$\iiint_M (x+y+z)\,dx\,dy\,dz$ over $M=\{(x,y,z)\in\mathbb{R^3}: 0≤z≤(x^2+y^2)^2≤81\}$

Question

$$\iiint_M (x+y+z)\,dx\,dy\,dz$$ 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.

Answer 1

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$$\int_0^{2\pi}\int_0^3\int_0^{\rho^4}\rho\bigl(\rho\cos(\theta)+\rho\sin(\theta)+z\bigr)\,\mathrm dz\,\mathrm d\rho\,\mathrm d\theta.$$

Answer 2

From $(x^2+y^2)^2\le81$ you get $x^2+y^2\le 0,$ so that's a disk of radius $3$ centered at $(0,0).$ By symmetry, $$ \iiint\limits_M (x+y) \,dx\,dy\,dz =0 $$ so you just have $$ \iiint\limits_M z \,dx\,dy\,dz. $$ If you use polar coordinates, then $(x^2+y^2)^2$ becomes $r^4$ and you have $$ \int_0^{2\pi} \left( \int_0^3 \left( \int_0^{r^4} z \, dz \right) r\,dr \right) \,d\theta. $$ Nothing inside the integral with respect to $\theta$ depends on $\theta,$ so you just get $2\pi$ times the inner double integral.

(As usually happens, the expression $r\,dr\,d\theta$ lends itself to computation, but from a geometric point of view it's really $(dr)(r\,d\theta).$)

Answer 3

In cylindrical coordinates,
$x = r \cos\theta, y = r \sin\theta, x^2 + y^2 = r^2$

The region is given by $0 \leq z \leq r^4 \leq 81$. Please note $r^4 \geq z \geq 0 \implies r \geq z^{1/4}$ and $r^4 \leq 81 \implies r \leq 3$

So the region is bound between surface $r^4 = z$ and cylinder $r = 3$.

If you integrate with respect to $dr$ first,

$z^{1/4} \leq r \leq 3$, $0 \leq z \leq 81$, $0 \leq \theta \leq 2\pi$

If you integrate with respect to $dz$ first,

$0 \leq z \leq r^4, 0 \leq r \leq 3$, $0 \leq \theta \leq 2\pi$

Another thing to notice is the symmetry of the region around x and y planes and as $x$ and $y$ are odd functions, their integral will be zero. So the integrand reduces to $z$.