1. **Problem Statement:** Evaluate the triple integral $$\iiint \frac{dx\,dy\,dz}{\sqrt{1 - x^2 - y^2 - z^2}}$$ where the integral is over all positive values of $x,y,z$ such that the expression under the square root is real. 2. **Understanding the region:** The domain is the portion of the unit ball $x^2 + y^2 + z^2 \leq 1$ in the first octant where $x,y,z \geq 0$. 3. **Change to spherical coordinates:** Recall the spherical coordinates: $$x = r \sin\phi \cos\theta, \quad y = r \sin\phi \sin\theta, \quad z = r \cos\phi$$ with $r \geq 0$, $0 \leq \phi \leq \pi$, $0 \leq \theta < 2\pi$. The volume element is: $$dx\,dy\,dz = r^2 \sin\phi \, dr \, d\phi \, d\theta$$ 4. **Limits for the first octant:** Since $x,y,z \geq 0$: - $\theta$ varies from $0$ to $\frac{\pi}{2}$ (first quadrant in $xy$-plane) - $\phi$ varies from $0$ to $\frac{\pi}{2}$ (from positive $z$-axis down to $xy$-plane) - $r$ varies from $0$ to $1$ (unit sphere radius) 5. **Rewrite the integral:** $$\iiint \frac{dx\,dy\,dz}{\sqrt{1 - x^2 - y^2 - z^2}} = \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \frac{r^2 \sin\phi}{\sqrt{1 - r^2}} dr \, d\phi \, d\theta$$ 6. **Separate the integral:** $$= \left( \int_0^{\pi/2} d\theta \right) \left( \int_0^{\pi/2} \sin\phi \, d\phi \right) \left( \int_0^1 \frac{r^2}{\sqrt{1 - r^2}} dr \right)$$ 7. **Evaluate angular integrals:** - $\int_0^{\pi/2} d\theta = \frac{\pi}{2}$ - $\int_0^{\pi/2} \sin\phi \, d\phi = [-\cos\phi]_0^{\pi/2} = 1$ 8. **Evaluate radial integral:** Let $$I = \int_0^1 \frac{r^2}{\sqrt{1 - r^2}} dr$$ Use substitution $r = \sin t$, so $dr = \cos t dt$, when $r=0$, $t=0$, when $r=1$, $t=\pi/2$. Then $$I = \int_0^{\pi/2} \frac{\sin^2 t}{\sqrt{1 - \sin^2 t}} \cos t dt = \int_0^{\pi/2} \frac{\sin^2 t}{\cos t} \cos t dt = \int_0^{\pi/2} \sin^2 t dt$$ 9. **Integral of $\sin^2 t$:** Use identity: $$\sin^2 t = \frac{1 - \cos 2t}{2}$$ So $$I = \int_0^{\pi/2} \frac{1 - \cos 2t}{2} dt = \frac{1}{2} \left[ t - \frac{\sin 2t}{2} \right]_0^{\pi/2} = \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}$$ 10. **Combine all results:** $$\text{Integral} = \frac{\pi}{2} \times 1 \times \frac{\pi}{4} = \frac{\pi^2}{8}$$ **Final answer:** $$\boxed{\frac{\pi^2}{8}}$$