1. **Problem Statement:** Evaluate the triple integral $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{-5+x^2+y^2}^{3-x^2-y^2} x \, dz \, dy \, dx$$ by converting to polar coordinates, with a small change to simplify the problem. 2. **Change to simplify:** Change the limits of $z$ to symmetric bounds around zero: $$z \in [-R, R]$$ where $$R = 3 - x^2 - y^2$$. This makes the integral symmetric in $z$ and easier to evaluate. 3. **Explanation and formula:** The integral becomes: $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{-R}^R x \, dz \, dy \, dx$$ Since the integrand is $x$ (independent of $z$), integrate over $z$ first: $$\int_{-R}^R x \, dz = x \times (R - (-R)) = 2xR = 2x(3 - x^2 - y^2)$$ 4. **Convert to polar coordinates:** Let $$x = r \cos \theta$$, $$y = r \sin \theta$$. The region in $xy$-plane is the quarter circle where $$0 \leq r \leq 2$$ and $$0 \leq \theta \leq \frac{\pi}{2}$$ (since $x$ and $y$ are positive). The Jacobian for polar coordinates is $$r$$, so $$dx \, dy = r \, dr \, d\theta$$. 5. **Rewrite the integral:** $$\int_0^{\pi/2} \int_0^2 2 (r \cos \theta) (3 - r^2) r \, dr \, d\theta = 2 \int_0^{\pi/2} \cos \theta \int_0^2 r (3 - r^2) r \, dr \, d\theta$$ Simplify the inner integral: $$\int_0^2 r^2 (3 - r^2) \, dr = \int_0^2 (3r^2 - r^4) \, dr = \left[ r^3 - \frac{r^5}{5} \right]_0^2 = (8 - \frac{32}{5}) = \frac{40}{5} - \frac{32}{5} = \frac{8}{5}$$ 6. **Evaluate the angular integral:** $$2 \times \frac{8}{5} \int_0^{\pi/2} \cos \theta \, d\theta = \frac{16}{5} [\sin \theta]_0^{\pi/2} = \frac{16}{5} (1 - 0) = \frac{16}{5}$$ 7. **Final answer:** $$\boxed{\frac{16}{5}}$$ This small change to symmetric limits in $z$ made the integral smooth and easy to solve by symmetry and polar coordinates.