Let $f$ be a function on $\mathbb{R}^n$ defined by $f(x)=|x|^{-a}\chi_{\{|x|<1\}}(x)$. Compute the integral using polar coordinates.
Set $r=|x|$, $r\in(0,1), \theta\in[0,2\pi]$. Then $$I=\int_0^{2\pi}\int_0^1\frac1{r^a}\,rdrd\theta=\int_0^{2\pi}\int_0^1r^{1-a}\,drd\theta=\frac{2\pi}{2-a}.$$
Is this correct? My doubts arise because the problem states that the function is on $\mathbb{R}^n$ and thus $f(x)=f(x_1,x_2,...,x_n)$, so would polar coordinates still work? (given that this method is usually used with when $n=2$).
Polar coordinates don't work, but since your function is radial, your integral is just as easy to compute using higher dimensional variants (spherical coordinates and beyond).
\begin{align}I=\int_{\mathbb B(0,1)}\frac{dx}{|x|^\alpha} &\overset{A}= \int_0^1\int_{\partial \mathbb B(0,r)}\frac{1}{r^\alpha}d\sigma_r dr \\&\overset{B}= \int_0^1\left(\int_{\partial \mathbb B(0,1)}d\sigma_1\right)\frac{r^{n-1}dr}{r^\alpha} \\&= C_n \int_0^1 r^{n-1-\alpha}dr \\&= \begin{cases}\frac{C_n}{n-\alpha} & \alpha < n,\\ \infty & \alpha \ge n.\end{cases}\end{align}
The equality $A$ is the required change of coordinates, where the surface measure $d\sigma_r$ of the boundary of the ball of radius $r$ is introduced. You can define it explicitly in terms of the $n-1$ angles with a lot of sines and cosines...but since your function is radial, if you are happy to not compute the constant $C_n=\int_{\partial \mathbb B(0,1)}d\sigma_1$ exactly (this is the length of the unit circle $C_2 = 2\pi$, or surface area of the unit sphere $C_3=4\pi$, etc.) you can avoid working with the trig functions.
Equality $B$ comes from rescaling the ball of radius $r$ to the unit ball; since the surface is $n-1$ dimensional, this introduces a factor of $r^{n-1}$.
The rest is basic 1D integration.