Find the volume of the solid bounded by the surfaces x² +

Find the volume of the solid bounded by the surfaces $x^2 + z^2 = 1$ and $y^2 + z^2 = 1$.

We slice the intersection of the two cylinders (a Steinmetz solid) into square cross-sections.

Step 1: Cross-section at height $z$

From $x^2+z^2=1$: $-\sqrt{1-z^2}\le x\le\sqrt{1-z^2}$. From $y^2+z^2=1$: $-\sqrt{1-z^2}\le y\le\sqrt{1-z^2}$. At each $z\in[-1,1]$ the cross-section is a square of side $2\sqrt{1-z^2}$.

Step 2: Area of the cross-section

\[A(z) = \left(2\sqrt{1-z^2}\right)^2 = 4(1-z^2)\]

Step 3: Integrate using symmetry

\[V = 2\int_0^1 4(1-z^2)\,dz = 8\left[z-\frac{z^3}{3}\right]_0^1 = 8\cdot\frac23 = \frac{16}{3}\]

\[\boxed{V = \frac{16}{3} \text{ cubic units}}\]

Step 4: Cross-check via octant symmetry

In the first octant: $V_{\text{oct}}=\int_0^1\int_0^{\sqrt{1-z^2}}\int_0^{\sqrt{1-z^2}}dy\,dx\,dz = \int_0^1(1-z^2)\,dz=\frac23$, so $V=8\times\frac23=\frac{16}{3}$. ✓

Step 5: Comparison

The inscribed sphere of radius $1$ has volume $\frac{4\pi}{3}\approx4.19$; the Steinmetz solid's volume $\frac{16}{3}\approx5.33$ is larger, but smaller than the circumscribing cube's volume of $8$.

Summary

\[\boxed{V = \frac{16}{3} \approx 5.333 \text{ cubic units}}\]

Key insight: Named for Charles Proteus Steinmetz, the intersection of two equal perpendicular cylinders has square (not circular) cross-sections, and its volume $\frac{16}{3}r^3$ is exactly twice the volume of the sphere inscribed in the same radius $r$, once $\frac{4\pi}{3}$ is replaced by its "squared-off" analogue.