Find the volume of the solid that is the common part

Find the volume of the solid that is the common part (intersection) of the two cylinders: $$ x^2+z^2=R^2,\qquad x^2+y^2=R^2. $$

Explanation of Method

This is the classical Steinmetz solid (bicylinder): two cylinders of equal radius $R$ with perpendicular axes. Slicing perpendicular to the shared $x$-axis, each cross-section is a square of side $2\sqrt{R^2-x^2}$.

Slice perpendicular to the $x$-axis and find the square cross-section.
Integrate the cross-sectional area over $x\in[-R,R]$.

Step 1 · Cross-sections

For fixed $x$, both $y$ and $z$ range over $\left[-\sqrt{R^2-x^2},\sqrt{R^2-x^2}\right]$, so the cross-section is a square of side $2\sqrt{R^2-x^2}$ and area $A(x)=4(R^2-x^2)$.

Step 2 · Integrate over $x$

$$ V = \int_{-R}^R4\big(R^2-x^2\big)\,dx = 8\int_0^R\big(R^2-x^2\big)\,dx = 8\left[R^2x-\frac{x^3}3\right]_0^R = 8\left(R^3-\frac{R^3}3\right) = \frac{16R^3}3. $$

Additional Notes

This is the classical Steinmetz solid result, matching the source exactly (also independently verifiable via the first-octant integral $V=8\int_0^R(R^2-x^2)\,dx$, the same computation).
The volume is exactly $\frac23$ of the volume of the circumscribing cube of side $2R$ (which is $8R^3$): $\frac{16R^3}3=\frac23\cdot8R^3$.

Result

The volume is:

$$ V = \frac{16R^3}3. $$