Finding the bounds of a solid for triple integrals Ok, so I have an answer, most likely the wrong one.
The question being asked is:
Using polar coordinates find the volume of the solid bounded below by the $xy–plane$ and above by the surface $x^2 +y^2 +z^6 =5$.
First off I found the bounds for $x,y$ and $z$. 
These are $x \space [0,\sqrt{5}], \space y \space[0, \sqrt{5}],  \space z \space[0, 5^{\frac{1}{6}}]$
I use these bounds and integrate the function $x^2 +y^2 +z^6 =5$
I realise this is not right, but I am unsure why? Could someone explain this to me.
I know I have to write the bounds in terms of the unused variables, otherwise it wouldn't ask for polar coordinates.
The best I could come up with was this:
$ y \space [0, \sqrt{5}], \space x \space[0, \sqrt{5-y^2-z^6}], \space z \space [0, (5-y^2-x^2)^{\frac{1}{6}} ]$
Afterwards we multiply the integral by 4.
I have 2 bounds with 2 variables and 1 bound with none. I know this can't be right, we should have 3,2,1.
But I am having trouble figuring out what exactly the bounds should be.
If someone could explain to me how I can find these bounds so I could use them for general questions that would be great!.
I'm also pretty sure I can integrate by converting to polar coordinates, I am just having trouble with the bounds, that is all.
Thanks for any help !
 A: Let's let $z$ be the function $z_C(x,y)=(5-(x^2+y^2))^{1/6}$ of $x$ and $y$ (the subscript $C$ designates Cartesian coordinates).  
We can also express $z$ as the function $z_P(\rho)=(5-\rho^2)^{1/6}$ of $\rho$ in polar coordinates $(\rho,\phi)$.
Note that the projection to the surface $z_C$ onto the $x-y$ plane is contained in the circle $x^2+y^2=5$, which can be written in polar coordinates $\rho^2=5$.  The volume between $z=0$ and $z=z_P$ is then given by
$$\begin{align}
\text{Volume}&=\int_V dz\,dx\,dy\\\\
&=\int_Vdz\,\rho d\rho\, d\phi\\\\
&=\int_0^{2\pi}\int_0^{\sqrt{5}}\int_0^{(5-\rho^2)^{1/6}}dz\,\rho d\rho \,d\phi\\\\
&=\int_0^{2\pi}\int_0^{\sqrt{5}}(5-\rho^2)^{1/6}\rho d\rho \,d\phi\\\\
&=2\pi\,\int_0^{\sqrt{5}}(5-\rho^2)^{1/6}\rho d\rho\\\\
&=\pi\frac{5^{7/6}}{7/6}
\end{align}$$
A: If we have $n$ dimensions (in this case, $n=3$) for integration, the process that I use to find Cartesian-coordinate integration limits over a shape is to gradually increase the dimensions one-by-one.  In particular, I start with all-but-one dimension at "$0$" (meaning, they all supply the minimum possible value to my function), and find the limits on the non-zero dimension.  For $x^2+y^2+z^6=5$ and bounded below by the $xy$ plane, I might not wish to deal with the sixth root of variables, so I would want the "constant" dimension limits to be placed on $z$, and consider my equation to be $z^6=5$; then the integration limits are as follows:
$$z\in[0,\sqrt[6]{5}]$$
Now I have limits on $z$, and I need to increase my dimensions by $1$; so I leave $y=0$ and revisit my initial equation, now as $x^2+z^6=5\implies x^2=5-z^6$, which I can then use to find limits on $x$ as
$$x\in[-\sqrt{5-z^6},\sqrt{5-z^6}]$$
Finally, we consider the $3$rd dimension $y$ with all the variables $x^2+y^2+z^6=5\implies y^2=5-x^2-z^6$, and we get our limits on $y$ as
$$y\in[-\sqrt{5-x^2-z^6}, \sqrt{5-x^2-z^6}]$$
This now corresponds to the integral
$$\int_0^{\sqrt[6]5}\int_{-\sqrt{5-z^6}}^{\sqrt{5-z^6}}\int_{-\sqrt{5-x^2-z^6}}^{\sqrt{5-x^2-z^6}}x^2+y^2+z^6\text dy\text dx\text dz$$
Of course, the question requests polar form, and with the obvious circular nature of the shape relative to $x,y$ and the helpful condition that one boundary is the $xy$ plane, it is better to refer to the other answer for how to obtain appropriate polar coordinates.
