I tried solving it by taking the $z = 4-x^2$ as the integrand then integrating w.r.t $x$ from $-2$ to $2$. Then w.r.t $y$ also from $-2$ to $2$ since the boundary equation was $x^2 +y^2 = 4$ which was the equation of a circle of radius $2$. So are my boundaries right or are they the boundaries of a square and not a circle? my result was $42.6$ unit$^3$
On
Observe that as $\;-2\le x\le2\;$ , we get that $\;-\sqrt{4-x^2}\le y\le\sqrt{y^2-4}\;$ , sin in rectangular coordinates you get
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(4-x^2)\,dy\,dx=\int_{-2}^2\left[8\sqrt{4-x^2}-2x^2\sqrt{4-x^2}\right]\,dx\;\;(**)$$
In the first case substitute $\;\sin t=\frac x2\implies dx=2\cos t\,dt\;$ , and in the second one do by parts:
$$\begin{cases}u=x,&u'=1\\{}\\ v'=-2x\sqrt{4-x^2}, &v=\frac23(4-x^2)^{3/2}\end{cases}$$
and we then get
$$(**)=16\int_{-\pi/2}^{\pi/2}2\cos^2t\,dt-\left(\left.\frac{2x}3(4-x^2)^{3/2}\right|_{-2}^2-\frac23\int_{-2}^2(4-x^2)^{3/2}\right)$$
You continue now (I'm already tired...), taking into account that the middle term above is zero, and $\;\int\cos^2x\,dx=\frac{x+\sin x\cos x}2\;$ and etc. You'll need another substitution for the rightmost integral, just as the one we did above...this really is ugly in rect. coordinates!
Your bounds currently represent a square, not a circle, and as @georg mentioned, using cylindrical coordinates is nice. Converting to them, we get $$\int_0^{2\pi} \int_0^2 (4-x^2) \ rdrd\theta = \int_0^{2\pi} \int_0^2 (4-(r\cos(\theta))^2) \ rdrd\theta$$ Can you finish from here?