Discussion
Let \(S\) be the solid of intersection of two solid spheres \(S_1, S_2\) given by
\(S_1: x^{2} + y^{2} + (z-1)^{2} \leq 4\) and \(S_2: x^{2} + y^{2} + (z+1)^{2} \leq 4\).
Which ONE of the following iterated integrals expresses the volume of \(S\)?
Answer: A
The volume is \(8\int_0^{\sqrt{3}}\int_0^{\sqrt{3-x^2}} (\sqrt{4-x^2-y^2} - 1)\,dy\,dx\).
Subtracting the two sphere equations gives \((z-1)^{2} = (z+1)^{2}\), so the spheres meet in the plane \(z = 0\), on the circle \(x^{2} + y^{2} = 3\). The solid is a lens, symmetric about \(z = 0\).
Above \(z = 0\) the binding surface is the sphere centred at \((0,0,-1)\): \(z \leq \sqrt{4 - x^{2} - y^{2}} - 1\), over the disc \(x^{2} + y^{2} \leq 3\). So
\[ \begin{aligned} V &= 2\iint_{x^2+y^2 \leq 3} (\sqrt{4-x^2-y^2} - 1)\,dA \\ &= 8 \times (\text{first-quadrant part}) \end{aligned} \]
The factor 8 counts the two halves of the lens and the four quadrants of the disc. The option with factor 2 over the same quadrant gives only a quarter of the volume. The two options over the disc of radius 2 include the ring \(\sqrt{3} \lt r \leq 2\), where \(\sqrt{4 - r^{2}} - 1 \lt 0\), so they subtract volume that is not there.
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