$\iiint_M (x+y+z)\,dx\,dy\,dz$ over $M=\{(x,y,z)\in\mathbb{R^3}: 0≤z≤(x^2+y^2)^2≤81\}$ $$\iiint_M (x+y+z)\,dx\,dy\,dz$$ over $M=\{(x,y,z)\in\mathbb{R^3}: 0≤z≤(x^2+y^2)^2≤81\}$. How would I express this with the correct bounds? Once I have the bounds I can continue on my own but I need the bounds, since this is the very first time encountering this type of boundering with the $M$. Any hints about the change of bounds would really help.

I think that it is more natural to do this in cylindrical coordinates. Then the conditions$$0\leqslant z\leqslant(x^2+y^2)^2\leqslant81$$become$$0\leqslant z\leqslant\rho^4\leqslant81.$$So, since $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, you have$$(x^2+y^2)^2=\bigl(\rho^2\cos^2\theta+\rho^2\sin^2(\theta)\bigr)=\rho^4$$and (since $\rho\geqslant0$)$$\rho^4\leqslant81\iff\rho^2\leqslant3^4\iff0\leqslant\rho\leqslant3.$$Since $M$ doesn't change when you rotate it around the $z$-axis, $\theta$ can take any value from $[0,2\pi]$. So, your integral becomes$$\int_0^{2\pi}\int_0^3\int_0^{\rho^4}\rho\bigl(\rho\cos(\theta)+\rho\sin(\theta)+z\bigr)\,\mathrm dz\,\mathrm d\rho\,\mathrm d\theta.$$
