# Finding the bounds of a solid for triple integrals

Ok, so I have an answer, most likely the wrong one.

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.

$y \space [0, \sqrt{5}], \space x \space[0, \sqrt{5-y^2-z^6}], \space z \space [0, (5-y^2-x^2)^{\frac{1}{6}} ]$

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 !

• One of the bounds on $x,z$ should only depend on $y$, otherwise your process is good... Note that you can choose any of the three as the "constant-bounded" one, so perhaps it would be better to bound $z$ by constants and have $x,y$ have more "normal" circle-type bounds... – abiessu May 4 '15 at 16:39
• Specifically, $z\in[0,\sqrt{5}],x\in[-\sqrt{5-z^6},\sqrt{5-z^6}],y\in[-\sqrt{5-x^2-z^6},\sqrt{5-x^2-z^6}]$. – abiessu May 4 '15 at 16:52
• Please let me know how I can improve my answer. I just want to give you the best answer I can. – Mark Viola May 4 '15 at 17:01
• @abiessu Simply let $z$ be the function $z(x,y)=(5-(x^2+y^2))^{1/6}$ of $x$ and $y$. The projection of the surface $z(x,y)$ onto the $x-y$ plane is the disc $x^2+y^2=5$. – Mark Viola May 4 '15 at 17:08
• @abiessu Could you explain how you only used z to determine where x will belong. I can't seem to comprehend this. Why is there no y variable used? This is my main question. – Gregory Peck May 4 '15 at 19:26

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}

• I think the best way you could improve this answer is by describing in some detail how you got to the integration bounds you have... – abiessu May 4 '15 at 17:04
• @abiessu Thanks! I edited accordingly. – Mark Viola May 4 '15 at 17:17

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:

$$z\in[0,\sqrt{5}]$$

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

$$\int_0^{\sqrt5}\int_{-\sqrt{5-z^6}}^{\sqrt{5-z^6}}\int_{-\sqrt{5-x^2-z^6}}^{\sqrt{5-x^2-z^6}}x^2+y^2+z^6\text dy\text dx\text dz$$

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.