Find the volume of the solid bounded by $z=\sqrt[4]{x^2+y^2}$, $z=1$ and $z=\sqrt{2}$

Question

Find the volume of the solid bounded by $z=\sqrt[4]{x^2+y^2}$, $z=1$ and $z=\sqrt{2}$. I have to calculate this using triple integral but the I'm stuck on finding the intervals.

Answer 1

The change of variables to cylindrical coordinates $\begin{cases} x:=r\cos\theta,\\y:=r\sin\theta,\\z:=z\end{cases}$ for $r\geqslant 0$, $0\leqslant \theta< 2\pi$ give the solid bounded by $$z=1,\quad z=\sqrt{2},\quad z^{2}=r$$ Since we have axial symmetry, then $$V=\iint_{R}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)dA,$$ where the region $R$ is the projection over the $rz-$plane, i.e., $$R=\{(r,z):1\leqslant z\leqslant \sqrt{2},\quad 0\leqslant r\leqslant z^{2} \}$$ Hence, the volume of the solid is given by \begin{align*}\iint_{V}1\cdot dV&=\iint_{R}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)dA\\&=\int_{z_{1}=1}^{z_{2}=\sqrt{2}}\int_{r_{1}=0}^{r_{2}=z^{2}}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)rdrdz\\&=\boxed{\frac{\pi}{5}\left(4\sqrt{2}-1\right)}\end{align*} or using two-integrals changing the order of integration \begin{align*}\iint_{R}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)dA&=\int_{r_{1}=0}^{r_{2}=1}\int_{z_{1}=1}^{z_{2}=\sqrt{2}}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)dzrdr\\&+\int_{r_{1}=1}^{r_{2}=2}\int_{z_{1}=r^{1/2}}^{z_{2}=\sqrt{2}}\left(\int_{\theta_{1}=0}^{\theta_{2}=2\pi}d\theta\right)dzrdr\\&=\boxed{\frac{\pi}{5}\left(4\sqrt{2}-1\right)}\end{align*}