Does anyone know how to solve this? The region is really confusing me.
Let $T$ be the region bounded by $(x,y,z) \in \mathbb{R}^3$ that satisfy the inequality: $$ 0<\sqrt{x^2+y^2+z^2}<1-\left|z\right| $$
Find the mass of $T$, where the density is given by: $$ \rho \left(x,y,z\right)=\left(x^2+y^2+z^2\right)^{-\frac{3}{4}} $$
What i have done so far is: $$ 0\le \sqrt{x^2+y^2+z^2}\le 1-\left|z\right|+\Rightarrow \left(x^2+y^2+z^2\right)\le \left(1-\left|z\right|\right)^2 $$ $$ \Rightarrow \:x^2+y^2\le 1-2\left|z\right|+z^2-z^2\Rightarrow \:x^2+y^2-1+\le 2\left|z\right| $$ $$ \Rightarrow \:\:\left|z\right|\le \frac{1}{2}\left(1-x^2-y^2\right)\Rightarrow \:z\le \frac{1}{2}\left(1-x^2-y^2\right)\:\vee \:z\ge \frac{1}{2}\left(x^2+y^2-1\right) $$ xy-plane get that: $y\pm\sqrt{1-x^2}$ such that: $$ \int _{-1}^1\int _{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int _{\frac{1}{2}\left(x^2+y^2-1\right)}^{\frac{1}{2}\left(1-x^2-y^2\right)\:}\left(x^2+y^2+z^2\right)^{-\frac{3}{4}}dzdydx\:\:\: $$ Is my solution so far correct? and what do i do now?
Notice that in the original description of the region, replacing any variable with its negative gives the same inequality. The same is also true of the density function, so we can use the property of even functions over symmetric regions to say that
$$\iiint_T \rho \:dV = 8 \iiint_{T_{O1}} \rho \:dV$$
where $T_{O1}$ is the portion of $T$ only in the first octant where $x,y,z>0$, which reduces our inequality for the region to simply be
$$\sqrt{x^2+y^2+z^2} < 1-z$$
Also notice that $z$ cannot be greater than $1$ because square root always returns a positive quantity. Now we can rearrange to figure out what this surface might be:
$$x^2+y^2+z^2 = 1 -2z+z^2 \implies 2z = 1 -x^2-y^2$$
which is a paraboloid facing down (because of the absolute value, it flips to being a paraboloid facing up below the $xy$ plane, but we don't have to worry about this anymore because of the symmetry we noted earlier).
Our integral now reads
$$8 \iiint_{T_{O1}} (x^2+y^2+z^2)^{-\frac{3}{4}}\:dV$$
which would be almost impossible to do in Cartesian coordinates. Our easiest options are Cylindrical and Spherical, the former making the bounds simple, and the latter making the integrand simple.
Ultimately what ends up being the easiest to do is Cylindrical with $dr$ first (but this can only be seen with trial and error, set up the other integrals to see what happens).
$$8\int_0^{\frac{\pi}{2}}\int_0^\frac{1}{2} \int_0^{\sqrt{1-2z}} \frac{r}{(r^2+z^2)^{\frac{3}{4}}}\:dr\:dz\:d\theta = 8\pi \int_0^\frac{1}{2} (1-2z+z^2)^{\frac{1}{4}}-(z^2)^{\frac{1}{4}}\:dz$$
$$= 8\pi \int_0^\frac{1}{2} \sqrt{1-z}-\sqrt{z}\:dz = -\frac{16\pi}{3}\Bigr[(1-z)^{\frac{3}{2}} - z^{\frac{3}{2}}\Bigr]_0^{\frac{1}{2}}=\frac{16\pi}{3}$$