Find the Area under $R$ Using Integrals

$R$ is a region in the first octant of $\mathbb{R^3}$ that is bounded by $x^2+y^2+z^2=4$ and $z=\sqrt{x^2+y^2}$. Set up the integral for the volume of this region $R$.

I think the first thing to do is to model this with a triple integral because there's 3 variables. The first integral should be $z$, from the above equation, I see that $\sqrt{x^2+y^2} \leq z \leq 4-x^2-y^2$. Next, getting rid of $z$, $x^2=y^2$ and $x^2+y^2=4$. Solving with respect to $y$, we get $\sqrt{x} \leq y \leq \sqrt{4-x^2}$. Then for $x$ it's simply $0 \leq x \leq 2$. So the final integral should be $\int_{0}^{2} \int_{\sqrt{x}}^{\sqrt{4-x^2}} \int_{\sqrt{x^2+y^2}}^{4-x^2-y^2}dzdydx$ right?

Also, how can I make this integral easier to solve? It looks reall intimidating and maybe converting it to cylindrical coordinates could help.

If you want to solve for the volume using integrals in a cartesian coordinate system, then you are on the right track, but quite a bit off. Your errors are mostly in solving for the bounds of $y$. From the constraint inequalities for $z$, we get $\sqrt{x^2+y^2}\leq 4-x^2-y^2 $, which can be solved as a 4th-degree polynomial in $y$. We cannot use $\sqrt{x^2+y^2}= 4-x^2-y^2 =0$ as you seem to have done.

As you correctly noted, it would be much easier to solve this integral in polar coordinates. (I certainly don't want to try to solve that 4th degree polynomial!) Thus we convert the boundary surfaces:

$$0\leq\sqrt{x^2+y^2}\leq z\leq4-x^2-y^2$$ $$\Rightarrow 0\leq r\leq z\leq 4-r^2$$

From this, we get a bound on $r$:

$$r^2+r-4\leq 0,\;0\leq r$$ $$\Rightarrow \frac{-1-\sqrt{17}}{2}\leq 0\leq r\leq\frac{-1+\sqrt{17}}{2}$$

To restrain our volume to the first octant, we set the boundaries of $\theta$:

$$0\leq \theta\leq\frac{\pi}{2}$$

Thus our volume is calculated as follows:

$$V=\iiint_R\ {\rm d}V=\int_0^\frac{\pi}{2}\int_0^\frac{-1+\sqrt{17}}{2}\int_r^{4-r^2}\ {\rm d}z\ {\rm d}r\ {\rm d}\theta$$ $$=\int_0^\frac{\pi}{2}\int_0^\frac{-1+\sqrt{17}}{2}4-r-r^2\ {\rm d}r\ {\rm d}\theta$$ $$=\int_0^\frac{\pi}{2}4\left(\frac{-1+\sqrt{17}}{2}\right)-\frac{\left(\frac{-1+\sqrt{17}}{2}\right)^2}{2}-\frac{\left(\frac{-1+\sqrt{17}}{2}\right)^3}{3}\ {\rm d}\theta$$ $$=\frac{\pi}{2}\left[4\left(\frac{-1+\sqrt{17}}{2}\right)-\frac{\left(\frac{-1+\sqrt{17}}{2}\right)^2}{2}-\frac{\left(\frac{-1+\sqrt{17}}{2}\right)^3}{3}\right]\approx 3.1231$$

Edit, I forgot to include the $r$ in the integrand as is necessary for cylindrical integration. The process should be otherwise correct