I have to solve this integral:
$\int_T(x^2+y^2)dx dy \ \ \ $ where $\ \ \ T=\{(x,y,z):x^2+y^2+z^2<1\}$
Now, what gets me confused is that i have to integrate only with respect to $x$ and $y$ but not $z$. The fact is that i have no idea on what this even means geometrically, is this a double integral on a 3d domain?
At first i didn't even noticed that and i managed to solve the integral (at least i believe) as if it was:
$\int_T(x^2+y^2)dx dy dz \ \ \ $ where $\ \ \ T=\{(x,y,z):x^2+y^2+z^2<1\}$
switching to spherical polar coordinates since $T$ represents a sphere of radius $1$:
$=\int_0^\pi d\theta\int_0^1 d\rho\int_0^{2\pi} \rho^2\sin^2(\theta)(\cos^2(\phi)+sin^2(\phi)) \ d\phi= \\ =\int_0^\pi d\theta\int_0^1 2\pi\rho^2\sin^2(\theta) \ d\rho= \\ =\int_0^\pi\frac{2}{3}\pi\sin^2(\theta) \ d\theta = \frac{\pi^2}{3}$
But the correct answer is $\frac{8\pi}{15}$ and i don't know how to get to that.
The integral is indeed $$\iiint (x^2+y^2) \, dxdydz$$ You made an error in computation. Note that $x^2+y^2$ is the distance to the $z$ axis, thus this can be thought of as the moment of inertia
$$I = \rho \iiint (x^2+y^2) \, dxdydz $$
where $\rho = \frac{M}{V}$ is the density of the sphere. But it is well known that
$$I = \frac{2}{5}MR^2 = \frac{2}{5}M$$ so, we are left with
$$\iiint (x^2+y^2) \, dxdydz = \frac{I}{\rho} = \frac{2}{5}\cdot\frac{4\pi}{3} = \frac{8\pi}{15}$$
I leave it to you to properly compute the integral, now that is clear you have made an error.