Fancy bowl

Use cylindrical/polar coordinates, since the region of integration in the xy-plane (from projecting the region onto the xy-plane) is bounded by $x^2 + y^2 = 1$ .

Since $x^2 + y^2 = r^2$ , we can rewrite $z = 1 - x^2 - y^2$ as $z = 1 - r^2$ .

So, the volume $\int\int\int1 dV$ equals :

$\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=0}^{z=1-r^2}1 * (r* dz*dr*d\theta)$

$= 2\pi\int_{r=0}^{r=1} r(1 - r^2) dr$

$= \pi\int_{r=0}^{r=1} (2r - 2r^3) dr$

$= \frac{\pi}{2}$

$= \frac{3.14}{2}$

$= \boxed{1.57}$

The volume is equivalent to the double integral of the suface over its projection onto the xy-plane, i.e. the region $R$ formed when $1-x^2-y^2\geq0 \Rightarrow x^2+y^2\leq1$ . Thus, the volume is $\int \!\!\! \int_R 1-x^2-y^2\,dx\,dy = \int_0^{2\pi} \!\!\! \int_0^1 (1-r^2)r \,dr\,d\theta = \left. 2\pi \left ( \frac{r^2}{2}-\frac{r^4}{4} \right ) \right|_0^1 = \frac{\pi}{2}.$