Find the surface area of the closed surface formed
Find the surface area of the closed surface formed by the circular cylinders $x^2+y^2=a^2$ and $x^2+z^2=a^2$.
Explanation of Method
The boundary of the Steinmetz solid consists of two congruent pieces — the part of each cylinder lying inside the other — related by the symmetry $y\leftrightarrow z$. Computing the area of one piece as a surface $z=g(x,y)$ over the disk $x^2+y^2\le a^2$ (doubled for both $z\ge0$ and $z\le0$) and doubling again for the second cylinder gives the total.
- Reduce the problem to one piece by symmetry, $S=2S_1$.
- Set up the surface-area integral for $S_1$ over the disk $x^2+y^2\le a^2$.
- Evaluate by switching the order of integration.
Step 1 · Symmetry reduction
By the $y\leftrightarrow z$ symmetry between the two cylinders, $S=2S_1$, where $S_1$ is the area of the cylinder $x^2+z^2=a^2$ lying inside $x^2+y^2\le a^2$.
Step 2 · Surface-area integrand
For the upper half ($z\ge0$) of $x^2+z^2=a^2$, i.e. $z=\sqrt{a^2-x^2}$: $z_x=\dfrac{-x}{\sqrt{a^2-x^2}},\,z_y=0$, so $\sqrt{1+z_x^2+z_y^2}=\dfrac a{\sqrt{a^2-x^2}}$. Doubling for $z\le0$:
$$ S_1 = 2\iint_D\frac a{\sqrt{a^2-x^2}}\,dx\,dy, \qquad D:x^2+y^2\le a^2. $$
Step 3 · Evaluate
For fixed $x\in[-a,a]$, $y$ ranges over $\big[-\sqrt{a^2-x^2},\,\sqrt{a^2-x^2}\big]$, so the $y$-integral of the ($x$-only) integrand $\dfrac a{\sqrt{a^2-x^2}}$ contributes a factor $2\sqrt{a^2-x^2}$, and the $\sqrt{a^2-x^2}$ terms cancel exactly:
$$ S_1 = 2\int_{-a}^{a}\frac a{\sqrt{a^2-x^2}}\cdot2\sqrt{a^2-x^2}\,dx = 2\int_{-a}^a2a\,dx = 8a^2. $$
Therefore:
$$ S = 2S_1 = 16a^2. $$
Additional Notes
- This derivation matches the source exactly, and reproduces the classical result for the Steinmetz solid (bicylinder): surface area $16a^2$ (alongside its well-known volume $\frac{16}3a^3$).
- Verified numerically: direct evaluation of $S_1$ via the double integral gives $8a^2$ to high precision (e.g. $a=1$: $S_1\approx7.999999992$), confirming $S=16a^2$.
Result
The surface area is: