1. **Problem Statement:** Evaluate the triple integral $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{-5+x^2+y^2}^{5-x^2-y^2} x \, dz \, dy \, dx$$ using polar coordinates, where the upper limit in the innermost integral is changed from $$3 - x^2 - y^2$$ to $$5 - x^2 - y^2$$ to simplify the problem. 2. **Understanding the integral:** The integral is over the volume bounded by: - $$x$$ from 0 to 2, - $$y$$ from 0 to $$\sqrt{4 - x^2}$$ (which describes a quarter circle of radius 2 in the first quadrant), - $$z$$ from $$-5 + x^2 + y^2$$ to $$5 - x^2 - y^2$$. 3. **Convert to polar coordinates:** Recall the transformations: $$x = r \cos \theta$$ $$y = r \sin \theta$$ $$dx \, dy = r \, dr \, d\theta$$ The region in the xy-plane is a quarter circle of radius 2 in the first quadrant, so: $$0 \leq r \leq 2$$ $$0 \leq \theta \leq \frac{\pi}{2}$$ 4. **Rewrite the limits and integrand:** - The integrand is $$x = r \cos \theta$$. - The limits for $$z$$ are from $$-5 + r^2$$ to $$5 - r^2$$. 5. **Set up the integral in polar coordinates:** $$\int_{\theta=0}^{\pi/2} \int_{r=0}^2 \int_{z=-5+r^2}^{5-r^2} r \cos \theta \, dz \, r \, dr \, d\theta$$ Note that $$dx \, dy = r \, dr \, d\theta$$, so the $$dy \, dx$$ part becomes $$r \, dr \, d\theta$$. 6. **Integrate with respect to $$z$$:** $$\int_{-5+r^2}^{5-r^2} r \cos \theta \, dz = r \cos \theta \times \left[(5 - r^2) - (-5 + r^2)\right] = r \cos \theta \times (10 - 2r^2)$$ 7. **Substitute back and simplify:** The integral becomes: $$\int_0^{\pi/2} \int_0^2 r \cos \theta (10 - 2r^2) r \, dr \, d\theta = \int_0^{\pi/2} \int_0^2 r^2 \cos \theta (10 - 2r^2) \, dr \, d\theta$$ 8. **Separate the integral:** $$\int_0^{\pi/2} \cos \theta \, d\theta \times \int_0^2 r^2 (10 - 2r^2) \, dr$$ 9. **Calculate the angular integral:** $$\int_0^{\pi/2} \cos \theta \, d\theta = \sin \theta \Big|_0^{\pi/2} = 1 - 0 = 1$$ 10. **Calculate the radial integral:** $$\int_0^2 r^2 (10 - 2r^2) \, dr = \int_0^2 (10r^2 - 2r^4) \, dr = 10 \int_0^2 r^2 \, dr - 2 \int_0^2 r^4 \, dr$$ Calculate each: - $$\int_0^2 r^2 \, dr = \frac{r^3}{3} \Big|_0^2 = \frac{8}{3}$$ - $$\int_0^2 r^4 \, dr = \frac{r^5}{5} \Big|_0^2 = \frac{32}{5}$$ So, $$10 \times \frac{8}{3} - 2 \times \frac{32}{5} = \frac{80}{3} - \frac{64}{5} = \frac{400}{15} - \frac{192}{15} = \frac{208}{15}$$ 11. **Final answer:** $$\boxed{\frac{208}{15}}$$ This is the value of the triple integral after the change and conversion to polar coordinates.