Volume of Viviani’s Window

Here is a detailed expalantion of @DanielWainfleet's idea. Let $\mathcal{V}$ denote the region. Then the the intersection of $\mathcal{V}$ and the plane $x = x_0$ is described by

$$ \mathcal{I}(x_0) \quad : \quad y^2 + z^2 \leq 4 - x_0^2 \quad \text{and} \quad y^2 \leq 1 - (1 - x_0)^2.$$

The area of $\mathcal{I}(x_0)$ can be computed by decomposing this into 4 congruent wedges and two congruent isosceles as in the following figure.

Indeed, note that the four corners of $\mathcal{I}(x_0)$ are given by $(\pm \sqrt{x_0(2-x_0)}, \pm \sqrt{2(2 - x_0)})$, which follows from solving the system of equations $y^2 + z^2 = 4-x_0^2$ and $y^2 = 1 - (1-x_0)^2$. From this, the angle of each of four wedges is given by $\arctan(\sqrt{x_0/2})$, and so, the area of $\mathcal{I}(x_0)$ is given by

$$ S(x_0) := 2(4 - x_0^2) \arctan(\sqrt{x_0 / 2}) + 2 \sqrt{2x_0}(2 - x_0) $$

Then the volume of $\mathcal{V}$ is obtained by integrating this with respect to $x_0$ over $[0, 2]$. Then,

\begin{align*} \text{[volume of $\mathcal{V}$]} &= \int_{0}^{2} S(x) \, \mathrm{d}x \\ &= \int_{0}^{2} \left[ 2(4 - x^2) \arctan\left(\sqrt{\frac{x}{2}}\right) + 2 \sqrt{2x}(2 - x) \right] \, \mathrm{d}x \\ &= \left[ \frac{2}{3} (4-x) (x+2)^2 \arctan\left(\sqrt{\frac{x}{2}}\right) - \frac{2}{9} \sqrt{2x} \left(3x^2 - 10x + 24\right) \right]_{0}^{2} \\ &= \frac{16 \pi}{3} - \frac{64}{9}. \end{align*}

The figure bellow shows the situation from above, in the $x,y-$plane.

Use the cylindrical coordinates
$$\begin{aligned}x&=r\cos t\\ y&=r\sin t\\ z&=z\\J&=r,\end{aligned}$$ where $J$ denotes the Jacobian. Now, we need to find the bounds for $t,r,z.$

• The arrows in the green circle (cylinder from above) indicate the movement of the angle $t$ counterclockwise from $-{\pi \over 2}$ to $\pi \over 2.$

• $(x-1)^2+y^2=1\;$ becomes $\;r =2\cos t,$ hence $r$ goes from $0$ to $2 \cos t.$

• $x^2+y^2+z^2\leq 4\;$ becomes $\;r^2+z^2\leq 4,$ from where the bounds for $z.$

The volume is given by the triple integral $$\int\limits_{-{\pi \over 2}}^{\pi \over 2}\int\limits_{0}^{2 \cos t}\int\limits_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}}r\,dz\,dr\,dt.$$