Evaluate the triple integral of z^2 over the region bounded by the sphere z = root(1 - x^2 - y^2) and the cone z = root(x^2 + y^2).

Step-by-step solution

1. 1

   Identify the solid. Squaring the sphere equation gives $x^2+y^2+z^2=1$ with $z\ge 0$: the upper unit hemisphere. The cone is $z=r$ in cylindrical coordinates. The region is the ice-cream-cone solid inside the hemisphere and above the cone.

2. 2

   Find where the two surfaces meet. Setting $r=\sqrt{1-r^2}$ gives $2r^2=1$, so $r=\tfrac{1}{\sqrt2}$ and $z=\tfrac{1}{\sqrt2}$. That fixes the radial range.

3. 3

   Set up in cylindrical coordinates. With $dV=r\,dz\,dr\,d\theta$, $\iiint_\Omega z^2\,dV=\int_0^{2\pi}\!\!\int_0^{1/\sqrt2}\!\!\int_r^{\sqrt{1-r^2}}z^2r\,dz\,dr\,d\theta.$ Cylindrical beats spherical here because the cone and sphere give clean $z$-limits.

4. 4

   Integrate in $z$ and drop $\theta$. $\int_r^{\sqrt{1-r^2}}z^2r\,dz=\frac{r}{3}\Bigl[(1-r^2)^{3/2}-r^3\Bigr],$ and the integrand has no $\theta$ dependence, so the $\theta$-integral contributes a factor $2\pi$.

5. 5

   Do the two radial integrals. With $u=1-r^2$, $\int_0^{1/\sqrt2}r(1-r^2)^{3/2}\,dr=\frac12\int_{1/2}^{1}u^{3/2}\,du=\frac15\left(1-\frac{1}{4\sqrt2}\right),$ and $\int_0^{1/\sqrt2}r^4\,dr=\frac15\left(\frac{1}{\sqrt2}\right)^{5}=\frac{1}{20\sqrt2}.$

6. 6

   Combine. The bracket is $\tfrac15-\tfrac{1}{20\sqrt2}-\tfrac{1}{20\sqrt2}=\tfrac15-\tfrac{1}{10\sqrt2}$, so $\iiint_\Omega z^2\,dV=\frac{2\pi}{3}\left(\frac15-\frac{1}{10\sqrt2}\right)=\frac{\pi}{15}\left(2-\frac{1}{\sqrt2}\right)=\frac{\pi(4-\sqrt2)}{30}.$ Numerically this is $0.270783$, which matches a direct numerical quadrature of the original integral.