A sphere of radius a centered at the origin

A sphere of radius $a$ centered at the origin and a circular cylinder of diameter $a$ tangent to the $yz$-plane at the origin (with axis parallel to the $z$-axis) intersect in a space curve. 1° Show that this curve is the intersection of the sphere $x^2+y^2+z^2=a^2$ with the cylinder $x^2+y^2=ax$. 2° Find the volume of the solid bounded by these two surfaces and the plane $z=0$, with $z>0$.

Step 1 · Identify the two surfaces (Part 1°)

A sphere of radius $a$ centered at the origin has equation $x^2+y^2+z^2=a^2$. A circular cylinder of diameter $a$, tangent to the $yz$-plane at the origin with axis parallel to the $z$-axis, has a circular cross-section of radius $\frac a2$ centered at $\left(\frac a2,0\right)$ in the $xy$-plane: $\left(x-\frac a2\right)^2+y^2=\left(\frac a2\right)^2$, which expands to $x^2+y^2=ax$. The space curve in which the two surfaces meet is exactly the intersection of $x^2+y^2+z^2=a^2$ and $x^2+y^2=ax$.

Step 2 · Set up the volume integral (Part 2°)

The solid lies over the disk $x^2+y^2\leqslant ax$, with height $z=\sqrt{a^2-x^2-y^2}$ above it. In polar coordinates the disk is $0\leqslant r\leqslant a\cos\theta$, $\theta\in\left[-\frac\pi2,\frac\pi2\right]$, so:

$$ V=\int_{-\pi/2}^{\pi/2}\int_0^{a\cos\theta}\sqrt{a^2-r^2}\,r\,dr\,d\theta. $$

Step 3 · Evaluate the inner integral

$$ \int_0^{a\cos\theta}\sqrt{a^2-r^2}\,r\,dr = \left[-\frac{(a^2-r^2)^{3/2}}3\right]_0^{a\cos\theta} = \frac{a^3-\left(a^2-a^2\cos^2\theta\right)^{3/2}}3 = \frac{a^3-\left(a^2\sin^2\theta\right)^{3/2}}3. $$

Here $\left(a^2\sin^2\theta\right)^{3/2}=a^3|\sin\theta|^3$ — the absolute value is essential, since $\sqrt{u^2}=|u|$, not $u$. So the inner integral equals $\dfrac{a^3\left(1-|\sin\theta|^3\right)}3$.

Step 4 · Integrate over $\theta$

Since $|\sin\theta|^3$ is even, $\displaystyle\int_{-\pi/2}^{\pi/2}|\sin\theta|^3\,d\theta = 2\int_0^{\pi/2}\sin^3\theta\,d\theta = 2\cdot\frac23=\frac43$. Therefore:

$$ V=\frac{a^3}3\int_{-\pi/2}^{\pi/2}\left(1-|\sin\theta|^3\right)d\theta = \frac{a^3}3\left(\pi-\frac43\right) = \frac{a^3(3\pi-4)}9. $$

Additional Notes

• A note on the source computation: the source's derivation for Part 2° trailed off without a complete, justified evaluation, ending in an unsupported final answer of $\dfrac{\pi a^3}6(2-\sqrt3)\approx0.1403\,a^3$ (for $a=1$). A full independent derivation, verified two ways — direct numerical sweep over $\theta$ and an independent direct Cartesian double integral over the disk $x^2+y^2\leqslant ax$ — both give $\approx0.60275\,a^3$ for $a=1$, matching the closed form $\dfrac{a^3(3\pi-4)}9$ derived above, not the source's guess (off by a factor of roughly $4.3$).
• This computation is the classic half-Viviani volume — exactly half of the full Viviani solid (both $z>0$ and $z<0$) computed independently in Problem 156, whose result $\dfrac{2a^3(3\pi-4)}9$ is exactly twice the value found here, as required by symmetry about $z=0$.
• As in Problem 156, the key subtlety is that $\sqrt{a^2\sin^2\theta}=a|\sin\theta|$; dropping the absolute value would incorrectly make the $|\sin\theta|^3$ term integrate to $0$ over a symmetric interval (since $\sin^3\theta$, without the absolute value, is odd), which would understate the true volume.