y=x^{2}$$ and the plane $$y=3x$$
I could draw this region in the space but, for me, there's no a maximum value for $z$. Since if $z=z$, then $$4x^{2}+4y^{2}=x^{2}+y^{2}\Rightarrow x=y=0$$ So, the only intersection between the paraboloids is the origin $(0,0,0)$. I think that it misses some data, cause in this way, I think the region has no maximum value in $z$, then its volume is infinity.
$y=x^2, y = 3x$ form a closed region above the $xy$ plane, forming the "lateral" bounds of your volume.
The paraboloids form the vertical bounds.
$V=\int_0^3\int_{x^2}^{3x}\int_{x^2+y^2}^{4(x^2+y^2)}\ dz\ dy\ dx$