What's The Cylindrical coordinates for this $\int_0^6\int_{-\sqrt{6x-x^2}}^{\sqrt{6x-x^2}}\int_0^{6x-x^2-y^2}\left(x^2+y^2\right)dzdydx$

Question

I want to convert this to cylindrical coordinates $$V=\int_0^6\int_{-\sqrt{6x-x^2}}^{\sqrt{6x-x^2}}\int_0^{6x-x^2-y^2}\left(x^2+y^2\right)dzdydx = 486π$$ I want to write it like this: $$V=\int_{\ }^{ }\int_{ }^{ }\int_{ }^{ }r^3dzdrd\theta$$ if you don't know how to do it at least vote please!

Answer 1

$z\in[0,6x-x^2-y^2]$ tells you the region is "vertically" bounded between the plane $z=0$ and the paraboloid $z=6x-x^2-y^2$. Complete the square to write the latter as $z=9-(x-3)^2-y^2$, which is a paraboloid whose vertex is located at $(3,0,9)$.

$y\in[-\sqrt{6x-x^2},\sqrt{6x-x^2}]$, or $y\in[-\sqrt{9-(x-3)^2},\sqrt{9-(x-3)^2}]$, tells you that for any $z$, $y$ is confined by the upper and lower halves of a circle with radius $3$ and centered at $(3,0,z)$. This describes a cylinder in $\mathbb R^3$.

$x\in[0,6]$ tells you the integration region covers the interior of the cylinder.

Now you can convert to a modified cylindrical coordinate system with

$$\begin{cases}x=r\cos\theta+3\\y=r\sin\theta\\z=z\end{cases}$$

and the region you're interested in (I'll call it $E$) is obtained with $0\le r\le3$, $0\le\theta\le2\pi$, and $0\le z\le9-r^2$. The Jacobian is the same as with the conversion to standard cylindrical coordinates. So the integral is

$$\begin{align*} \iiint_E(x^2+y^2)\,\mathrm dV&=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=0}^{z=9-r^2}\left((r\cos\theta+3)^2+(r\sin\theta)^2\right)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta\\[1ex] &=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{z=0}^{z=9-r^2}(r^2+6r\cos\theta+9)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta \end{align*}$$

Answer 2

Hint: the limits give the set: $$0\le x\le 6,$$ $$y^2\le 6x - x^2,$$ $$0\le z\le 6x - x^2 - y^2.$$ Now, do $x = r\cos\theta$, $y = r\sin\theta$...