Volume Common to Intersecting Spheres
\[x^2+y^2=16\]
and \[x^2+y^2=8x\]
using Volumes of Revolution. The first sphere has centre \[(0,0)\]
and radius 4, and the second, which we can write \[(x-4)^2+y^2=16\]
has centre \[(4,0)\]
and radius 4. There is a line of symmetry halfway between the centres of the spheres. The equation of this line is \[x=2\]
, and by substituting \[x=2\]
into the equation of the first sphere we obtain \[y= \pm 2 \sqrt{3}\]
and the spheres intersect at the points \[(2, \pm 2 \sqrt{3} )\]
(the line between these points is a diameter of a circle of intersection drawn on each sphere), so points on the first sphere between the points of intersection of the two spheres have coordinates \[(x,y)\]
, the volume required is\[\begin{equation} \begin{aligned} V &= \pi \int^{2 \sqrt{3}}_{-2 \sqrt{3}} (2-x)^2 dx \\ &=2 \pi \int^{2 \sqrt{3}}_0 (2-x)^2 dx \\ &=2 \pi \int^{2 \sqrt{3}}_0 (2-\sqrt{16-y^2} )^2dy \\ &= 2 \pi \int^{2 \sqrt{3}}_0 (4-4 \sqrt{16-y^2} +16-y^2)dy \\ &= 2 \pi \int^{2 \sqrt{3}}_0 (20-4 \sqrt{16-^2}-y^2)dy \\ &= 2 \pi [20y - \frac{y^3}{3} -2y \sqrt{16-y^2}-32 sin^{-1}( \frac{y}{4})]^{2 \sqrt{3}}_0 \\ &= 2 \pi (40 \sqrt{3}- \frac{8 \times 3 \sqrt{3}}{3}- 4 \sqrt{3} \times 2- 32 sin^{-1}( \frac{\sqrt|{3}}{2}) \\ &= 48 \pi \sqrt{3} - \frac{64 \pi^2}{3}\end{aligned} \end{equation}\]
