Find the volume of the smaller of the two regions into which the plane $$Ax + By + Cz = D$$ divides the interior of the ellipsoid $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$

# [Math] Volume of the smaller region of ellipsoid cut by plane

volume

## Best Answer

The region $R$ inside the ellipsoid $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ and above the plane $$Ax + By + Cz = D$$ is transformed by the change of variables $$x = au,\,\,\,y = bv,\,\,\,z = cw$$ to the region $S$ inside the sphere $${u^2} + {v^2} + {w^2} = 1$$ and above the plane $$(Aa)u + (Bb)v + (Cc)w = D$$ The distance from the origin to this plane is $$d = \frac{{\left| {(Aa){u_0} + (Bb){v_0} + (Cc){w_0} + D} \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }} = \frac{{\left| D \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }}\left| {_{({u_0},{v_0},{w_0}) = (0,0,0)}} \right.$$ so, by symmetry, the volume of $S$ is equal to the volume inside the sphere and above the plane $$w = \frac{{\left| D \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }}$$ $${\text{Let }}\xi {\text{ = }}\frac{{\left| D \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }},\,\,\,\xi \ge 0,{\text{ then }}$$ $${V_S} = \iiint\limits_S {dV} = \int\limits_0^{2\pi } {\int\limits_0^{\sqrt {1 - {\xi ^2}} } {\int\limits_\xi ^{\sqrt {1 - {r^2}} } {r\,dz\,dr\,d\theta } } } = \frac{\pi }{3}\left( {{\xi ^3} - 3\xi + 2} \right)$$ Since $$J = \left| {\frac{{d(x,y,z)}}{{d(u,v,w)}}} \right| = \left| {\begin{array}{*{20}{c}} a&0&0 \\ 0&b&0 \\ 0&0&c \end{array}} \right| = abc\,du\,dv\,dw$$ the volume of $R$ is $${V_R} = abc \cdot {V_S} = \frac{{abc\pi }}{3}\left( {{{\left( {\frac{{\left| D \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }}} \right)}^3} - 3\left( {\frac{{\left| D \right|}}{{\sqrt {{{(Aa)}^2} + {{(Bb)}^2} + {{(Cc)}^2}} }}} \right) + 2} \right)$$