Volume of Interesting Structure

We can start by integrating in the z-direction, given $0\le z \le 1-x^2-y^2$ : $\int_{0}^{1-x^2-y^2} \,\mathrm{d}z= 1-x^2-y^2$ (Ignoring constants of integration.) Then, given $x^2+y^2\le x$ , we can see that $y\le \sqrt{x-x^2}$ . Now we integrate in the y-direction: \begin{aligned}\int_{0}^{\sqrt{x-x^2}} (1-x^2-y^2) \,\mathrm{d}y &= \left[y-x^2 y-\frac{1}{3} y^3\right]_0^\sqrt{x-x^2} \\ &= \sqrt{x-x^2} -x^2 \sqrt{x-x^2} -\frac{1}{3}\sqrt{x-x^2}^3 \\ &= \left(-\frac{2}{3}x^2 -\frac{1}{3}x+1\right) \sqrt{x-x^2} \end{aligned} Next, given that $x\ge0$ , for $0\le y\le \sqrt{x-x^2}$ , then $x\le1$ . So we can finally integrate in the x-direction: $\int_{0}^{1} \left(\left(-\frac{2}{3}x^2 -\frac{1}{3}x+1\right) \sqrt{x-x^2}\right)\,\mathrm{d}x = \frac{5\pi}{64} \approx \boxed{0.2454}$