Integrate the following 3 dimensional field's moment of inertia

I'm having trouble writing down the integral for the following field. I've tried to transform it to spherical coordinates and cylinder coordinates, but no success.

So our field is the field which is bounded by the following objects:

EDIT: aschepler pointed out, that the $$z=xy$$ isn't a plane, but a saddle shape. I translated the document and there it was labeled as a plane so I think it's probably the $$z=0$$ plane instead. Otherwise, the density/mass function of the field could give negative values for some $$z\lt0$$ values.

1. The plane, $$z=0$$
2. The sphere, $$x^2+y^2+z^2=4$$
3. The cone, $$z=\sqrt{\frac{x^2+y^2}{3}}$$

Let it be $$A$$, as our integral's range. A little help is given by saying that the $$z=xy$$ plane bounds our field from the "bottom", the sphere bounds it from the "sides" and the cone bounds it from the "above". We need to calculate the moment of inertia for the axis $$z$$ and our mass function is simply $$\phi(x,y,z)=z$$.

So basically, the exercise is to calculate the following integral, keeping in mind, that our distance function from the z axis is $$\sqrt{x^2+y^2}$$: $$\iiint_A{\sqrt{x^2+y^2}^2z}\,dx\,dy\,dz=\iiint_A{(x^2+y^2)z\,dx\,dy\,dz}$$

Here is where I'm beginning to have troubles, transforming the integral to spherical or cylinder coordinates didn't give me much success. I realized, that this field is symmetrical to the $$x=0$$ plane, so if I could give a proper range with $$x\geq0$$ in mind, for one side of the field, let this be $$A'$$, then we could get the following: $$\iiint_A{(x^2+y^2)z\,dx\,dy\,dz}=2\iiint_{A'}{(x^2+y^2)z\,dx\,dy\,dz}$$ Unfortunately I'm still having troubles, to give a proper interval or non-rectangular region to integrate this over.

Any help is appreciated!

• $z = xy$ does not describe a plane, it describes a saddle shape. Did you maybe mean the $xy$-plane, which is described by $z=0$? – aschepler Sep 29 '18 at 10:21
• That's true, I was translating the text from a different language and there it was labeled as a plane. Thanks for pointing that out! I'm not sure though then whether it's the plane or the saddle shape. The notation used in the document never mentioned a plane like that, so I'm thinking it's the saddle shape, but then the density function could be negative. – Levente Kovács Sep 29 '18 at 10:24

With aschepler pointing out, that one of the borders could be the $$z=0$$ plane instead of the $$z=xy$$ saddle shape the problem became much easier and I could solve the integral.
My approach was to transform to cylinder coordinates. We have: $$x=rcos(\alpha)$$ $$y=rsin(\alpha)$$ $$z=z$$ transformation, and by transforming the integral we get that $$r\in[0,2]$$, $$\alpha\in[0,2\pi]$$ and $$z$$ has to be determined. We can conclude from the $$z=\sqrt{\frac{x^2+y^2}{3}}$$ and $$x^2+y^2+z^2=4$$ equations (while transforming the coordinates) that $$z=\frac{r}{\sqrt{3}}$$ and $$z=\sqrt{4-r^2}$$. Solving the $$\frac{r}{\sqrt{3}}=\sqrt{4-r^2}$$ equation we get that $$r=\sqrt{3}$$.
So our range for the integral for $$z$$ changes at the $$r=\sqrt{3}$$ value. So for $$r\in[0,\sqrt{3}]$$ we get that $$z\in[0,\frac{r}{\sqrt{3}}]$$ and for $$r\in[\sqrt{3},2]$$ we have $$z\in[0, \sqrt{4-r^2}]$$. With this we can modify the integral by applying the transformation to something easier: $$\iiint_A(x^2+y^2)z\,dx\,dy\,dz=\int_0^{\sqrt{3}}\int_0^{2\pi}\int_0^{\frac{r}{\sqrt{3}}}r^3z\,dz\,d\alpha\,dr +\int_{\sqrt{3}}^2\int_0^{2\pi}\int_0^{\sqrt{4-r^2}}r^3z\,dz\,d\alpha\,dr$$
Solving the integrals we get: $$\int_0^{\sqrt{3}}\int_0^{2\pi}\int_0^{\frac{r}{\sqrt{3}}}r^3z\,dz\,d\alpha\,dr=\frac{3\pi}{2}$$, and $$\int_{\sqrt{3}}^2\int_0^{2\pi}\int_0^{\sqrt{4-r^2}}r^3z\,dz\,d\alpha\,dr=\frac{5\pi}{6}$$ Adding them together yields the final result: $$\iiint_A(x^2+y^2)z\,dx\,dy\,dz=\frac{3\pi}{2}+\frac{5\pi}{6}=\frac{7\pi}{3}$$