\(dz\, dy\, dx\text{:}\) \(\ds\int_{0}^1\int_{0}^{1-x^2}\int_{0}^{\sqrt{1-y}}\, dz\, dy\, dx\)
\(dz\, dx\, dy\text{:}\) \(\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dz\, dx\, dy\)
\(dy\, dz\, dx\text{:}\) \(\ds\int_{0}^1\int_0^{x}\int_{0}^{1-x^2}\, dy\, dz\, dx + \int_{0}^1\int_x^{1}\int_{0}^{1-z^2}\, dy\, dz\, dx\)
\(dy\, dx\, dz\text{:}\) \(\ds\int_0^1\int_{0}^{z}\int_{0}^{1-z^2}\, dy\, dx\, dz+\int_0^1\int_{z}^{1}\int_{0}^{1-x^2}\, dy\, dx\, dz\)
\(dx\, dz\, dy\text{:}\) \(\ds\int_{0}^1\int_{0}^{\sqrt{1-y}}\int_{0}^{\sqrt{1-y}}\, dx\, dz\, dy\)
\(dx\, dy\, dz\text{:}\) \(\ds\int_0^1\int_{0}^{1-z^2}\int_{0}^{\sqrt{1-y}}\, dx\, dy\, dz\) Answers will vary. Neither order is particularly “hard.” The order \(dz\, dy\, dx\) requires integrating a square root, so powers can be messy; the order \(dy\, dz\, dx\) requires two triple integrals, but each uses only polynomials.